POJ - 1679 The Unique MST

[align=center]The Unique MST[/align]

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 31842		Accepted: 11506

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

题意：
问是否存在次小生成树，如果存在输出'Not Unique!'.  否则输出最小生成树的权值和

思路：

用prime算法，每找到一条路就从重新开始判断走过的路中是否存在多条可以到此点（代码中p点）的路，并且距离相等，存在说明有次小生成树，否则没有。

代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f

int n,m;
int dis[105][105],dist[105];
int vis[105];

void prime(){
int sum=0;
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++){
dist[i]=dis[1][i];
}
vis[1]=1;
m=n-1;
while(m--){
int minx=INF;
int p=0;
for(int i=1;i<=n;i++){
if(!vis[i] && dist[i]<minx){
minx=dist[i];
p=i;
}
}
if(p==0)break;

int k=0;
for(int i=1;i<=n;i++){//重新查找走过的路中是否还有此距离的最短路
if(vis[i] && dis[i][p]==minx){
k++;//每找到一条总数加一
}
}
if(k>1){//当超过一条时表示存在多条路
printf("Not Unique!\n");
return ;
}
vis[p]=1;
sum+=minx;

for(int i=1;i<=n;i++){
if(!vis[i] && dist[i]>dis[i][p])
dist[i]=dis[i][p];
}
}
printf("%d\n",sum);
}

int main(){
int t;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
int a,b,c;
memset(dis,INF,sizeof(dis));
for(int i=0;i<m;i++){
scanf("%d%d%d",&a,&b,&c);
dis[a][b]=dis[b][a]=min(dis[a][b],c);
}
prime();
}
return 0;
}