BZOJ 1051: [HAOI2006]受欢迎的牛 强连通缩点

题目链接：

http://www.lydsy.com/JudgeOnline/problem.php?id=1051

题解：

强连通缩点得到DAG图，将图转置一下，对入度为零的点跑dfs看看能不能访问到所有的点。

代码：

#include<iostream>
#include<cstdio>
#include<vector>
#include<stack>
#include<algorithm>
#include<cstring>
using namespace std;

const int maxn = 10000 + 10;
const int INF = 0x3f3f3f3f;

vector<int> G[maxn],G2[maxn];
int pre[maxn], lowlink[maxn], sccno[maxn], dfs_clock, scc_cnt;
int ind[maxn],siz[maxn],vis[maxn];
stack<int> S;

int n,m;

void dfs(int u) {
    pre[u] = lowlink[u] = ++dfs_clock;
    S.push(u);
    for (int i = 0; i < G[u].size(); i++) {
        int v = G[u][i];
        if (!pre[v]) {
            dfs(v);
            lowlink[u] = min(lowlink[u], lowlink[v]);
        }
        else if (!sccno[v]) {
            lowlink[u] = min(lowlink[u], pre[v]);
        }
    }
    if (lowlink[u] == pre[u]) {
        scc_cnt++;
        int cnt = 0;
        for (;;) {
            cnt++;
            int x = S.top(); S.pop();
            sccno[x] = scc_cnt;
            if (x == u) break;
        }
        siz[scc_cnt] = cnt;
    }
}

void find_scc(int n) {
    dfs_clock = scc_cnt = 0;
    memset(sccno, 0, sizeof(sccno));
    memset(pre, 0, sizeof(pre));
    for (int i = 0; i < n; i++) if (!pre[i]) dfs(i);
    //for (int i = 0; i < n; i++) printf("sccno[%d]:%d
", i, sccno[i]);
}

void dfs2(int u) {
    if (vis[u]) return;
    vis[u] = 1;
    for (int i = 0; i < G2[u].size(); i++) {
        int v = G2[u][i];
        dfs2(v);
    }
}

void init() {
    for (int i = 0; i < n; i++) G[i].clear(), G2[i].clear();
    memset(ind, 0,sizeof(ind));
    memset(vis, 0, sizeof(vis));
}

int main() {
    while (scanf("%d%d", &n,&m) == 2 && n) {
        init();
        for (int i = 0; i < m; i++) {
            int u, v;
            scanf("%d%d", &u, &v); u--, v--;
            G[u].push_back(v);
        }
        find_scc(n);
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < G[i].size(); j++) {
                int v = G[i][j];
                if (sccno[i] != sccno[v]) {
                    G2[sccno[v]].push_back(sccno[i]);
                    ind[sccno[i]]++;
                }
            }
        }
        int rt = -1;
        for (int i = 1; i <= scc_cnt; i++) {
            if (ind[i] == 0) rt = i;
        }
        dfs2(rt);
        int su = 1;
        for (int i = 1; i <= scc_cnt; i++) {
            if (!vis[i]) {
                su = 0;
                break;
            }
        }
        if (su) {
            printf("%d
", siz[rt]);
        }
        else {
            printf("0
");
        }
    }
    return 0;
}