2019 ICPC NanChang Regional Online Contest

有点毒的比赛。

题目链接：https://www.jisuanke.com/contest/3870?view=challenges

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B：

添加一个点连接所有救火点，路径设为0，就变成了跑两次dijkstra的题。

 1 /* Contest nanchang_2019_online
 2  * Problem B
 3  * Team: Make One For Us
 4  */
 5 #include 
 6 
 7 using namespace std;
 8 
 9 const long long int INF = 10000000000000L;
10 
11 struct edge {
12     int u, v;
13     long long int w;
14 };
15 
16 struct node {
17     int u;
18     long long int d;
19     bool operator < (const node &n) const {
20         return d > n.d;
21     }
22 };
23 
24 long long int read(void) {
25     char ch;
26     do {
27         ch = getchar();
28     } while (!isdigit(ch));
29     long long int ret = 0;
30     while (isdigit(ch)) {
31         ret *= 10;
32         ret += ch - '0';
33         ch = getchar();
34     }
35     return ret;
36 }
37 
38 void dijkstra(vector int>> &graph, vector  &edges, vector <long long int> &dis, int s) {
39     priority_queue  pq;
40     pq.push({s, dis[s] = 0});
41     vector <int> vis(graph.size());
42     while (!pq.empty()) {
43         auto tmp = pq.top();
44         pq.pop();
45         if (vis[tmp.u]) {
46             continue;
47         }
48         vis[tmp.u] = true;
49         for (unsigned int i = 0; i < graph[tmp.u].size(); i++) {
50             edge &e = edges[graph[tmp.u][i]];
51             if (dis[e.v] > dis[e.u] + e.w) {
52                 pq.push((node) {e.v, dis[e.v] = dis[e.u] + e.w});
53             }
54         }
55     }
56 }
57 
58 int main(void) {
59     ios::sync_with_stdio(false);
60     cin.tie(nullptr);
61     int T;
62     T = read();
63     while (T--) {
64         int n = read(), m = read(), s = read(), k = read(), c = read();
65         int ptr = 0;
66         vector  edges(2 * m + k);
67         vector int>> graph(n + 1);
68         auto add_edge = [&] (int u, int v, long long int w) {
69             graph[u].emplace_back(ptr);
70             edges[ptr++] = ((edge) {u, v, w});
71         };
72         for (int i = 0; i < k; i++) {
73             int v = read();
74             add_edge(0, v, 0);
75         }
76         for (int i = 0; i < m; i++) {
77             int u = read(), v = read();
78             long long int w = read();
79             add_edge(u, v, w);
80             add_edge(v, u, w);
81         }
82 
83         vector <long long int> dis(n + 1, INF);
84         dijkstra(graph, edges, dis, s);
85         auto ans_s = *max_element(dis.begin() + 1, dis.end());
86         for (int i = 0; i <= n; i++) {
87             dis[i] = INF;
88         }
89         dijkstra(graph, edges, dis, 0);
90         auto ans_k = *max_element(dis.begin() + 1, dis.end());
91 
92         if (ans_k * c < ans_s) {
93             cout << ans_k << endl;
94         } else {
95             cout << ans_s << endl;
96         }
97     }
98     return 0;
99 }

View Code

C：

一开始以为是莫队后来发现不对，就以为是字符串乱搞，结果是线段树维护dp矩阵区间合并。原题是Codeforces goodbye 2016 E。(tourist都没做出来的题这么多人做出来，有点假啊)

 

E：

一个n有4e7的约瑟夫问题居然是模拟，死都不信，垃圾数据。

G：

签到。

 1 /* basic header */
 2 #include 
 3 /* define */
 4 #define ll long long
 5 #define dou double
 6 #define pb emplace_back
 7 #define mp make_pair
 8 #define sot(a,b) sort(a+1,a+1+b)
 9 #define rep1(i,a,b) for(int i=a;i<=b;++i)
10 #define rep0(i,a,b) for(int i=a;i11 #define eps 1e-8
12 #define int_inf 0x3f3f3f3f
13 #define ll_inf 0x7f7f7f7f7f7f7f7f
14 #define lson (curpos<<1)
15 #define rson (curpos<<1|1)
16 /* namespace */
17 using namespace std;
18 /* header end */
19 
20 int t;
21 
22 int main() {
23     scanf("%d", &t);
24     while (t--) {
25         ll n; scanf("%lld", &n);
26         if (n == 1) puts("18000");
27         else puts("0");
28     }
29     return 0;
30 }

View Code

H：

二分矩阵快速幂+记忆化。

 1 #include
 2 typedef long long ull;
 3 
 4 using namespace std;
 5 
 6 const int maxn = 1e6 + 10;
 7 const ull mod = 998244353;
 8 
 9 struct Matrix {
10     ull mem[2][2];
11     void init() {
12         memset(mem, 0, sizeof(mem));
13     }
14     Matrix operator*(const Matrix &rhs)const {
15         Matrix ret; ret.init();
16         for (int i = 0; i < 2; i++) {
17             for (int j = 0; j < 2; j++) {
18                 for (int k = 0; k < 2; k++)
19                     ret.mem[i][j] += mem[i][k] * rhs.mem[k][j];
20                 ret.mem[i][j] %= mod;
21             }
22         }
23         return ret;
24     }
25 } mAns, mBase;
26 unordered_map M;
27 
28 int tot = 0;
29 Matrix solve(ull n) {
30     if (M.count(n)) return M[n];
31     if (n == 1) {
32         M[n] = mBase;
33         return mBase;
34     }
35     Matrix ret = solve(n / 2ull);
36     ret = ret * ret;
37     if (n % 2ull == 1ull) ret = ret * mBase;
38     return M[n] = ret;
39 }
40 
41 int main() {
42     ull q, n;
43     cin >> q >> n;
44     ull lastans = 0ull;
45     ull ans = 0;
46     for (int i = 1ull; i <= q; i++) {
47         mAns.mem[0][0] = 1, mAns.mem[0][1] = 0;
48         mBase.mem[0][0] = 3, mBase.mem[0][1] = 1, mBase.mem[1][0] = 2;
49         if (i >= 2ull) n = (n ^ (ans * ans));
50         Matrix val = solve(n % (mod - 1));
51         ans = val.mem[0][1];
52         lastans = (lastans ^ ans);
53     }
54     cout << lastans << endl;
55     return 0;
56 }

View Code

L：

三位偏序，cdq分治。待补。