POJ 1679 The Unique MST
The Unique MST
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 17785 | Accepted: 6176 |
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'.
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!
题意：有n个点，其中m条边可以连通。求最小生成树是否唯一
这道题一开始没什么特别好的想法，就是想着把最小生成树当中的每一条边都去掉，然后看看生成树的最小路径是否还相同
这道题要虽然是枚举，不过要注意几个方面：
1. 在枚举的过程中，去掉边之后可能就无法得到完整的树。
2. 我在用Prime做的时候，一开始没有考虑清楚一条边的两个端点。wa了好多次。
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#define INF 1000000
//相当于求次小生成树 从最小生成树当中每次删去一边，然后判断生成树是否和最小值相同
//要注意删边之后可能会连通不了
using namespace std;
int n,m,p,ans;
int map[105][105];
int way[105][2];//用prime算法做的时候，要记录每条边比较麻烦。
//每次加入集合的是点，并不代表最后两个加入的两点之间的那条边就是所取边
//因为这个wa了好多次。。
int prime()
{
int low[105][2]= {0},visit[105]= {0};
int i,j,ans=0,k;
p=0;
for (i=0; i<=n; i++)
{
low[i][0]=map[1][i];
low[i][1]=1;
}
visit[1]=1;
way[p][0]=1;
for (i=1; i<n; i++)
{
int max=INF;
k=-1;
for (j=1; j<=n; j++)
if (!visit[j] && low[j][0]<max)
{
max=low[j][0];
k=j;
}
if (k==-1) return 0;
way[p][0]=low[k][1];
way[p++][1]=k;
visit[k]=1;
ans+=low[k][0];
for (j=1; j<=n; j++)
if(!visit[j] && low[j][0]>map[k][j])
{
low[j][0]=map[k][j];
low[j][1]=k;
}
}
return ans;
}
int find()
{
int low[105]= {0},visit[105]= {0};
int i,j,s=0,k;
for (i=0; i<=n; i++)
low[i]=map[1][i];
visit[1]=1;
for (i=1; i<n; i++)
{
int max=INF;
k=-1;
for (j=1; j<=n; j++)
if (!visit[j] && low[j]<max)
{
max=low[j];
k=j;
}
if (k==-1) return -1;
visit[k]=1;
s+=low[k];
for (j=1; j<=n; j++)
if(!visit[j] && low[j]>map[k][j])
low[j]=map[k][j];
}
return s;
}
int main ()
{
int t,i,j;
cin>>t;
while(t--)
{
cin>>n>>m;
for (i=0; i<=n; i++)
{
way[i][0]=way[i][1]=0;
for (j=0; j<=n; j++)
map[i][j]=INF;
}
for (i=0; i<m; i++)
{
int a,b,c;
cin>>a>>b>>c;
map[a][b]=map[b][a]=c;
}
ans=prime();
int flag=0;
int x,y,temp,s;
i=0;
while(i<p)
{
x=way[i][0];
y=way[i][1];
temp=map[x][y];
map[x][y]=map[y][x]=INF;
s=find();
if (s==ans)
{
flag=1;
break;
}
map[x][y]=map[y][x]=temp;
i++;
}
if (flag) cout<<"Not Unique!"<<endl;
else cout<<ans<<endl;
}
return 0;
}