poj 2762 Popular Cows + 强连通 + 重构图 + 拓扑排序

/*
本题要判断图中两两可达， 在用强联通缩点后， 二次构图，进行拓扑排序， 在排序过程中若出现一个以上入度为零的点， 这次图不能满足
图中点两两可达，
*/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>

using namespace std;

const int M = 2005;
const int N = 6005;

struct node {

    int to;
    int next;

}num[N*2], num1[N];

int ins[M];
int dfn[M];
int low[M];
int rdu[M];
int cdu[M];
int sccf[M];
int vist[M];
int head1[M];
int head[M];
stack<int>s;
int index, cont;
int ans, flag;
int T, n, m, e, e1;

void init() {

    e = 0;
    e1 = 0;
    index = 1;
    cont = 0;
    flag = true;
    memset(num, 0, sizeof(num));
    memset(head, -1, sizeof(head));
    memset(head1, -1, sizeof(head1)); //开始忘记head1初始化，wa了很多次， 血的教训啊！
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(sccf, 0, sizeof(sccf));
    memset(rdu, 0, sizeof(rdu));
    memset(cdu,0, sizeof(cdu));

}
void add(int a, int b) {

    num[e].to = b;
    num[e].next = head[a];
    head[a] = e++;
}

void add1(int a, int b) {

    num1[e1].to = b;
    num1[e1].next = head1[a];
    head1[a] = e1++;
}

void Tanjian(int u) {  //缩点，

    int v;
    dfn[u] = low[u] = index++;
    ins[u] = 1;
    s.push(u);
    for(int k = head[u]; k != -1; k = num[k].next) {

        v = num[k].to;
        if(!dfn[v]) {

            Tanjian(v);
            low[u] = min(low[u], low[v]);
        }
        else if(ins[v]) {

            low[u] = min(low[u], low[v]);
        }
    }
    if(low[u] == dfn[u]) {

        cont++;
        do {

            v = s.top();
            s.pop();
            sccf[v] = cont;
            ins[v] = 0;

        }while(u != v);
    }
}

void conggougraph() {

    int v;
    for(int i = 1; i <= n; i++) {

        for(int k = head[i]; k != -1; k = num[k].next) {

            v = num[k].to;
            if(sccf[i] != sccf[v]) {    //使用缩点后编号构图，

                add1(sccf[i], sccf[v]);
                rdu[sccf[v]]++;
            }
        }
    }
   // printf("here\n");
    memset(vist, 0, sizeof(vist));
    for(int k = 1; k < cont; k++) {
       int coum = 0;
       int sign = 0;
    for(int i = 1; i <= cont; i++) {

        if(rdu[i] == 0 && !vist[i]) { //计算入度为零个数，
            sign = i;
            vist[i] = 1;  //标记此点已被删除，
            coum++;
          
        }
    }
    if(coum > 1) {  //判断入度为零个数，大于1 标记不满足，

            flag = false;
        }
    else if(flag) {   
         // printf("ssss%d\n", sign);
        for(int j = head1[sign]; j != -1; j = num1[j].next) {  //把要删除点到达的点的入度减一；
            int v = num1[j].to;
            rdu[v]--;

        }
    }
 }
}

int main()
{
     int a, b;
    scanf("%d", &T);
     while(T--) {

    scanf("%d%d", &n, &m);
     init();
    for(int i = 0; i < m; i++) {

        scanf("%d%d", &a, &b);
           add(a, b);
    }
    for(int i = 1; i <= n; i++) {  //Tanjian() 缩点模板；

        if(!dfn[i]) {

            Tanjian(i);
        }
    }
   // printf("%d\n", cont);
    conggougraph(); //二次构图；
    if(flag)
        printf("Yes\n");
    else
        printf("No\n");

  }
    return 0;
}