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POJ1679 The Unique MST 【次小生成树】

2024-07-16 21:49:31   218人阅读

The Unique MST
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 20421 Accepted: 7183

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!

Source

POJ Monthly--2004.06.27 srbga@POJ

这题做得好开心，一次AC~

题意：给定一个联通图，判断最小生成树是否唯一。

题解：求出最小生成树后再求次小生成树，若次小生成树的长度与最小生成树相等就说明不唯一，否则唯一。

这是我的第一道次小生成树题，在这里总结下这个算法：

利用一个矩阵max【】【】表示最小生成树中任意两点路径上的最长边权值（关键！！），在求最小生成树时将已经选上的边标记为已用，求完后，遍历剩下未用的边，这条边若添加到最小树中必定构成回路，所以此时需要去掉原来树中那条回路中的最大值，也就是max矩阵存储的值，所以问题转换成找到一条未用的边，使得它跟对应于max矩阵里的边差值最小，遍历之后，次小生成树的值即为原最小生成树的值加上这个最小的差值。

#include <stdio.h>
#include <string.h>
#include <limits.h>
#define maxn 102
#define maxm (maxn * maxn) >> 1

int head[maxn], max[maxn][maxn];
struct Node{
	int u, v, cost, next;
	bool vis;
} E[maxm];
bool vis[maxn];

int mini(int a, int b){
	return a < b ? a : b;
}

int prim(int n, int m)
{
	int u, i, tmp, j, len = 0, count = 0;
	memset(max, 0x7f, sizeof(max));
	memset(vis, 0, sizeof(vis));
	vis[1] = 1;
	while(count < n - 1){
		for(i = 1, tmp = INT_MAX; i <= n; ++i){
			if(!vis[i]) continue;
			for(j = head[i]; j != -1; j = E[j].next){
				if(vis[E[j].v]) continue;
				if(E[j].cost < tmp){
					tmp = E[j].cost; u = j;
				}
			}
		}
		++count; len += tmp;
		for(i = 1; i <= n; ++i){
			if(!vis[i]) continue;
			max[i][E[u].v] = max[E[u].v][i] = E[u].cost;
		}
		vis[E[u].v] = 1; E[u].vis = 1;
	}
	return len;
}

int getSecLen(int n, int m)
{
	int min = INT_MAX, u, v, w;
	for(int i = 0; i < m; ++i){
		if(E[i].vis) continue;
		u = E[i].u; v = E[i].v;
		w = E[i].cost;
		min = mini(min, w - max[u][v]);
		if(min == 0) return 0;
	}
	return min;
}

int main()
{
	int t, n, m, i, minLen, secLen;
	scanf("%d", &t);
	while(t--){
		scanf("%d%d", &n, &m);
		memset(head, -1, sizeof(head));
		for(i = 0; i < m; ++i){
			scanf("%d%d%d", &E[i].u, &E[i].v, &E[i].cost);
			E[i].vis = 0; E[i].next = head[E[i].u];
			head[E[i].u] = i;
		}
		minLen = prim(n, m);
		secLen = getSecLen(n, m);
		if(secLen == 0) printf("Not Unique!\n");
		else printf("%d\n", minLen);
	}
	return 0;
}


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