1679 The Unique MST //次小生成树

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本文介绍了一种算法，用于判断给定的加权无向图是否具有唯一最小生成树（MST）。通过Prim算法实现，并考虑了边权重可能存在的重复情况，确保能够正确判断最小生成树的唯一性。

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The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 7518		Accepted: 2526

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!

Source

POJ Monthly--2004.06.27 srbga@POJ

#include<stdio.h>

#include<string.h>

#define MAXN 1005

#define inf 1000000000

int maxe[MAXN][MAXN],g[MAXN][MAXN];

bool inmst[MAXN][MAXN];

int min(int a,int b)

{

if(a<b) return a;

return b;

}

int used[MAXN],pre[MAXN],d[MAXN];

int prim(int n,int g[][MAXN])

{

int r,minc,res=0;

memset(used,0,sizeof(used));

for(int i=0;i<n;i++) d[i]=inf;

pre[0]=0;

d[0]=0;

for(int k=0;k<n;k++)

{

minc=inf;

for(int i=0;i<n;i++)

if(!used[i]&&d[i]<minc) { minc=d[i]; r=i; }

res+=minc;

inmst[pre[r]][r]=inmst[r][pre[r]]=1;

for(int i=0;i<n;i++)

{

if(!used[i]&&g[r][i]<d[i])

{

d[i]=g[r][i];

pre[i]=r;

}

if(used[i]) maxe[i][r]=maxe[r][i]=minc;

}

used[r]=1;

}

return res;

}

int main()

{

int T,n,m;

scanf("%d",&T);

while(T--)

{

scanf("%d%d",&n,&m);

for(int i=0;i<n;i++)

for(int j=0;j<n;j++)

{

g[i][j]=inf;

maxe[i][j]=inf;

}

memset(inmst,0,sizeof(inmst));

while(m--)

{

int x,y,c;

scanf("%d%d%d",&x,&y,&c);

x--;y--;

g[x][y]=g[y][x]=min(g[x][y],c);//防止重边

}

int mst=prim(n,g);

int mst2=inf;

for(int i=0;i<n;i++)

for(int j=0;j<n;j++)

if(!inmst[i][j]&&g[i][j]<inf&&maxe[i][j]<inf)

{

if(mst+g[i][j]-maxe[i][j]<mst2)

mst2=mst+g[i][j]-maxe[i][j];

}

if(mst==mst2) printf("Not Unique!/n");

else printf("%d/n",mst);

}

return 0;

}