HDU 1827 Summer Holiday（强连通）

Posted 2020-10-24 zhchoutai

HDU 1827 Summer Holiday

题目链接

题意：中文题

思路：强连通缩点，每一个点的权值为强连通中最小值，然后入度为0的点就是答案

代码：

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

const int N = 1005;
const int INF = 0x3f3f3f3f;

int n, m, val[N];
vector<int> g[N];
int pre[N], lowlink[N], dfs_clock, sccno[N], scc_cnt, scc_val[N];
stack<int> S;

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 (pre[u] == lowlink[u]) {
		scc_cnt++;
		scc_val[scc_cnt] = INF;
		while (1) {
			int x = S.top(); S.pop();
			scc_val[scc_cnt] = min(scc_val[scc_cnt], val[x]);
			sccno[x] = scc_cnt;
			if (x == u) break;
		}
	}
}

void find_scc() {
	dfs_clock = scc_cnt = 0;
	memset(pre, 0, sizeof(pre));
	memset(sccno, 0, sizeof(sccno));
	for (int i = 1; i <= n; i++)
		if (!pre[i]) dfs(i);
}

int in[N];

int main() {
	while (~scanf("%d%d", &n, &m)) {
		for (int i = 1; i <= n; i++) {
			g[i].clear();
			scanf("%d", &val[i]);
		}
		int u, v;
		while (m--) {
			scanf("%d%d", &u, &v);
			g[u].push_back(v);
		}
		find_scc();
		memset(in, 0, sizeof(in));
		for (int u = 1; u <= n; u++) {
			for (int j = 0; j < g[u].size(); j++) {
				int v = g[u][j];
				if (sccno[u] != sccno[v])
					in[sccno[v]]++;
			}
		}
		int ans = 0, cnt = 0;
		for (int i = 1; i <= scc_cnt; i++)
			if (!in[i]) {
				ans += scc_val[i];
				cnt++;
			}
		printf("%d %d\n", cnt, ans);
	}
	return 0;
}