2016 ACM-ICPC World Finals - Phuket

Ceiling Function

签到题，暴力

 1 #include 
 2 using namespace std;
 3 const int maxn = 1e5 + 10;
 4 int rt[maxn], ls[maxn], rs[maxn], val[maxn], cnt;
 5 
 6 void add(int x, int v) {
 7     if(v < val[x]) {
 8         if(ls[x]) add(ls[x], v);
 9         else {
10             ls[x] = ++cnt;
11             val[cnt] = v;
12             return;
13         }
14     }
15     else {
16         if(rs[x]) add(rs[x], v);
17         else {
18             rs[x] = ++cnt;
19             val[cnt] = v;
20             return;
21         }
22     }
23 }
24 
25 bool check(int x, int y) {
26     bool ret = true;
27     if(ls[x] && !ls[y] || !ls[x] && ls[y]) return false;
28     if(ls[x]) ret &= check(ls[x], ls[y]);
29     if(rs[x] && !rs[y] || !rs[x] && rs[y]) return false;
30     if(rs[x]) ret &= check(rs[x], rs[y]);
31     return ret;
32 }
33 
34 int main() {
35     int n, k;
36     scanf("%d %d", &n, &k);
37     for(int i = 1; i <= n; ++i) {
38         int x;
39         scanf("%d", &x);
40         rt[i] = ++cnt, val[cnt] = x;
41         for(int j = 2; j <= k; ++j) {
42             scanf("%d", &x);
43             add(rt[i], x);
44         }
45         for(int j = 1; j < i; ++j) {
46             if(check(rt[i], rt[j])) {
47                 i--; n--; break;
48             }
49         }
50     }
51     printf("%d\n", n);
52     return 0;
53 }

Aguin

 

Swap Space

EOJ大鱼吃小鱼

 1 #include 
 2 using namespace std;
 3 const int maxn = 1e6 + 10;
 4 typedef long long ll;
 5 int a[maxn], b[maxn], id[maxn];
 6 
 7 bool cmp(int i, int j) {
 8     if(a[i] <= b[i] && a[j] > b[j]) return true;
 9     if(a[i] > b[i] && a[j] <= b[j]) return false;
10     if(a[i] <= b[i]) return a[i] < a[j];
11     return b[i] > b[j];
12 }
13 
14 int main() {
15     int n;
16     scanf("%d", &n);
17     for(int i = 1; i <= n; ++i) {
18         scanf("%d %d", a + i, b + i);
19         id[i] = i;
20     }
21     sort(id + 1, id + 1 + n, cmp);
22     ll ex = 0, ans = 0;
23     for(int i = 1; i <= n; ++i) {
24         int o = id[i];
25         if(a[o] > ex) ans += a[o] - ex, ex += a[o] - ex;
26         ex -= a[o] - b[o];
27     }
28     printf("%lld\n", ans);
29     return 0;
30 }

Aguin

 

Forever Young

枚举1e6内基数，否则不会超过四位可以解三次方程，系数都是正的单调可以二分

 1 #include 
 2 using namespace std;
 3 typedef long long ll;
 4 
 5 inline double dcal(int i, int j, int p, int q, ll x) {
 6     double y = 0. + x;
 7     return i * y * y * y + j * y * y + p * y + q;
 8 }
 9 
10 inline ll cal(int i, int j, int p, int q, ll x) {
11     return i * x * x * x + j * x * x + p * x + q;
12 }
13 
14 int main() {
15     ll y, l, ans = 10;
16     scanf("%lld %lld", &y, &l);
17     for(int i = 0; i <= 9; ++i) {
18         for(int j = 0; j <= 9; ++j) {
19             for(int p = 0; p <= 9; ++p) {
20                 for(int q = 0; q <= 9; ++q) {
21                     if(1000 * i + 100 * j + 10 * p + q < l) continue;
22                     ll lb = 11, rb = 1e18;
23                     while(lb < rb) {
24                         ll mid = (lb + rb) / 2;
25                         if(dcal(i, j, p, q, mid) > y + 1 || cal(i, j, p, q, mid) >= y) rb = mid;
26                         else lb = mid + 1;
27                     }
28                     if(cal(i, j, p, q, rb) == y) ans = max(ans, rb);
29                 }
30             }
31         }
32     }
33     for(ll b = 11; b <= 1000000; ++b) {
34         int ok = 1;
35         ll t = y, x = 1, s = 0;
36         while(t) {
37             ll r = t % b;
38             if(r >= 10) {ok = 0; break;}
39             t /= b;
40             s += x * r;
41             x *= 10;
42         }
43         if(ok && s >= l) ans = max(ans, b);
44     }
45     printf("%lld\n", ans);
46     return 0;
47 }

Aguin

 

Oil

枚举端点极角排序扫一遍

 1 #include 
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 2222;
 5 ll x[2][maxn], y[maxn];
 6 int vis[maxn];
 7 
 8 struct point {
 9     ll x, y, id;
10     point(ll _x = 0, ll _y = 0, ll _id = 0) {
11         x = _x, y = _y, id = _id;
12     }
13 };
14 
15 bool cmp(point A, point B) {
16     return A.x * B.y > A.y * B.x;
17 }
18 
19 int main() {
20     int n;
21     scanf("%d", &n);
22     for(int i = 1; i <= n; ++i) {
23         scanf("%lld %lld %lld", x[0] + i, x[1] + i, y + i);
24         if(x[0][i] > x[1][i]) swap(x[0][i], x[1][i]);
25     }
26     ll ans = 0;
27     for(int i = 1; i <= n; ++i) {
28         for(int oi = 0; oi < 2; ++oi) {
29             ll xo = x[oi][i], yo = y[i];
30             vector p;
31             for(int j = 1; j <= n; ++j) {
32                 if(j == i) continue;
33                 for(int oj = 0; oj < 2; ++oj) {
34                     ll xj = x[oj][j] - xo, yj = y[j] - yo;
35                     if(yj == 0) continue;
36                     if(yj < 0) xj = -xj, yj = -yj;
37                     p.emplace_back(point(xj, yj, j));
38                 }
39             }
40             sort(p.begin(), p.end(), cmp);
41             memset(vis, 0, sizeof(vis));
42             int sz = int(p.size());
43             ll tmp = x[1][i] - x[0][i];
44             ans = max(ans, tmp);
45             for(int j = 0; j < sz; ++j) {
46                 int k = j;
47                 while(k + 1 < sz && !cmp(p[k], p[k + 1])) ++k;
48                 for(int r = j; r <= k; ++r) {
49                     ll id = p[r].id;
50                     if(!vis[id]) {
51                         vis[id] = 1;
52                         tmp += x[1][id] - x[0][id];
53                         p[r].id = 0;
54                     }
55                 }
56                 ans = max(ans, tmp);
57                 for(int r = j; r <= k; ++r) {
58                     ll id = p[r].id;
59                     if(vis[id]) {
60                         vis[id] = 0;
61                         tmp -= x[1][id] - x[0][id];
62                     }
63                 }
64                 j = k;
65             }
66         }
67     }
68     printf("%lld\n", ans);
69     return 0;
70 }

Aguin

 

String Theory

合法的一定可以看成是中间一堆单个的形式

 1 #include 
 2 using namespace std;
 3 int a[111];
 4 
 5 int main() {
 6     int n, sum = 0;
 7     scanf("%d", &n);
 8     for(int i = 1; i <= n; ++i) {
 9         scanf("%d", a + i);
10         sum += a[i];
11     }
12     if(sum & 1) puts("no quotation");
13     else {
14         int ans = 1;
15         for(int i = 100; i > 1; --i) {
16             int l = 1, r = n, usedl = 0, usedr = 0, ok = 1;
17             for(int j = i; j >= 1; --j) {
18                 if(l == r) {
19                     if(usedl + 2 * j > a[l]) {ok = 0; break;}
20                     usedl = usedr += 2 * j;
21                 }
22                 else {
23                     if(usedl + j > a[l]) {ok = 0; break;}
24                     usedl += j;
25                     if(usedl == a[l]) {
26                         l++;
27                         if(l == r) usedl = usedr;
28                         else usedl = 0;
29                     }
30                     if(usedr + j > a[r]) {ok = 0; break;}
31                     usedr += j;
32                     if(l == r) usedl += j;
33                     if(r > l && usedr == a[r]) r--, usedr = 0;
34                 }
35             }
36             if(ok) {ans = i; break;}
37         }
38         if(ans == 1 && sum != 2) puts("no quotation");
39         else printf("%d\n", ans);
40     }
41     return 0;
42 }

Aguin

 

Branch Assignment

决策单调性于是分治

 1 #include 
 2 using namespace std;
 3 typedef long long ll;
 4 vector<int> g[2][5555], d[2][5555];
 5 ll sum[5555], f[5555][5555];
 6 
 7 const int inf = 1e9;
 8 typedef pair<int, int> pii;
 9 int dist[5555], sumd[5555];
10 void dijkstra(vector<int> g[], vector<int> d[], int s) {
11     priority_queue pq;
12     for(int i = 0; i < 5555; ++i) dist[i] = inf;
13     dist[s] = 0;
14     pq.push(pii(0, s));
15     while (!pq.empty()) {
16         pii tmp = pq.top();
17         pq.pop();
18         int x = tmp.second;
19         if (dist[x] < -tmp.first) continue;
20         for (int i = 0; i < g[x].size(); ++i) {
21             int to = g[x][i], cost = d[x][i];
22             if (dist[to] > dist[x] + cost) {
23                 dist[to] = dist[x] + cost;
24                 pq.push(pii(-dist[to], to));
25             }
26         }
27     }
28 }
29 
30 void solve(int i, int l, int r, int pl, int pr) {
31     int p = -1;
32     int m = (l + r) / 2;
33     for(int j = pl; j <= min(pr, m - 1); ++j) {
34         ll tmp = f[i - 1][j] + (m - j - 1) * (sum[m] - sum[j]);
35         if(tmp < f[i][m]) f[i][m] = tmp, p = j;
36     }
37     if(m > l) solve(i, l, m - 1, pl, p);
38     if(m < r) solve(i, m + 1, r, p, pr);
39 }
40 
41 int main() {
42     int n, b, s, r;
43     scanf("%d %d %d %d", &n, &b, &s, &r);
44     for(int i = 1; i <= r; ++i) {
45         int u, v, w;
46         scanf("%d %d %d", &u, &v, &w);
47         g[0][u].push_back(v), d[0][u].push_back(w);
48         g[1][v].push_back(u), d[1][v].push_back(w);
49     }
50     int S = b + 1;
51     for(int i = 0; i < 2; ++i) {
52         dijkstra(g[i], d[i], S);
53         for(int j = 1; j <= b; ++j) sumd[j] += dist[j];
54     }
55     sort(sumd + 1, sumd + 1 + b);
56     for(int i = 1; i <= b; ++i) sum[i] = sum[i - 1] + sumd[i];
57     for(int i = 0; i < 5555; ++i)
58         for(int j = 0; j < 5555; ++j)
59             f[i][j] = 1e18;
60     f[0][0] = 0;
61     for(int i = 1; i <= s; ++i)
62         solve(i, i, b, i - 1, b - 1);
63     printf("%lld\n", f[s][b]);
64     return 0;
65 }

Aguin

 

Balanced Diet

把不能吃，能吃但不必须吃，必须吃的边界搞出来，必须吃个数必须小于等于1，如果没有必须吃就从能吃里拿一个最左的，没有能吃的就gg了

 1 #include 
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 1e5 + 10;
 5 ll a[maxn], b[maxn], num[maxn];
 6 
 7 typedef pair pll;
 8 set cant, can;
 9 set<int> must;
10 
11 int main() {
12     int m, k, c, sum = 0;
13     scanf("%d %d", &m, &k);
14     for(int i = 1; i <= m; ++i) scanf("%lld", a + i), sum += a[i];
15     for(int i = 1; i <= k; ++i) {
16         scanf("%lld", b + i);
17         num[b[i]] += 1;
18     }
19     for(int i = 1; i <= m; ++i) {
20         // n * (a[i] / sum) - 1 < num[i] < n * (a[i] / sum) + 1
21         // can't eat
22         // (n + 1) * (a[i] / sum) - 1 < num[i] < (n + 1) * (a[i] / sum) + 1
23         // (num[i] - 1) * sum / a[i] < n + 1 <= num[i] * sum / a[i]
24         // can eat
25         // (n + 1) * (a[i] / sum) - 1 < num[i] + 1 < (n + 1) * (a[i] / sum) + 1
26         // num[i] * sum / a[i] < n + 1 < (num[i] + 1) * sum / a[i]
27         // must eat
28         // (num[i] + 1) * sum / a[i] <= n + 1 < (num[i] + 2) * sum / a[i]
29         if(k + 1 <= (num[i] * sum) / a[i]) cant.insert(pll((num[i] * sum) / a[i], i));
30         else if(k + 1 < ((num[i] + 1) * sum + a[i] - 1) / a[i]) can.insert(pll(((num[i] + 1) * sum + a[i] - 1) / a[i], i));
31         else must.insert(i);
32     }
33     for(c = k + 1; c <= k + sum; ++c) {
34         while(!cant.empty()) {
35             auto it = cant.begin();
36             if(c > (*it).first) {
37                 int x = (*it).second;
38                 cant.erase(*it);
39                 can.insert(pll(((num[x] + 1) * sum + a[x] - 1) / a[x], x));
40             }
41             else break;
42         }
43         while(!can.empty()) {
44             auto it = can.begin();
45             if(c >= (*it).first) {
46                 int x = (*it).second;
47                 can.erase(pll(((num[x] + 1) * sum + a[x] - 1) / a[x], x));
48                 must.insert(x);
49             }
50             else break;
51         }
52         if(must.size() > 1) break;
53         if(must.empty()) {
54             if(can.empty()) break;
55             auto it = can.begin();
56             int x = (*it).second;
57             can.erase(pll(((num[x] + 1) * sum + a[x] - 1) / a[x], x));
58             must.insert(x);
59         }
60         auto it = must.begin();
61         int x = *it;
62         must.erase(*it);
63         num[x] += 1;
64         cant.insert(pll((num[x] * sum) / a[x], x));
65     }
66     if(c == k + sum + 1) puts("forever");
67     else printf("%d\n", c - k - 1);
68     return 0;
69 }

Aguin

 

Clock Breaking

快乐画图每一天

  1 #include 
  2 using namespace std;
  3 char ti[24 * 60][7][22];
  4 char g[111][7][22];
  5 
  6 char ept[7][5] = {
  7         "....",
  8         "....",
  9         "....",
 10         "....",
 11         "....",
 12         "....",
 13         "...."
 14 };
 15 
 16 char col[7][4] = {
 17         "...",
 18         "...",
 19         ".X.",
 20         "...",
 21         ".X.",
 22         "...",
 23         "..."
 24 };
 25 
 26 char non[7][22] = {
 27         ".XX...XX.....XX...XX.",
 28         "X..X.X..X...X..X.X..X",
 29         "X..X.X..X.X.X..X.X..X",
 30         ".XX...XX.....XX...XX.",
 31         "X..X.X..X.X.X..X.X..X",
 32         "X..X.X..X...X..X.X..X",
 33         ".XX...XX.....XX...XX."
 34 };
 35 
 36 char digit[10][7][5] = {
 37 
 38         // 0
 39         ".XX.",
 40         "X..X",
 41         "X..X",
 42         "....",
 43         "X..X",
 44         "X..X",
 45         ".XX.",
 46 
 47         // 1
 48         "....",
 49         "...X",
 50         "...X",
 51         "....",
 52         "...X",
 53         "...X",
 54         "....",
 55 
 56         // 2
 57         ".XX.",
 58         "...X",
 59         "...X",
 60         ".XX.",
 61         "X...",
 62         "X...",
 63         ".XX.",
 64 
 65         // 3
 66         ".XX.",
 67         "...X",
 68         "...X",
 69         ".XX.",
 70         "...X",
 71         "...X",
 72         ".XX.",
 73 
 74         // 4
 75         "....",
 76         "X..X",
 77         "X..X",
 78         ".XX.",
 79         "...X",
 80         "...X",
 81         "....",
 82 
 83         // 5
 84         ".XX.",
 85         "X...",
 86         "X...",
 87         ".XX.",
 88         "...X",
 89         "...X",
 90         ".XX.",
 91 
 92         // 6
 93         ".XX.",
 94         "X...",
 95         "X...",
 96         ".XX.",
 97         "X..X",
 98         "X..X",
 99         ".XX.",
100 
101         // 7
102         ".XX.",
103         "...X",
104         "...X",
105         "....",
106         "...X",
107         "...X",
108         "....",
109 
110         // 8
111         ".XX.",
112         "X..X",
113         "X..X",
114         ".XX.",
115         "X..X",
116         "X..X",
117         ".XX.",
118 
119         // 9
120         ".XX.",
121         "X..X",
122         "X..X",
123         ".XX.",
124         "...X",
125         "...X",
126         ".XX."
127 };
128 
129 inline void draw(char g[][22], int c, int w, char fig[][5]) {
130     for(int i = 0; i < 7; ++i) {
131         for(int j = c; j < c + w; ++j) {
132             g[i][j] = fig[i][j - c];
133         }
134     }
135 }
136 
137 inline void drawcol(char g[][22]) {
138     for(int i = 0; i < 7; ++i) {
139         for(int j = 9; j < 12; ++j) {
140             g[i][j] = col[i][j - 9];
141         }
142     }
143 }
144 
145 bool on[7][22], off[7][22], work[7][22];
146 int cnt_on[7][22], cnt_off[7][22];
147 bool check(char g[][22], char fig[][22]) {
148     for(int i = 0; i < 7; ++i) {
149         for(int j = 0; j < 21; ++j) {
150             if(fig[i][j] == '.' && g[i][j] == 'X') on[i][j] = true;
151             if(fig[i][j] == 'X' && g[i][j] == '.') off[i][j] = true;
152         }
153     }
154 }
155 
156 int main() {
157     for(int h = 0; h < 24; ++h) {
158         for(int m = 0; m < 60; ++m) {
159             int t = 60 * h + m;
160             if(h < 10) draw(ti[t], 0, 4, ept);
161             else draw(ti[t], 0, 4, digit[h / 10]);
162             drawcol(ti[t]);
163             draw(ti[t], 5, 4, digit[h % 10]);
164             draw(ti[t], 12, 4, digit[m / 10]);
165             draw(ti[t], 17, 4, digit[m % 10]);
166         }
167     }
168     int n, cnt = 0;
169     scanf("%d", &n);
170     for(int i = 0; i < n; ++i)
171         for(int j = 0; j < 7; ++j)
172             scanf("%s", g[i][j]);
173     for(int st = 0; st < 24 * 60; ++st) {
174         memset(on, 0, sizeof(on));
175         memset(off, 0, sizeof(off));
176         for(int i = 0; i < n; ++i) {
177             check(g[i], ti[(st + i) % (24 * 60)]);
178         }
179         int ok = 1;
180         for(int i = 0; i < n; ++i) {
181             for(int j = 0; j < 7; ++j) {
182                 for(int k = 0; k < 21; ++k) {
183                     if(on[j][k] && g[i][j][k] == '.') ok = 0;
184                     if(off[j][k] && g[i][j][k] == 'X') ok = 0;
185                 }
186             }
187         }
188         if(ok) {
189             cnt += 1;
190             for(int i = 0; i < 7; ++i) {
191                 for(int j = 0; j < 21; ++j) {
192                     if(on[i][j]) cnt_on[i][j] += 1;
193                     if(off[i][j]) cnt_off[i][j] += 1;
194                 }
195             }
196         }
197     }
198     for(int i = 1; i < n; ++i) {
199         for(int j = 0; j < 7; ++j) {
200             for(int k = 0; k < 21; ++k) {
201                 if(g[i][j][k] != g[i - 1][j][k]) work[j][k] = 1;
202             }
203         }
204     }
205     if(!cnt) puts("impossible");
206     else {
207         for(int i = 0; i < 7; ++i) {
208             for(int j = 0; j < 21; ++j) {
209                 if(non[i][j] == '.') putchar('.');
210                 else if(cnt_on[i][j] == cnt) putchar('1');
211                 else if(cnt_off[i][j] == cnt) putchar('0');
212                 else if(work[i][j] && !cnt_on[i][j] && !cnt_off[i][j]) putchar('W');
213                 else putchar('?');
214             }
215             puts("");
216         }
217     }
218     return 0;
219 }

Aguin

 

Longest Rivers

距离从大到小枚举叶子，如果一个子树里最短的距离都比这个大，至少有一个河会更长，就放弃这条河直接把海水引过来

 1 #include 
 2 using namespace std;
 3 typedef long long ll;
 4 const ll inf = 1e18;
 5 const int maxn = 1e6 + 10;
 6 vector<int> g[maxn];
 7 char s[maxn][22];
 8 int f[maxn];
 9 ll d[maxn];
10 
11 ll dis[maxn], mi[maxn];
12 void dfs(int x) {
13     mi[x] = inf;
14     for(auto y: g[x]) {
15         dis[y] = dis[x] + d[y];
16         dfs(y), mi[x] = min(mi[x], mi[y]);
17     }
18     if(mi[x] == inf) mi[x] = 0;
19     mi[x] += d[x];
20 }
21 
22 bool cmpnd(int i, int j) {
23     return mi[i] > mi[j];
24 }
25 
26 bool cmplf(int i, int j) {
27     return dis[i] > dis[j];
28 }
29 
30 int lf[maxn], nd[maxn], vis[maxn], used[maxn], ans[maxn];
31 int main() {
32     int n, m;
33     scanf("%d %d", &n, &m);
34     for(int i = 1; i <= n; ++i) {
35         scanf("%s %d %lld", s[m + i], f + m + i, d + m + i);
36         g[f[m + i]].push_back(m + i);
37         lf[m + i] = nd[m + i] = m + i;
38     }
39     for(int i = 1; i <= m; ++i) {
40         scanf("%d %lld", f + i, d + i);
41         g[f[i]].push_back(i);
42         nd[i] = i;
43     }
44     dfs(0);
45     sort(nd + 1, nd + n + m + 1, cmpnd);
46     sort(lf + m + 1, lf + n + m + 1, cmplf);
47     int p = 0, tmp = 1;
48     for(int i = m + 1; i <= n + m; ++i) {
49         while(p < n + m && mi[nd[p + 1]] > dis[lf[i]]) {
50             int t = nd[p + 1];
51             while(t && !vis[t]) {
52                 vis[t] = 1;
53                 if(!used[t]) tmp++;
54                 if(vis[f[t]] && !used[f[t]]) tmp--;
55                 used[f[t]] = 1;
56                 t = f[t];
57             }
58             ++p;
59         }
60         ans[lf[i]] = tmp;
61     }
62     for(int i = m + 1; i <= n + m; ++i) printf("%s %d\n", s[i], ans[i]);
63     return 0;
64 }

Aguin

 

Road Times

把每条路时间当变量就是线性规划

  1 #include 
  2 using namespace std;
  3 int g[33][33];
  4 
  5 const double eps = 1e-8;
  6 int cmp(const double &x) {
  7     return x < -eps ? -1 : x > eps;
  8 }
  9 const int maxn = 444, maxm = 888, inf = 1e9;
 10 double a[maxm][maxn], b[maxm];
 11 double ca[maxm][maxn], cb[maxm];
 12 int idn[maxn], idm[maxm];
 13 int n, m;
 14 void pivot(int x,int y) {
 15     swap(idm[x], idn[y]);
 16     double k = a[x][y];
 17     for (int j = 1; j <= n; ++j)a[x][j] /= k;
 18     b[x] /= k;
 19     a[x][y] = 1 / k;
 20     for (int i = 0; i <= m; ++i)
 21         if (i != x) {
 22             k = a[i][y];
 23             if (!cmp(k))continue;
 24             b[i] -= k * b[x];
 25             a[i][y] = 0;
 26             for (int j = 1; j <= n; ++j)a[i][j] -= a[x][j] * k;
 27         }
 28 }
 29 void simplex() {
 30     idn[0] = inf;
 31     while (1) {
 32         int y = 0;
 33         for (int j = 1; j <= n; ++j)
 34             if (cmp(a[0][j]) > 0 && idn[j] < idn[y])y = j;
 35         if (!y)break;
 36         int x = 0;
 37         for (int i = 1; i <= m; ++i)
 38             if (cmp(a[i][y]) > 0) {
 39                 if (!x)x = i;
 40                 else {
 41                     int t = cmp(b[i] / a[i][y] - b[x] / a[x][y]);
 42                     if (t < 0 || (t == 0 && idm[i] < idm[x]))x = i;
 43                 }
 44             }
 45         if (!x) {
 46             puts("Unbounded");
 47             exit(0);
 48         }
 49         pivot(x, y);
 50     }
 51 }
 52 void init() {
 53     idm[0] = inf;
 54     idn[0] = inf;
 55     while (1) {
 56         int x = 0;
 57         for (int i = 1; i <= m; ++i)
 58             if (cmp(b[i]) < 0 && idm[i] < idm[x])x = i;
 59         if (!x)break;
 60         int y = 0;
 61         for (int j = 1; j <= n; ++j)
 62             if (cmp(a[x][j]) < 0 && idn[j] < idn[y])y = j;
 63         if (!y) {
 64             puts("Infeasible");
 65             exit(0);
 66         }
 67         pivot(x, y);
 68     }
 69 }
 70 
 71 int N;
 72 vector<int> get_path(int s, int t) {
 73     if(s == t) return vector<int>(1, s);
 74     for(int i = 0; i < N; ++i) {
 75         if(g[s][i] && g[i][t] && g[s][i] + g[i][t] == g[s][t]) {
 76             vector<int> pre = get_path(s, i);
 77             vector<int> suf = get_path(i, t);
 78             for(int j = 1; j < suf.size(); ++j) pre.push_back(suf[j]);
 79             return pre;
 80         }
 81     }
 82     return vector<int>{s, t};
 83 }
 84 
 85 int id[33][33];
 86 int main() {
 87     int R, Q;
 88     scanf("%d", &N);
 89     for(int i = 0; i < N; ++i) {
 90         for(int j = 0; j < N; ++j) {
 91             scanf("%d", &g[i][j]);
 92             if(i != j && g[i][j] != -1) {
 93                 id[i][j] = ++n;
 94                 ++m;
 95                 a[m][n] = 1, b[m] = g[i][j] + g[i][j];
 96                 ++m;
 97                 a[m][n] = -1, b[m] = -g[i][j];
 98             }
 99             if(g[i][j] == -1) g[i][j] = inf;
100         }
101     }
102     for(int k = 0; k < N; ++k)
103         for(int i = 0; i < N; ++i)
104             for(int j = 0; j < N; ++j)
105                 g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
106     scanf("%d", &R);
107     for(int i = 1; i <= R; ++i) {
108         int s, t, d;
109         scanf("%d %d %d", &s, &t, &d);
110         vector<int> path = get_path(s, t);
111         ++m;
112         for(int j = 0; j < path.size() - 1; ++j) {
113             a[m][id[path[j]][path[j + 1]]] = 1;
114         }
115         b[m] = d;
116         ++m;
117         for(int j = 0; j < path.size() - 1; ++j) {
118             a[m][id[path[j]][path[j + 1]]] = -1;
119         }
120         b[m] = -d;
121     }
122     memcpy(ca, a, sizeof(ca));
123     memcpy(cb, b, sizeof(cb));
124     scanf("%d", &Q);
125     while(Q--) {
126         int s, t;
127         scanf("%d %d", &s, &t);
128         printf("%d %d ", s, t);
129         vector<int> path = get_path(s, t);
130         memcpy(a, ca, sizeof(a));
131         memcpy(b, cb, sizeof(b));
132         for(int j = 1; j <= n; ++j) idn[j] = j;
133         for(int i = 1; i <= m; ++i) idm[i] = n + i;
134         for(int i = 0; i < path.size() - 1; ++i) a[0][id[path[i]][path[i + 1]]] = -1;
135         init();
136         simplex();
137         printf("%.9f ", b[0]);
138         memcpy(a, ca, sizeof(a));
139         memcpy(b, cb, sizeof(b));
140         for(int j = 1; j <= n; ++j) idn[j] = j;
141         for(int i = 1; i <= m; ++i) idm[i] = n + i;
142         for(int i = 0; i < path.size() - 1; ++i) a[0][id[path[i]][path[i + 1]]] = 1;
143         init();
144         simplex();
145         printf("%.9f\n", -b[0]);
146     }
147     return 0;
148 }

Aguin

 

转载于:https://www.cnblogs.com/Aguin/p/10615957.html