POJ 3160 Father Christmas flymouse(强连通+DP)

POJ 3160 Father Christmas flymouse

题目链接

题意：给定一个有向图，每个点有权值（可能为负），现在要求一条路径，走过每个点可以选择获得或或得该点权值，一个点最多获得一次，问一条最大值的路

思路：图有环，所以先缩点，然后就是DAG最长路了，dp搞搞就可以了

代码：

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

const int N = 30005;

int n, m, val[N];
vector<int> g[N], scc[N];

#define MP(a,b) make_pair(a,b)
typedef pair<int, int> pii;
map<pii, int> vis;
stack<int> S;

int pre[N], dfn[N], dfs_clock, sccn, sccno[N], scc_val[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++;
		int sum = 0;
		while (1) {
			int x = S.top(); S.pop();
			if (val[x] > 0)
				sum += val[x];
			sccno[x] = sccn;
			if (x == u) break;
		}
		scc_val[sccn] = sum;
	}
}

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 dp[N];

int dfs(int u) {
	if (dp[u] != -1) return dp[u];
	dp[u] = 0;
	for (int i = 0; i < scc[u].size(); i++) {
		int v = scc[u][i];
		dp[u] = max(dp[u], dfs(v));
	}
	dp[u] += scc_val[u];
	return dp[u];
}

int main() {
	while (~scanf("%d%d", &n, &m)) {
		vis.clear();
		for (int i = 0; i < n; i++) {
			scanf("%d", &val[i]);
			g[i].clear();
		}
		int u, v;
		while (m--) {
			scanf("%d%d", &u, &v);
			g[u].push_back(v);
		}
		find_scc();
		for (int i = 1; i <= sccn; i++) scc[i].clear();
		for (int u = 0; u < n; u++) {
			for (int i = 0; i < g[u].size(); i++) {
				int v = g[u][i];
				pii tmp = MP(u, v);
				if (sccno[u] == sccno[v] || vis.count(tmp)) continue;
				vis[tmp] = 1;
				scc[sccno[u]].push_back(sccno[v]);
			}
		}
		int ans = 0;
		memset(dp, -1, sizeof(dp));
		for (int i = 1; i <= sccn; i++)
			ans = max(ans, dfs(i));
		printf("%d\n", ans);
	}
	return 0;
}