POJ - 1679 The Unique MST （次小生成树）

Description

Given a connected undirected graph, tell if its minimum spanning tree is unique.

Definition 1 (Spanning Tree): Consider a connected, undirected graph G = (V, E). A spanning tree of G is a subgraph of G, say T = (V‘, E‘), with the following properties:

1. V‘ = V.

2. T is connected and acyclic.

Definition 2 (Minimum Spanning Tree): Consider an edge-weighted, connected, undirected graph G = (V, E). The minimum spanning tree T = (V, E‘) of G is the spanning tree that has the smallest total cost. The total cost of T means the sum of the weights on all
the edges in E‘.

Input

The first line contains a single integer t (1 <= t <= 20), the number of test cases. Each case represents a graph. It begins with a line containing two integers n and m (1 <= n <= 100), the number of nodes and edges. Each of the
following m lines contains a triple (xi, yi, wi), indicating that xi and yi are connected by an edge with weight = wi. For any two nodes, there is at most one edge connecting them.

Output

For each input, if the MST is unique, print the total cost of it, or otherwise print the string ‘Not Unique!‘.

Sample Input

2
3 3
1 2 1
2 3 2
3 1 3
4 4
1 2 2
2 3 2
3 4 2
4 1 2

Sample Output

3
Not Unique!

题意：求最小生成树是否唯一

思路：次小生成树的应用，先求出最小生成树，然后枚举删除的边，再求最小生成树，如果值一样，代表不唯一

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
const int maxn = 105;
const int inf = 0x3f3f3f3f;

struct Edge {
	int from, to, dist;
} edge[maxn*maxn];
int n, m;
int mst, fa[maxn];
int uni;

int cmp(Edge a, Edge b) {
	return a.dist < b.dist;
}

int find(int x) {
	if (x != fa[x])
		fa[x] = find(fa[x]);
	return fa[x];
}

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

void kruskal(){
	int num[maxn], cnt = 0;
	init();
	sort(edge, edge+m, cmp);
	mst = 0;
	for (int i = 0; i < m; i++) {
		int x = find(edge[i].from);
		int y = find(edge[i].to);
		if (x != y) {
			fa[y] = x;
			mst += edge[i].dist;
			num[cnt++] = i;
		}
	}
	uni = 1;
	int ans, tmp;
	for (int k = 0; k < cnt; k++) {
		init();
		ans = 0;
		tmp = 0;
		for (int i = 0; i < m; i++) {
			if (i == num[k])
				continue;
			int x = find(edge[i].from);
			int y = find(edge[i].to);
			if (x != y) {
				fa[y] = x;
				ans += edge[i].dist;
				tmp++;
			}
		}
		if (tmp != cnt)
			continue;
		if (ans == mst) {
			uni = 0;
			return;
		}
	}
}

int main() {
	int t, i, j;
	scanf("%d", &t);
	while(t --){
		scanf("%d%d", &n, &m);
		for(i = 0; i < m; i ++)
			scanf("%d%d%d", &edge[i].from, &edge[i].to, &edge[i].dist);
		kruskal();
		if(!uni)
			printf("Not Unique!\n");
		else printf("%d\n", mst);
	}
	return 0;
}

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