POJ 1679 The Unique MST （次小生成树）

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!

题意

给定一张图，求它的最小生成树是否唯一，若唯一输出最小生成树权值和，否则输出 Not Unique! 。

思路

直接求给定图的次小生成树，然后枚举所有不在 MST 中的边，替换掉最大边权的边以后若此时的结果仍是原 MST 边权和，说明最小生成树不唯一，否则唯一。

AC 代码

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<cmath>
#include<algorithm>
#include<iostream>
using namespace std;

const int maxn = 110;
const int inf = 0x3f3f3f;
bool vis[maxn];
int lowc[maxn];
int pre[maxn];
int Max[maxn][maxn];
bool used[maxn][maxn];
int cost[maxn][maxn];

int prim(int n)
{
    int ans=0;
    memset(vis,false,sizeof(vis));
    memset(Max,0,sizeof(Max));
    memset(used,false,sizeof(used));
    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]&&j!=p)
                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;
}

void init(int n)
{
    memset(cost,inf,sizeof(cost));
    for(int i=0; i<n; i++)
        cost[i][i]=0;
}

void solve(int n)
{
    int ans=prim(n);    // 最小生成树权值和
    int ss=inf;
    for(int i=0; i<n; i++)  // 枚举两点之间的边
        for(int j=i+1; j<n; j++)
            if(cost[i][j]!=inf&&!used[i][j])    // 如果 (i,j) 间存在边并且不在MST中
                ss=min(ss,ans+cost[i][j]-Max[i][j]);
    if(ss==ans)
        printf("Not Unique!\n");
    else printf("%d\n",ans);
}

int main()
{
    int T,n,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        init(n);
        for(int i=0; i<m; i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            cost[a-1][b-1]=cost[b-1][a-1]=c;
        }
        solve(n);
    }
    return 0;
}