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POJ1679 K - The Unique MST (最小生成树是否唯一)

程序员文章站 2024-03-21 17:36:58
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题目传送

判断最小生成树是否唯一。
先找出最小生成树,并记录组成,然后循环去掉一个组成,判断剩下的边能否构成最小生成树。

#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 110;
const int maxm = maxn * maxn;

int par[maxn];
int ans[maxn];

struct edgs {
	int from, to, cost;
	bool operator < (const edgs& a) const {
		return cost < a.cost;
	}
}edg[maxm];

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

int main(void)
{
	int t;
	scanf("%d", &t);
	while (t--) {
		int n, m, k = 1, cnt = 0, tot = 0, flag = 1;
		scanf("%d%d", &n, &m);
		for (int i = 1; i <= n; i++) par[i] = i;
		for (int i = 1; i <= m; i++) {
			int a, b, c;
			scanf("%d%d%d", &a, &b, &c);
			edg[k].from = a;
			edg[k].to = b;
			edg[k++].cost = c;
		}
		sort(edg + 1, edg + k);
		for (int i = 1; i < k; i++) {
			int rf = find(edg[i].from);
			int rt = find(edg[i].to);
			if (rf == rt) continue;
			par[rf] = rt;
			ans[++cnt] = i;
			tot += edg[i].cost;
			if (cnt == n - 1) break;
		}
		for (int i = 1; i <= cnt; i++) {
			int sum = 0, p = 0;
			for (int j = 1; j <= n; j++) par[j] = j;
			for (int j = 1; j < k; j++) {
				if (ans[i] == j) continue;
				int rf = find(edg[j].from);
				int rt = find(edg[j].to);
				if (rf == rt) continue;
				par[rf] = rt;
				sum += edg[j].cost;
				if (++p == n - 1) break;
			}
			if (p == n - 1 && sum == tot) {
				printf("Not Unique!\n");
				flag = 0;
				break;
			}
		}
		if (flag) printf("%d\n", tot);
	}
	return 0;
}

相关标签: 生成树