POJ 2186 Popular Cows

这道题求被其他牛欢迎的牛的个数。原图是一个有向图，将其缩点后，每一个强连通分量是

一个点，找出出度为1的点的个数，如果大于1，那么无解。如果等于1，则输出这个强连通

分量中点的个数。

/*Accepted    736K    110MS    C++    1999B    2012-07-30 11:49:48*/
#include<cstdio>
#include<cstring>
#include<cstdlib>

const int MAXN = 10010;
const int MAXM = 50050;
int first[MAXN], next[MAXM], v[MAXM], cnt, top, col, e, N, M, ans;
int dfn[MAXN], color[MAXN], s[MAXN], ins[MAXN], low[MAXN], outdgr[MAXN];

void tarjan(int cur)
{
    int i;
    dfn[cur] = low[cur] = ++ cnt;
    s[top ++] = cur, ins[cur] = 1;
    for(i = first[cur]; i != -1; i = next[i])
    {
        if(!dfn[v[i]])
        {
            tarjan(v[i]);
            if(low[v[i]] < low[cur])
                low[cur] = low[v[i]];
        }
        else if(dfn[v[i]] < low[cur] && ins[v[i]])
            low[cur] = dfn[v[i]];
    }
    if(low[cur] == dfn[cur])
    {
        ++ col;
        for(s[top] = -1; s[top] != cur; )
        {
            color[s[-- top]] = col, ins[s[top]] = 0;
        }
    }
}

void cal()
{
    int i, j, zeroout = 0;
    col = cnt = top = 0;
    memset(dfn, 0, sizeof dfn);
    memset(ins, 0, sizeof ins);
    memset(outdgr, 0, sizeof outdgr);
    for(i = 1; i <= N; i ++)
        if(!dfn[i])
            tarjan(i);
    for(i = 1; i <= N; i ++)
        for(j = first[i]; j != -1; j = next[j])
            if(color[i] != color[v[j]])
                ++ outdgr[color[i]];
    for(i = 1; i <= col; i ++)
    {
        if(outdgr[i] == 0)
        {
            zeroout ++;
            j = i;
        }
    }
    ans = 0;
    if(zeroout > 1) return;
    else{
        for(i = 1; i <= N; i ++)
        {
            if(color[i] == j)
                ans ++;
        }
    }
}

void addedge(int x, int y)
{
    v[e] = y;
    next[e] = first[x], first[x] = e ++;
}

void ReadGraph()
{
    int i, x, y;
    e = 0;
    memset(first, -1, sizeof first);
    while(M --)
    {
        scanf("%d%d", &x, &y);
        addedge(x, y);
    }
}

int main()
{
    while(scanf("%d%d", &N, &M) == 2)
    {
        ReadGraph();
        cal();
        printf("%d\n", ans);
    }
    return 0;
}