POJ-1679 The Unique MST (次小生成树)

最新推荐文章于 2022-01-21 19:00:37 发布

原创最新推荐文章于 2022-01-21 19:00:37 发布 · 391 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#ACM #并查集 #POJ #次小生成树

图论-最小生成树专栏收录该内容

11 篇文章

订阅专栏

本文介绍了一种算法，用于判断给定的无向图是否存在唯一的最小生成树(MST)。通过构造图并进行特定的遍历来确定是否有不同的MST具有相同的总权重。

The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 32996		Accepted: 12007

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

Sample Output

3
Not Unique!

#include <cstdio>  
#include <cmath>  
#include <algorithm> 
#include <vector>
#include <string.h>
using namespace std;
struct edge{
	int from, to, val;
	bool operator < (const edge& x) const{
		return val < x.val;
	}
}e[10005];
int p[101], dis[101][101], cnt[10003];
int findset(int x){
	return p[x] == x ? x : p[x] = findset(p[x]); 
}
vector<pair<int, int> > g[101];
void dfs(int rt, int x, int fa, int d){
	dis[rt][x] = d;
	int cur;
	for(int i = 0; i < g[x].size(); ++i){
		cur = g[x][i].first;
		if(cur == fa) continue;
		dfs(rt, cur, x, max(d, g[x][i].second));
	}
}
int main(){
	int T;
	scanf("%d", &T);
	while(T--){
		memset(cnt, 0, sizeof(cnt));
		memset(dis, 0, sizeof(dis));
		int n, m, u, v;
		scanf("%d %d", &n, &m);
		for(int i = 1; i <= n; ++i){
			p[i] = i;
		}
		for(int i = 1; i <= n; ++i){
			g[i].clear();
		}
		for(int i = 1; i <= m; ++i){
			scanf("%d %d %d", &e[i].from, &e[i].to, &e[i].val);
		}
		sort(e + 1, e + 1 + m);
		int ans = 0;
		for(int i = 1; i <= m; ++i){
			u = findset(e[i].from);
			v = findset(e[i].to);
			if(u != v){
				p[u] = v;
				cnt[i] = 1;
				ans += e[i].val;
				g[e[i].from].push_back(make_pair(e[i].to, e[i].val));
				g[e[i].to].push_back(make_pair(e[i].from, e[i].val));
			}
		}
		for(int i = 1; i <= n; ++i){
			dfs(i, i, 0, 0);
		}
		int flag = 0;
		for(int i = 1; i <= m; ++i){
			if(cnt[i] == 0){
				u = e[i].from;
				v = e[i].to;
				if(dis[u][v] == e[i].val){
					flag = 1;
					break;
				}
			}
		}
		if(flag){
			printf("Not Unique!\n");
		}
		else{
			printf("%d\n", ans);
		}
	}
}

/*
题意：
100个点，给出一些边，问其最小生成树是否唯一。

思路：
求次小生成树，看代价是否一样即可。求次小生成树时，先对最小生成树建图，
然后n^2求一下该图上两点间路劲中最大路径。枚举所有最小生成树之外的点，尝试
将它们加进树中，同时在生成的环中减去一条边，显然应该减的就是那条最大的边，
看这颗生成树的代价是否和最小代价一样即可。
*/