poj 2762 Going from u to v or from v to u?【强连通分量缩点+拓扑排序】

Going from u to v or from v to u?

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15812		Accepted: 4194

Description

In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each time, Wind choose two rooms x and y, and ask one of their little sons go from one to the other. The son can either go from x to y, or from y to x. Wind promised that her tasks are all possible, but she actually doesn't know how to decide if a task is possible. To make her life easier, Jiajia decided to choose a cave in which every pair of rooms is a possible task. Given a cave, can you tell Jiajia whether Wind can randomly choose two rooms without worrying about anything?

Input

The first line contains a single integer T, the number of test cases. And followed T cases.

The first line for each case contains two integers n, m(0 < n
< 1001,m < 6000), the number of rooms and corridors in the cave.
The next m lines each contains two integers u and v, indicating that
there is a corridor connecting room u and room v directly.

Output

The output should contain T lines. Write 'Yes' if the cave has the property stated above, or 'No' otherwise.

Sample Input

1
3 3
1 2
2 3
3 1

Sample Output

Yes
题目大意：n个节点 m条单向边。构成一个图，问你是否可以保证任选2个点u、v，这两个点之间是连通的？(能从x到达y，或者能从y到达x既可，不一定是x与y是互通
的) 保证则输出：Yes 否则：No    【强连通缩点， 到最后我们只要进行一次拓扑排序，看看是否存在一条一次可以走完的路径既可】
强连通分量算法将图化简缩点后，进行拓扑排序。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <vector>
#include <stack>
#include <algorithm>
#define N 1002

using namespace std;

struct node
{
    int e, val;
}item;
vector<node>g[N];
stack<int>st;

int dfn[N], low[N];
int id[N]; //标记i节点属于哪个连通分量
int mat2[N][N];
int n2;
int counter, n, scnt;

void Tarjan(int e)
{
    int t, i;
    int mm=low[e]=dfn[e]=counter++;
    st.push(e);//入栈
    for(i=0; i<g[e].size(); i++){
        t=g[e][i].e;
        if(dfn[t]==-1)
            Tarjan(t);
        if(low[t]<mm)
            mm=low[t];
    }
    if(mm<low[e]){//有回边
        low[e]=mm; return;
    }
    do{
        id[t=st.top()]=scnt; low[t]=n;
        st.pop();
    }while(t!=e);
    scnt++;
}

void Search(int n)
{
    int i;
    memset(dfn, -1, sizeof(dfn));
    counter=0; scnt=0;
    for(i=0; i<n; i++)
        if(dfn[i]==-1)
            Tarjan(i);
}

void base_vertex()
{
    int i, j, t;
    Search(n); //调用求强连通分量
    n2=scnt;
    memset(mat2,0,sizeof(mat2));
    for(i=0; i<n; i++){
        for(j=0; j<g[i].size(); j++){
            t=g[i][j].e; mat2[id[i]][id[t]]=1;
        }
    }
}

int topo_sort(int n, int mat[][N], int *topo)
{//拓扑排序的n是强连通分量的个数
    int d[N], i, j, k;
    for(i=0; i<n; i++){
        d[i]=0;
        for(j=0; j<n; j++)
            d[i]=d[i]+mat[j][i];//入度数
    }
    for(k=0; k<n; k++){
        for(i=0; d[i]&&i<n; i++);
        if(i==n) return 0;
        d[i]=-1;
        for(j=0; j<n; j++)
            d[j]-=mat[i][j];
        topo[k]=i;
    }
    return 1;
}
int main()
{
    int tg; scanf("%d", &tg);
    int m, u, v;
    int i, j;
    int topo[N];
    while(tg--)
    {
        scanf("%d %d", &n, &m);
        for(i=0; i<=n; i++)
            g[i].clear();
        for(i=0; i<m; i++){
            scanf("%d %d", &u, &v);
            item.e=v-1; item.val=1;
            g[u-1].push_back(item);
        }//建立单向图
        base_vertex();
        topo_sort(n2, mat2, topo);
        int flag=1;
        for(i=0; i<n2-1; i++){
            if(!mat2[topo[i]][topo[i+1]] ){
                flag=0; break;
            }
        }
        if(flag) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}

/*
    下面的数据会访问冲突崩溃掉 但poj可以Accepted 不懂~~~
1
8 11
1 2
2 3
2 5
2 6
3 5
4 3
5 2
5 4
6 7
6 8
7 6

*/