hdu3639Hawk-and-Chicken(无向图缩点+dfs)

#include<bits/stdc++.h>
using namespace std;
const int maxn = 50100;
typedef long long ll;
struct edge {
    int f, t, nxt;
}e[maxn << 1];
int hd[maxn], tot;
void add(int f, int t) {
    e[++tot] = { f,t,hd[f] };
    hd[f] = tot;
}
int n, m;
int low[maxn], dfn[maxn], cnt;
int stk[maxn], sk, instk[maxn];
int col[maxn], colnum;
int du[maxn], num[maxn], vis[maxn];;
int sa[maxn];
void init() {
    memset(hd, 0, sizeof(hd));
    memset(dfn, 0, sizeof(dfn));
    memset(col, 0, sizeof(col));
    memset(du, 0, sizeof(du));
    memset(num, 0, sizeof(num));
    memset(vis, 0, sizeof(vis));
    memset(sa, 0, sizeof(sa));
    cnt = sk = colnum = tot = 0;
}
void tarjan(int u) {//有向图缩点模板
    low[u] = dfn[u] = ++cnt;
    stk[++sk] = u; instk[u] = 1;
    for (int i = hd[u]; i; i = e[i].nxt) {
        int v = e[i].t;
        if (!dfn[v]) {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if (instk[v]) {
            low[u] = min(low[u], dfn[v]);
        }
    }
    if (low[u] == dfn[u]) {
        colnum++;
        while (1) {
            int v = stk[sk--];
            instk[v] = 0;
            num[colnum]++;
            //该连通分量中的节点个数，然后后面可以直接用来dfs
            col[v] = colnum;//标记属于那个连通分量
            if (v == u)break;
        }
    }
}
void dfs(int u,int rt) {
    if (vis[u])return;
    vis[u] = 1;
    sa[rt] += num[u];
    for (int i = hd[u]; i; i = e[i].nxt) {
        int v = e[i].t;
        dfs(v,rt);
    }
}
int main() {
    //freopen("test.txt", "r", stdin);
    int t; scanf("%d", &t);
    for (int cse = 1; cse <= t; cse++) {
        scanf("%d%d", &n, &m);
        init();
        for (int i = 1; i <= m; i++) {
            int a, b; scanf("%d%d", &a, &b);
            add(b, a);
        }
        for (int i = 0; i < n; i++) {
            if (!dfn[i]) {
                tarjan(i);
            }
        }
        memset(hd, 0, sizeof(hd)); tot = 0;
        //直接用之前的数组来存,难得多开了。
        for (int i = 1; i <= m; i++) {//缩点
            int u = e[i].f, v = e[i].t;
            if (col[u] != col[v]) {
                add(col[u], col[v]);
                du[col[v]]++;
            }
        }//缩点部分应该都没错,我一直用的这个板子
        int ans = 0;
        for (int i = 1; i <= colnum; i++) {//缩点后的DAG图找入读为0的点dfs下去
            if (du[i] == 0) {
                memset(vis, 0, sizeof(vis));
                dfs(i,i);
                sa[i]--;//减去自己
                ans = max(ans, sa[i]);
            }
        }
        printf("Case %d: %d\n", cse, ans);
        set<int>rescol;
        for (int i = 1; i <= colnum; i++) {//找的票与最大值相同的分组(强连通分量)
            if (sa[i]== ans) {
                rescol.insert(i);
            }
        }
        bool h = 0;
        for (int i = 0; i < n; i++) {//如果自己所在的分组是的票最多的就输出
            if (rescol.count(col[i])) {
                if (h)printf(" ");
                printf("%d", i);
                h = 1;
            }
        }
        printf("\n");
    }
    return 0;
}