The Unique MST POJ - 1679 (次小生成树)

时间：2018-07-08 16:40:47 阅读：184 评论：0 收藏：0 [点我收藏+]

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!


求次小生成树 看与最小生成树是否相同 
prime求次小生成树

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#define mem(a, b) memset(a, b, sizeof(a))
using namespace std;
const int maxn = 10010, INF = 0x7fffffff;
typedef long long LL;
int graph[510][510], d[maxn], vis[maxn], maxd[510][510], pre[maxn];
int n, m;

int prime(int s)
{
    int temp, sum = 0;
    mem(vis, 0);
    for(int i=1; i<=n; i++) d[i] = graph[s][i], pre[i] = s;
    vis[s] = 1;
    d[s] = 0;
    for(int i=1; i<n; i++)
    {
        int mincost = INF;
        for(int j=1; j<=n; j++)
        {
            if(!vis[j] && mincost > d[j])
                mincost = d[j], temp = j;
        }
        for(int j=1; j<=n; j++)
            if(vis[j]) maxd[temp][j] = maxd[j][temp] = max(mincost, maxd[pre[temp]][j]);
        vis[temp] = 1;
        sum += mincost;
        for(int j=1; j<=n; j++)
        {
            if(!vis[j] && d[j] > graph[temp][j])
                d[j] = graph[temp][j], pre[j] = temp;
        }
    }
//    for(int i=1; i<=n; i++)
//        sum += d[i];
    return sum;
}


int main()
{
    int T;
    cin>> T;
    while(T--)
    {
        cin>> n >> m;
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                if(i == j) graph[i][j] = 0;
                else graph[i][j] = graph[j][i] = INF;
        for(int i=0; i<m; i++)
        {
            int u, v, w;
            cin>> u >> v >> w;
            graph[u][v] = graph[v][u] = w;
        }
        int sum = prime(1);
        int lsum = INF;
        for(int i=1; i<=n; i++)
            for(int j=i+1; j<=n; j++)
            {
            if(i != pre[j] && j != pre[i]  && graph[i][j] != INF)
                if(sum - maxd[i][j] + graph[i][j] < lsum)
                    lsum = sum - maxd[i][j] + graph[i][j];
            }

        if(lsum == sum)
            cout<< "Not Unique!" <<endl;
        else
            cout<< sum <<endl;

    }



    return 0;
}

View Code

The Unique MST POJ - 1679 (次小生成树)

标签：cin one repr ++ div class sum -- mst

原文地址：https://www.cnblogs.com/WTSRUVF/p/9280124.html

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