poj2186

题意：给出一个有向图，求一共有多少个点，满足这样的条件：所有其它的点都可以到达这个点。

分析：强连通分支+缩点，然后统计每个强连通分支的出度，如果只有一个为0，则输出其内部点的个数，如果有多个为0，说明没有答案。

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
usingnamespace std;

#define maxn 10005
#define maxm 50005

struct Edge
{
    int v, next;
} edge[maxm], opedge[maxm];
int n, m, head[maxn], ophead[maxn], ncount, nowtime, pos[maxn *2], sig[maxn], signnum, in[maxn];
int pointnum[maxn];
bool flag[maxn];
void addedge(int a, int b)
{
    edge[ncount].v = b;
    edge[ncount].next = head[a];
    head[a] = ncount;
    opedge[ncount].v = a;
    opedge[ncount].next = ophead[b];
    ophead[b] = ncount;
    ncount++;
}
void input()
{
    scanf("%d%d", &n, &m);
    for (int i =0; i < m; i++)
    {
        int a, b;
        scanf("%d%d", &a, &b);
        a--;
        b--;
        addedge(a, b);
    }
}
void dfs(int a)
{
    flag[a] =true;
    pos[nowtime] = a;
    nowtime++;
    for (int i = head[a]; i !=-1; i = edge[i].next)
        if (!flag[edge[i].v])
            dfs(edge[i].v);
    pos[nowtime] = a;
    nowtime++;
}
void rdfs(int a)
{
    flag[a] =true;
    sig[a] = signnum;
    pointnum[sig[a]]++;
    for (int i = ophead[a]; i !=-1; i = opedge[i].next)
        if (!flag[opedge[i].v])
            rdfs(opedge[i].v);
}

int main()
{
    //freopen("t.txt", "r", stdin);
    ncount =0;
    memset(head, -1, sizeof(head));
    memset(ophead, -1, sizeof(ophead));
    memset(flag, 0, sizeof(flag));
    input();
    nowtime =1;
    for (int i =0; i < n; i++)
        if (!flag[i])
            dfs(i);
    memset(flag, 0, sizeof(flag));
    memset(pointnum, 0, sizeof(pointnum));
    signnum =0;
    for (int i =2* n; i >0; i--)
        if (!flag[pos[i]])
        {
            rdfs(pos[i]);
            signnum++;
        }
    memset(in, 0, sizeof(in));
    ncount =0;
    for (int i =0; i < n; i++)
        for (int j = head[i]; j !=-1; j = edge[j].next)
            if (sig[i] != sig[edge[j].v])
                in[sig[i]]++;
    int ans =-1;
    bool ok =true;
    for (int i =0; i < signnum; i++)
        if (in[i] ==0)
        {
            if (ans !=-1)
            {
                ok =false;
                break;
            }
            ans = i;
        }
    if (ok)
        printf("%d\n", pointnum[ans]);
    else
        printf("0\n");
    return0;
}