洛谷P3242 接水果

关于矩形与点其实有两种关系。

一种是每个矩形包含多少点。一种是每个点被多少矩形包含。

 

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解：因为可以离线所以直接套整体二分。关键是考虑如何能够被覆盖。

我一开始都是想的树上操作...其实是转化成DFS序。分链和有lca两种情况。

考虑每个盘子能接住的水果，两端DFS序满足的性质。发现是二维平面上的矩形。

一个水果就是询问一个点被多少矩形覆盖(能被多少盘子接)。于是整体二分里面扫描线，片改点查用树状数组。

 

  1 #include 
  2 
  3 const int N = 80010;
  4 
  5 struct Edge {
  6     int nex, v;
  7 }edge[N << 1]; int tp;
  8 
  9 int e[N], pos[N], ed[N], num, n, fa[N][20], pw[N], d[N]; /// tree
 10 int X[N], xx, ans[N];
 11 
 12 struct Node {
 13     int id, x, y, k, z;
 14     Node(int ID = 0, int X = 0, int Y = 0, int K = 0, int Z = 0) {
 15         id = ID;
 16         x = X;
 17         y = Y;
 18         k = K;
 19         z = Z;
 20     }
 21     inline bool operator <(const Node &w) const {
 22         if(id != w.id) return id < w.id;
 23         return z < w.z;
 24     }
 25 }node[N], t1[N], t2[N], stk[N << 2];
 26 
 27 namespace ta {
 28     int ta[N];
 29     inline void add(int x, int v) {
 30         for(int i = x; i <= n + 1; i += (i & (-i))) {
 31             ta[i] += v;
 32         }
 33         return;
 34     }
 35     inline int ask(int x) {
 36         int ans = 0;
 37         for(int i = x; i >= 1; i -= (i & (-i))) {
 38             ans += ta[i];
 39         }
 40         return ans;
 41     }
 42     inline void del(int x) {
 43         for(int i = x; i <= n + 1; i += (i & (-i))) {
 44             ta[i] = 0;
 45         }
 46         return;
 47     }
 48 }
 49 
 50 inline void add(int x, int y) {
 51     tp++;
 52     edge[tp].v = y;
 53     edge[tp].nex = e[x];
 54     e[x] = tp;
 55     return;
 56 }
 57 
 58 void DFS(int x, int f) {
 59     pos[x] = ++num;
 60     fa[x][0] = f;
 61     d[x] = d[f] + 1;
 62     for(int i = e[x]; i; i = edge[i].nex) {
 63         int y = edge[i].v;
 64         if(y == f) continue;
 65         DFS(y, x);
 66     }
 67     ed[x] = num;
 68     return;
 69 }
 70 
 71 inline void prework() {
 72     for(int i = 2; i <= n; i++) pw[i] = pw[i >> 1] + 1;
 73     for(int j = 1; j <= pw[n]; j++) {
 74         for(int i = 1; i <= n; i++) {
 75             fa[i][j] = fa[fa[i][j - 1]][j - 1];
 76         }
 77     }
 78     return;
 79 }
 80 
 81 inline int lca(int x, int y) {
 82     if(d[x] > d[y]) std::swap(x, y);
 83     int t = pw[n];
 84     while(t >= 0 && d[y] > d[x]) {
 85         if(d[fa[y][t]] >= d[x]) {
 86             y = fa[y][t];
 87         }
 88         t--;
 89     }
 90     if(x == y) return x;
 91     t = pw[n];
 92     while(t >= 0 && fa[x][0] != fa[y][0]) {
 93         if(fa[x][t] != fa[y][t]) {
 94             x = fa[x][t];
 95             y = fa[y][t];
 96         }
 97         t--;
 98     }
 99     return fa[x][0];
100 }
101 
102 inline int getPos(int x, int y) {
103     int t = pw[n];
104     while(t >= 0 && fa[y][0] != x) {
105         if(d[fa[y][t]] > d[x]) {
106             y = fa[y][t];
107         }
108         t--;
109     }
110     return y;
111 }
112 
113 void Div(int L, int R, int l, int r) {
114     if(L > R) return;
115     if(l == r) {
116         for(int i = L; i <= R; i++) {
117             if(node[i].id) ans[node[i].id] = r;
118         }
119         return;
120     }
121 
122     int mid = (l + r) >> 1, top1 = 0, top2 = 0, top = 0;
123     for(int i = L; i <= R; i++) {
124         int x = node[i].x, y = node[i].y, z = node[i].z;
125         if(!node[i].id) { /// change  |  this is a Matrix
126             if(node[i].k > mid) {
127                 t2[++top2] = node[i];
128                 continue;
129             }
130             t1[++top1] = node[i];
131             if(z) { /// line
132                 stk[++top] = Node(1, pos[y], ed[y], 1, 0);
133                 stk[++top] = Node(pos[z], pos[y], ed[y], -1, 0);
134                 stk[++top] = Node(pos[y], ed[z] + 1, n, 1, 0);
135                 stk[++top] = Node(ed[y] + 1, ed[z] + 1, n, -1, 0);
136             }
137             else { /// lca
138                 stk[++top] = Node(pos[x], pos[y], ed[y], 1, 0);
139                 stk[++top] = Node(ed[x] + 1, pos[y], ed[y], -1, 0);
140             }
141         }
142         else { /// ask  |  this is a Point
143             stk[++top] = Node(pos[x], 0, pos[y], i, 1);
144         }
145     }
146     std::sort(stk + 1, stk + top + 1);
147     for(int i = 1; i <= top; i++) {
148         if(stk[i].z == 0) { /// change
149             ta::add(stk[i].x, stk[i].k);
150             ta::add(stk[i].y + 1, -stk[i].k);
151         }
152         else { /// ask
153             int t = ta::ask(stk[i].y), id = stk[i].k;
154             if(node[id].k <= t) {
155                 t1[++top1] = node[id];
156             }
157             else {
158                 node[id].k -= t;
159                 t2[++top2] = node[id];
160             }
161         }
162     }
163     for(int i = 1; i <= top; i++) { /// clear
164         if(stk[i].z == 0) {
165             ta::del(stk[i].x);
166             ta::del(stk[i].y + 1);
167         }
168     }
169     memcpy(node + L, t1 + 1, top1 * sizeof(Node));
170     memcpy(node + L + top1, t2 + 1, top2 * sizeof(Node));
171     Div(L, L + top1 - 1, l, mid);
172     Div(L + top1, R, mid + 1, r);
173     return;
174 }
175 
176 int main() {
177     int m, q;
178     scanf("%d%d%d", &n, &m, &q);
179     for(int i = 1, x, y; i < n; i++) {
180         scanf("%d%d", &x, &y);
181         add(x, y); add(y, x);
182     }
183     DFS(1, 0);
184     prework();
185     for(int i = 1, x, y, z; i <= m; i++) {
186         scanf("%d%d%d", &x, &y, &X[i]);
187         if(pos[x] > pos[y]) std::swap(x, y);
188         z = lca(x, y);
189         node[i] = Node(0, x, y, X[i], (z == x) ? getPos(x, y) : 0);
190     }
191     std::sort(X + 1, X + n + 1);
192     xx = std::unique(X + 1, X + n + 1) - X - 1;
193     for(int i = 1; i <= m; i++) {
194         node[i].k = std::lower_bound(X + 1, X + xx + 1, node[i].k) - X;
195     }
196     for(int i = 1, x, y, k; i <= q; i++) {
197         scanf("%d%d%d", &x, &y, &k);
198         if(pos[x] > pos[y]) std::swap(x, y);
199         node[m + i] = Node(i, x, y, k, 0);
200     }
201     Div(1, m + q, 1, xx);
202     for(int i = 1; i <= q; i++) printf("%d\n", X[ans[i]]);
203     return 0;
204 }

AC代码

 

转载于:https://www.cnblogs.com/huyufeifei/p/10485234.html