PAT (Top Level)Practice 1001 Battle Over Cities - Hard Version （35 分）

题目链接：https://pintia.cn/problem-sets/994805148990160896/problems/994805156657348608

主要考连通性  ，简单来说就是枚举删除每个点的最小生成树的最大的花费是多少，注意如果不能联通那么一定是最大的，最后在遍历一遍输出答案即可  

算法：kustal最小生成树

code：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<algorithm>
using namespace std;
#define inf 1000000
#define N 550
#define MAX 300000
int n, m;
int p[N];
struct road {
	int s;
	int e;
	int vis;
};
road use[MAX], nuse[MAX];
void init() {
	for (int i = 1; i <= n; i++) {
		p[i] = i;
	}
}
int find(int x) {
	if (x != p[x]) {
		p[x] = find(p[x]);
	}
	else
		return x;
}
int main() {
	int x, y, c, s;
	scanf("%d%d", &n, &m);
	int ans1 = 0, ans2 = 0;
	for (int i = 0; i < m; i++) {
		scanf("%d%d%d%d", &x, &y, &c, &s);
		if (s == 1) {
			use[ans1].s = x;
			use[ans1].e = y;
			use[ans1++].vis = c;
		}
		else {
			nuse[ans2].s = x;
			nuse[ans2].e = y;
			nuse[ans2++].vis = c;
		}
	}
	int mx = 0;
	int sum[N];
	memset(sum, 0, sizeof(sum));
	sort(nuse, nuse + ans2);
	for (int i = 1; i <= n; i++) {
		init();
		int num = n - 2;
		for (int j = 0; j < ans1; j++) {
			if (use[j].s == i || use[j].e == i)
				continue;
			int xx = find(use[j].s);
			int yy = find(use[j].e);
			if (xx != yy) {
				p[xx] = yy;
				num--;
			}
		}
		for (int k = 0; k < ans2; k++) {
			if (nuse[k].s == i || nuse[k].e == i)
				continue;
			int xx = find(nuse[k].s);
			int yy = find(nuse[k].e);
			if (xx != yy) {
				p[xx] = yy;
				num--;
				sum[i] += nuse[k].vis;
			}
		}
		if (num > 0) {
			sum[i] = inf;
		}
		mx = max(mx, sum[i]);
	}
	if (mx == 0) {
		cout << mx;
		return 0;
	}
	int t = 0;
	for (int i = 1; i <= n; i++) {
		if (sum[i] == mx) {
			if (t == 0)
			{
				t = 1;
				cout << i;
			}
			else
				cout << " " << i;
		}
	}
	return 0;
}