POJ 1679(次小生成树裸题)

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本文介绍了一种求解次小生成树的问题，通过Prim算法寻找最小生成树，并利用额外的数据结构记录最大边权，进而求得次小生成树。提供了完整的代码实现及样例输入输出。

L - 次小生成树

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u

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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!

模板题目，没啥可说的。

/****************************
* author:crazy_石头
* date：2014/01/18
* algorithm:次小生成树
* Pro：POJ 1679
* 次小生成树
* 求最小生成树时，用数组Max[i][j]来表示MST中i到j最大边权
* 求完后，直接枚举所有不在MST中的边，替换掉最大边权的边，更新答案
* 点的编号从0开始
***************************/
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <string>
#include <queue>
#include <vector>

using namespace std;

#define INF 1<<29
#define eps 1e-8
#define A system("pause")
#define rep(i,h,n) for(int i=(h);i<=(n);i++)
#define ms(a,b) memset((a),(b),sizeof(a))

const int maxn=100+10;
const int maxm=3000+10;

bool vis[maxn],used[maxn][maxn];
int lowc[maxn],pre[maxn],Max[maxn][maxn],ans,cost[maxn][maxn];
int test,n,m;

inline int Prim(int cost[][maxn],int n)
{
	int ans=0;
	ms(vis,0),ms(Max,0),ms(used,0);
	vis[0]=true,pre[0]=-1;
	for(int i=1;i<n;i++)
	{
		lowc[i]=cost[0][i];
		pre[i]=0;
	}
	lowc[0]=0;
	for(int i=1;i<n;i++)
	{
		int minc=INF;
		int p=-1;
		for(int j=0;j<n;j++)
			if(!vis[j]&&minc>lowc[j])
			{
				minc=lowc[j];
				p=j;
			}
		if(minc==INF) return -1;
		ans+=minc;
		vis[p]=true;
		used[p][pre[p]]=used[pre[p]][p]=true;
		for(int j=0;j<n;j++)
		{
			if(vis[j])
				Max[j][p]=Max[p][j]=max(Max[j][pre[p]],lowc[p]);
			if(!vis[j]&&lowc[j]>cost[p][j])
			{
				lowc[j]=cost[p][j];
				pre[j]=p;
			}
		}
	}
	return ans;
}

inline int smst(int cost[][maxn],int n)
{
	int Min=INF;
	for(int i=0;i<n;i++)
		for(int j=i+1;j<n;j++)
		if(cost[i][j]!=INF&&!used[i][j])
			Min=min(Min,ans+cost[i][j]-Max[i][j]);
	if(Min==INF)  return -1;
	return Min;
}

int main()
{
	scanf("%d",&test);
	while(test--)
	{
		int u,v,w;
		scanf("%d%d",&n,&m);
		rep(i,0,n-1) rep(j,0,n-1) i==j?cost[i][j]=0:cost[i][j]=INF;
		while(m--)//0开始 
		{
			scanf("%d%d%d",&u,&v,&w);
			u--,v--;
			cost[u][v]=cost[v][u]=w;
		}
		ans=Prim(cost,n);
		if(ans==-1)
		{
			printf("Not Unique!\n");
			continue;
		}
		if(ans==smst(cost,n))   printf("Not Unique!\n");
		else    printf("%d\n",ans);
	}
	return 0;
}