题目:给定一个有向图(可能含环)及其各点权值,求一条经过点权值和最大的路径(重复经过的点只计算一次)

个人思路是先用tarjan找强联通分量然后锁点变成无环图再dfs,不知道为啥MLE了

时间限制:1000ms,内存限制:256MB

点数不超过$10^4$,边数不超过$10^5$,各点权值不超过$10^3$

#include<bits/stdc++.h>
using namespace std;
int n = 0, m = 0;
vector<int> g[10001], g2[10001];
int a[10001] = {0}, a2[10001] = {0};
int low[10001] = {0}, dfn[10001] = {0};
stack<int> s;
vector<pair<int, int> > edge;
int newp[10001] = {0}, tail = 1;
int vis[10001] = {0};
void tarjan(int x) {
	s.push(x);
	for(int i = 0; i < g[x].size(); i++) {
		int y = g[x][i];
		if(dfn[y]) {
			low[x] = min(low[x], dfn[y]);
		} else {
			dfn[y] = low[y] = dfn[x] + 1;
			tarjan(y);
			low[x] = min(low[x], low[y]);
		}
	}
	if(dfn[x] == low[x]) {
		while(s.top() != x) {
			newp[s.top()] = tail;
			a2[tail] += a[s.top()];
			s.pop();
		}
		newp[x] = tail;
		a2[tail] += a[x];
		tail++;
		s.pop();
	}
}
int dfs(int x, int s) {
	int dis = s;
	for(int i = 0; i < g2[x].size(); i++) {
		int y = g[x][i];
		dis = max(dis, dfs(y, s + a2[y]));
	}
	return dis;
}
int main() {
	cin >> n >> m;
	for(int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	for(int i = 0; i < m; i++) {
		int u = 0, v = 0;
		cin >> u >> v;
		g[u].push_back(v);
		edge.push_back(make_pair(u, v));
	}
	for(int i = 1; i <= n; i++) {
		if(!dfn[i]) {
			tarjan(i);
		}
	}
	for(int i = 0; i < m; i++) {
		int u = edge[i].first, v = edge[i].second;
		u = newp[u];
		v = newp[v];
		g2[u].push_back(v);
	}
	int ans = 0;
	for(int i = 1; i < tail; i++) {
		ans = max(ans, dfs(i, a2[i]));
	}
	cout << ans;
	return 0;
}