[poj][2186][Popular Cows]

题目：http://poj.org/problem?id=2186

题意：给一张有向图，求一些点的个数，这些点满足其他所有点都能直接或间接指向它。

方法：tarjan+缩点

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>

#define Max(A,B) ((A)>(B)?(A):(B))
#define Min(A,B) ((A)<(B)?(A):(B))
#define clr(a,b) memset(a, b, sizeof(a))
using namespace std;

const int MAXV = 10000+10;
const int inf = 0x3f3f3f3f;

const int N = 10010;
const int M = 50010;
int e, ev[M], nxt[M], head[N];
bool instack[N];
int dfn[N], low[N], dindex, q[N], ed;
int belong[N], bcnt;

int cou[N];
bool out[N];

void addedge(int u, int v)
{
    ev[e] = v; nxt[e] = head[u]; head[u] = e++;
}

void tarjan(int u)
{
    int i, v;
    dfn[u] = low[u] = ++dindex;
    instack[u] = 1;
    q[ed++] = u;
    for (i = head[u]; ~i; i = nxt[i])
    {
        v = ev[i];
        if (!dfn[v])
        {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if (instack[v])
        {
            low[u] = min(low[u], dfn[v]);
        }
    }
    if (dfn[u] == low[u])
    {
        do{
            belong[v = q[--ed]] = bcnt;
            instack[v] = 0;
        }while (v != u);
        bcnt++;
    }
}

int main()
{
    //freopen("D:/a.txt", "r", stdin);
    int n, m;
    e = 0;
    clr(head, -1);
    scanf("%d%d", &n, &m);
    for (int i=0; i<m; i++)
    {
        int u, v;
        scanf("%d%d", &u, &v);
        addedge(u-1, v-1);
    }
    ed = bcnt = dindex = 0;
    clr(dfn, 0);
    clr(cou, 0);
    clr(out, false);
    clr(instack, 0);
    for (int i=0; i<n; i++)
    {
        if (!dfn[i]) tarjan(i);
    }
    for (int i=0; i<n; i++)
    {
        cou[belong[i]]++;
        for (int j=head[i]; ~j; j = nxt[j])
        {
            if (belong[i]!=belong[ev[j]])
            {
                out[belong[i]] = true;
            }
        }
    }
    int number = 0, ans = 0;
    for (int i=0; i<bcnt; i++)
    {
        if (!out[i])
        {
            number++;
            ans = cou[i];
        }
    }
    if (number == 1) printf("%d\n", ans);
    else printf("0\n");
    return 0;
}