HDU 6201 transaction transaction transaction(最短路)

解法：由于每个点有一个商品，花费为w，我们想卖出去，但是路也有花费，如果在某个点卖出去，相当于赚回wi元。那么我们建图，一个起点向每个点连接一个边权为-wi的边，每个点向终点连接一个边权为+wi的边，原图中的边也加进去，从起点向终点跑一边spfa即可。

代码如下：

#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<utility>
#include<stack>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<set>
#include<map>
using namespace std;
typedef pair<int, int> pii;
const int maxn = 1e5 + 5;
const int INF = 0x7fffffff;
int n;
int a[maxn];

vector <pii> edge[maxn];

int dis[maxn];
bool flag[maxn];
void spfa(int s, int t) {
	fill(dis, dis + t + 1, -INF);
	dis[s] = 0;
	memset(flag, 0, sizeof(flag));
	queue <int> q;
	q.push(s);
	while(!q.empty()) {
		int u = q.front();
		q.pop();
		flag[u] = 0;
		for(int i = 0; i < edge[u].size(); i++) {
			int v = edge[u][i].second;
			int d = edge[u][i].first + dis[u];
			if(dis[v] < d) {
				dis[v] = d;
				if(!flag[v]) {
					flag[v] = 1;
					q.push(v);
				}
			}
		}
	}
}

int main() {
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);
#endif
	int T;
	scanf("%d", &T);
	while(T--) {
		scanf("%d", &n);
		for(int i = 1; i <= n; i++)
			scanf("%d", &a[i]);
		for(int i = 1, u, v, val; i < n; i++) {
			scanf("%d%d%d", &u, &v, &val);
			edge[u].push_back(pii(-val, v));
			edge[v].push_back(pii(-val, u));
		}
		for(int i = 1; i <= n; i++) {
			edge[0].push_back(pii(-a[i], i));
			edge[i].push_back(pii(a[i], n + 1));
		}
		spfa(0, n + 1);
		printf("%d\n", dis[n + 1]);
		for(int i = 0; i <= n + 1; i++)
			edge[i].clear();
	}
	return 0;
}