Blow up the city（HDU-6604）
Problem Description
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.
Input
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).
Output
For each query output a line with one integer, means the number of plan that makes the mission of country A fail.
Sample Input
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 1Sample Output
4
3
2
2
题意：t 组数据，每组数据给出 n 个点 m 条边构成一个 DAG 图，然后给出 q 组查询，每组询问从点 x 到点 y 之间有多少个点被删掉后，使得这两个点中的任何一个无法到达另一个
思路：支配树的 DAG 图模版题
首先进行反向拓扑，然后按照拓扑序建立支配树
由于某点的支配树上的父亲就是该点在 DAG 中的所有出点在支配树上的 LCA，因此对于询问 query(x,y)，结果也就是 x 的深度与 y 的深度的和减去 LCA(x,y) 的深度
Source Program
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include