HDU 3639 Hawk-and-Chicken

题目链接

题意：就是在一个有向图上，满足传递关系，比如a->b, b->c，那么c可以得到2的支持，问得到支持最大的是谁，并且输出这些人

思路：先强连通的缩点，然后逆向建图，对于每个出度为0的点，进行dfs求哪些点可达这个点

代码：

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

const int N = 5005;

int t, n, m;
int sccn, dfs_clock, sccno[N], pre[N], dfn[N];
stack<int> S;
vector<int> g[N], save[N], scc[N];

void dfs_scc(int u) {
	pre[u] = dfn[u] = ++dfs_clock;
	S.push(u);
	for (int i = 0; i < g[u].size(); i++) {
		int v = g[u][i];
		if (!pre[v]) {
			dfs_scc(v);
			dfn[u] = min(dfn[u], dfn[v]);
		} else if (!sccno[v]) dfn[u] = min(dfn[u], pre[v]);
	}
	if (pre[u] == dfn[u]) {
		++sccn;
		save[sccn].clear();
		while (1) {
			int x = S.top(); S.pop();
			sccno[x] = sccn;
			save[sccn].push_back(x);
			if (x == u) break;
		}
	}
}

void find_scc() {
	dfs_clock = sccn = 0;
	memset(pre, 0, sizeof(pre));
	memset(sccno, 0, sizeof(sccno));
	for (int i = 0; i < n; i++)
		if (!pre[i]) dfs_scc(i);
}

int out[N];
int ans[N], an, dp[N], vis[N];

int dfs(int u) {
	vis[u] = 1;
	int ans = save[u].size();
	for (int i = 0; i < scc[u].size(); i++) {
		int v = scc[u][i];
		if (vis[v]) continue;
		ans += dfs(v);
	}
	return ans;
}

int main() {
	int cas = 0;
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d", &n, &m);
		for (int i = 0; i < n; i++) g[i].clear();
		int u, v;
		while (m--) {
			scanf("%d%d", &u, &v);
			g[u].push_back(v);
		}
		find_scc();
		memset(out, 0, sizeof(out));
		for (int i = 1; i <= sccn; i++) scc[i].clear();
		for (int u = 0; u < n; u++) {
			for (int j = 0; j < g[u].size(); j++) {
				int v = g[u][j];
				if (sccno[u] != sccno[v]) {
					scc[sccno[v]].push_back(sccno[u]);
					out[sccno[u]]++;
				}
			}
		}
		int Max = 0;
		for (int i = 1; i <= sccn; i++)
			if (!out[i]) { 
				memset(vis, 0, sizeof(vis));
				dp[i] = dfs(i);
				Max = max(Max, dp[i]);
			}
		an = 0;
		for (int i = 1; i <= sccn; i++) {
			if (!out[i] && dp[i] == Max) {
				for (int j = 0; j < save[i].size(); j++) {
					ans[an++] = save[i][j];
				}
			}
		}
		sort(ans, ans + an);
		printf("Case %d: %d\n", ++cas, Max - 1);
		for (int i = 0; i < an; i++)
			printf("%d%c", ans[i], i == an - 1 ? '\n' : ' ');
	}
	return 0;
}

HDU 3639 Hawk-and-Chicken(强连通）