Blow up the city
时间限制: 1 Sec 内存限制: 128 MB题目描述
Country A and B are at war. Country A needs to organize transport teams to deliver supplies toward some command center cities.
In order to ensure the delivery works efficiently, all the roads in country A work only one direction. Therefore, map of country A can be regarded as DAG( Directed Acyclic Graph ). Command center cities only received supplies and not send out supplies.
Intelligence agency of country B is credibly informed that there will be two cities carrying out a critical transporting task in country A.
As long as **any** one of the two cities can not reach a command center city, the mission fails and country B will hold an enormous advantage. Therefore, country B plans to destroy one of the n cities in country A and all the roads directly connected. (If a city carrying out the task is also a command center city, it is possible to destroy the city to make the mission fail)
Now country B has made q hypotheses about the two cities carrying out the critical task.
Calculate the number of plan that makes the mission of country A fail.
In order to ensure the delivery works efficiently, all the roads in country A work only one direction. Therefore, map of country A can be regarded as DAG( Directed Acyclic Graph ). Command center cities only received supplies and not send out supplies.
Intelligence agency of country B is credibly informed that there will be two cities carrying out a critical transporting task in country A.
As long as **any** one of the two cities can not reach a command center city, the mission fails and country B will hold an enormous advantage. Therefore, country B plans to destroy one of the n cities in country A and all the roads directly connected. (If a city carrying out the task is also a command center city, it is possible to destroy the city to make the mission fail)
Now country B has made q hypotheses about the two cities carrying out the critical task.
Calculate the number of plan that makes the mission of country A fail.
输入
The first line contains a integer T (1≤T≤10), denoting the number of test cases.
In each test case, the first line are two integers n,m, denoting the number of cities and roads(1≤n≤100,000,1≤m≤200,000).
Then m lines follow, each with two integers u and v, which means there is a directed road from city u to v (1≤u,v≤n,u≠v).
The next line is a integer q, denoting the number of queries (1≤q≤100,000)
And then q lines follow, each with two integers a and b, which means the two cities carrying out the critical task are a and b (1≤a,b≤n,a≠b).
A city is a command center if and only if there is no road from it (its out degree is zero).
In each test case, the first line are two integers n,m, denoting the number of cities and roads(1≤n≤100,000,1≤m≤200,000).
Then m lines follow, each with two integers u and v, which means there is a directed road from city u to v (1≤u,v≤n,u≠v).
The next line is a integer q, denoting the number of queries (1≤q≤100,000)
And then q lines follow, each with two integers a and b, which means the two cities carrying out the critical task are a and b (1≤a,b≤n,a≠b).
A city is a command center if and only if there is no road from it (its out degree is zero).
输出
For each query output a line with one integer, means the number of plan that makes the mission of country A fail.
样例输入
2
8 8
1 2
3 4
3 5
4 6
4 7
5 7
6 8
7 8
2
1 3
6 7
3 2
3 1
3 2
2
1 2
3 1
样例输出
4
3
2
2
题意：有一个DAG，求有多少方案，使得割掉图中一个点，图上特定两个点不能都到达出度为0的点。
思路：反向建边，连接超级源点到原图中出度为0的点。那么题目就变成了：有一个DAG，求有多少种方案，使得割掉图中一个点，从超级源点不能都到达两个特定点。这就是支配树问题。支配树是一个有向图抽象出来的树，它满足某一点到根节点的路径上的点，都是根结点在原图中到达这个点的必经点。
#include<bits/stdc++.h> #define N 100005 using namespace std; struct ss { int u,v,next; }; ss edg[2*N]; int head[N],sum_edge=0; void addedge(int u,int v) { edg[sum_edge]=(ss){u,v,head[u]}; head[u]=sum_edge++; } int grand[N][25]={0}; int depth[N],DEPTH; void deal(int u,int v) { grand[v][0]=u; depth[v]=depth[u]+1; for(int i=1;i<=DEPTH;i++) { grand[v][i]=grand[grand[v][i-1]][i-1]; } } int lca(int a,int b) { if(a==0||b==0)return a+b; if(depth[a]>depth[b])swap(a,b); for(int i=DEPTH;i>=0;i--) if(depth[a]<depth[b]&&depth[grand[b][i]]>=depth[a])b=grand[b][i]; for(int i=DEPTH;i>=0;i--) if(grand[a][i]!=grand[b][i]) { a=grand[a][i]; b=grand[b][i]; } if(a!=b) { return grand[a][0]; } return a; } void init(int n) { DEPTH=floor(log(n + 0.0) / log(2.0)); memset(grand[0],0,sizeof(grand[0])); depth[0]=0; sum_edge=0; memset(head,-1,sizeof(head)); } void getControl_tree(int n,int s)//从s出发可以到达图的所有点,s为支配树的根 { int fa[N]={0}; int rd[N]={0}; stack<int>Stack; for(int i=1;i<=n;i++) for(int j=head[i];j!=-1;j=edg[j].next)rd[edg[j].v]++; Stack.push(s); while(!Stack.empty()) { int x=Stack.top(); Stack.pop(); deal(fa[x],x);//建支配树 for(int i=head[x];i!=-1;i=edg[i].next) { int v=edg[i].v; fa[v]=lca(fa[v],x); rd[v]--; if(!rd[v])Stack.push(v); } } } int main() { int t; scanf("%d",&t); while(t--) { int n,m; scanf("%d %d",&n,&m); init(n); int rd[N]={0}; while(m--) { int u,v; scanf("%d %d",&u,&v); addedge(v,u); rd[u]++; } for(int i=1;i<=n;i++)if(!rd[i])addedge(n+1,i); getControl_tree(n+1,n+1); scanf("%d",&m); while(m--) { int u,v; scanf("%d %d",&u,&v); int LCA=lca(u,v); printf("%d\n",depth[u]+depth[v]-depth[LCA]-1-1+1); } } return 0; }
支配树参考博客：https://www.cnblogs.com/fenghaoran/p/dominator_tree.html