POJ 2186 Popular Cows

 

大意：牛牛之间互相喜欢，而且这种喜欢具有传递性，要求你求出最受欢迎的牛牛们的个数（A single integer that is the number of cows who are considered popular by every other cow. ）

思路：通过“缩点”之后，然后求强连通分量出度的个数，如果为一，那么求出这个“缩点”里所有牛牛的个数。如果大于一，没有符合条件的，手推一遍即可证实。

CODE：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
using namespace std;

#define MAXN 10010
#define MAXM 100010 

struct Edge
{
    int v, w, next;
}edge[MAXM];

int first[MAXN], stack[MAXN], ins[MAXN], dfn[MAXN], low[MAXN];
int belong[MAXM];
int outd[MAXN];

int n, m;
int cnt;
int scnt, top, tot;

void init()
{
    cnt = 0;
    scnt = top = tot = 0;
    memset(first, -1, sizeof(first));
    memset(dfn, 0, sizeof(dfn));
}

void read_graph(int u, int v)
{
    edge[cnt].v = v;
    edge[cnt].next = first[u];
    first[u] = cnt++;
}

void dfs(int u)
{
    int t;
    low[u] = dfn[u] = ++tot;
    ins[u] = 1;
    stack[top++] = u;
    for(int e = first[u]; e != -1; e = edge[e].next)
    {
        int v = edge[e].v;
        if(!dfn[v])
        {
            dfs(v);
            low[u] = min(low[u], low[v]);
        }
        else if(ins[v])
        {
            low[u] = min(low[u], dfn[v]);
        }
    }
    if(low[u] == dfn[u])
    {
        ++scnt;
        do
        {
            t = stack[--top];
            belong[t] = scnt;
            ins[t] = 0;
        }while(t != u);
    }
}

void Tarjan()
{
    for(int v = 1; v <= n; v++) if(!dfn[v])
        dfs(v);
}

void solve()
{
    Tarjan(); //Tarjan();
    memset(outd, 0, sizeof(outd));
    for(int u = 1; u <= n; u++)
    {
        for(int e = first[u]; e != -1; e = edge[e].next)
        {
            int v = edge[e].v;
            if(belong[u] != belong[v])    outd[belong[u]]++; //强连通分量的出度增加，即缩点的出度增加。
        }
    }
    int index = 0, flag = 0;
    for(int i = 1; i <= scnt; i++) if(!outd[i]) //如果强连通的出度为0
    {
        index++;
        flag = i; //标记所在强连通
    }
    if(index > 1) printf("0\n"); //出度为0的强连通不只一个 
    else
    {
        index = 0;
        for(int i = 1; i <= n; i++) if(belong[i] == flag) index++; //所有的顶点属于出度为0的强连通分量点的数目 
        printf("%d\n", index);
    }
}

int main()
{
    while(~scanf("%d%d", &n, &m))
    {
        init();
        while(m--)
        {
            int u, v;
            scanf("%d%d", &u, &v);
            read_graph(u, v);
        }
        solve();
    }
    return 0;
}