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;

}