蓝桥杯 - 安慰奶牛 （最小生成树）

题目传送：蓝桥杯 - 安慰奶牛

思路：先算好边的权值，为本来的边的权值的两倍加上两个点的权值，再进行kruskal，因为边数较大，不宜采用prim

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

#define LL long long
#define INF 0x7fffffff

const int maxn = 100005;

int pa[maxn];
int val[maxn];
int n, m, sum;

struct node {
	int u, v, val;
	bool operator < (const node& a) const {
		return val < a.val;
	}
}edge[maxn];

int find(int x) {
	return x == pa[x] ? x : pa[x] = find(pa[x]);
}

void add(int u, int v, int pri) {
	edge[sum].u = u;
	edge[sum].v = v;
	edge[sum].val = val[u] + val[v] + pri * 2;
	sum ++;
}

void kruscal() {
	for(int i = 1; i <= n; i++) {
		pa[i] = i;
	}

	int ans = 0;
	for(int i = 0; i < sum; i++) {
		int u = edge[i].u;
		int v = edge[i].v;
		int pu = find(u), pv = find(v);
		if(pu == pv) continue;
		ans += edge[i].val;
		pa[pu] = pv;
	}
	int mi = INF;
	for(int i = 1; i <= n; i++) {
		mi = min(mi, val[i]);
	}
	printf("%d\n", ans + mi);
}

int main() {
	while(scanf("%d %d", &n, &m) != EOF) {
		sum = 0;
		for(int i = 1; i <= n; i++) {
			scanf("%d", &val[i]);
		}

		for(int i = 0; i < m; i++) {
			int u, v, pri;
			scanf("%d %d %d", &u, &v, &pri);
			add(u, v, pri);
		}

		sort(edge, edge + sum);
		/*for(int i = 0; i < sum; i++) {
			printf("%d %d %d\n", edge[i].u, edge[i].v, edge[i].val);
		}*/

		kruscal();
	}
	return 0;
}