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 =  + ;

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 = ; 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 = ;

        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 = ;

    memset(sccno, , sizeof(sccno));

    memset(pre, , sizeof(pre));

    for (int i = ; i < n; i++) if (!pre[i]) dfs(i);

    //for (int i = 0; i < n; i++) printf("sccno[%d]:%d\n", i, sccno[i]);

}

void dfs2(int u) {

    if (vis[u]) return;

    vis[u] = ;

    for (int i = ; i < G2[u].size(); i++) {

        int v = G2[u][i];

        dfs2(v);

    }

}

void init() {

    for (int i = ; i < n; i++) G[i].clear(), G2[i].clear();

    memset(ind, ,sizeof(ind));

    memset(vis, , sizeof(vis));

}

int main() {

    while (scanf("%d%d", &n,&m) ==  && n) {

        init();

        for (int i = ; i < m; i++) {

            int u, v;

            scanf("%d%d", &u, &v); u--, v--;

            G[u].push_back(v);

        }

        find_scc(n);

        for (int i = ; i < n; i++) {

            for (int j = ; 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 = -;

        for (int i = ; i <= scc_cnt; i++) {

            if (ind[i] == ) rt = i;

        }

        dfs2(rt);

        int su = ;

        for (int i = ; i <= scc_cnt; i++) {

            if (!vis[i]) {

                su = ;

                break;

            }

        }

        if (su) {

            printf("%d\n", siz[rt]);

        }

        else {

            printf("0\n");

        }

    }

    return ;

}