ZOJ 1586 QS Network 最小生成树

题意：给n个节点，两个节点连接起来需要两个路由器和一条电线，每个节点有偏爱路由器（和这个节点相连的电线必须选这个路由器），给出所有节点相连电线的费用，和每个节点偏爱路由的费用，求出要将所有节点连接起来的最小费用
思路：可以看出一条边的费用相当于电线的费用加上电线两边相连节点偏爱路由器的费用，用这个作为权值求最小生成树，即可得到最小费用

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<map>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn = 1e4+5;
const int inf = 0x3f3f3f3f;
int n, T, u[maxn], q[maxn], d[maxn];
struct node {
	int u, v, w;
	node(int u = 0, int v = 0, int w = 0) : u(u), v(v), w(w) {}
};
vector<node> edge;
bool cmp(node x, node y)
{
	return x.w < y.w;
}
int ufind(int x)
{
	if (u[x] != x)
		u[x] = ufind(u[x]);
	return u[x];
} 
void unite(int x, int y)
{
	u[ufind(x)] = ufind(y);
}
int main()
{
	scanf("%d", &T);
	for (int t = 0; t < T; t++) {
		scanf("%d", &n);
		memset(d, 0, sizeof(d));
		edge.clear();
		for (int i = 1; i <= n; i++) {
			scanf("%d", &q[i]);
			u[i] = i;
		}
		int k = 0, res = 0;
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++) {
				int t;
				scanf("%d", &t);
				if (i != j) {
					edge.push_back(node(i, j, t+q[i]+q[j]));
				}
			}
		}
		sort(edge.begin(), edge.end(), cmp); 
		for (int i = 0; i < edge.size(); i++) {
			int x = edge[i].u, y = edge[i].v;
			if (ufind(x) != ufind(y)) {
				unite(x, y);
				d[x]++;
				d[y]++;
				k++;
				res += edge[i].w;
			}
			if (k == n-1)
				break;
		}
		printf("%d\n", res);
	}
 	return 0;
}