poj 1679 The Unique MST

The Unique MST
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 32373 Accepted: 11743
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：给你一个图，判断其最小生成树是否唯一

求其次小生成树，如果次小生成树与最小生成树总权值相同，则说明不唯一，否则唯一

次小生成树求法：先用prime求出最小生成树，在求的过程中记录下来从点i到点j路径中最大的那条边的权值，然后再枚举没在最小生成树中的边，进行替换最小生成树中的边，如果能替换并且替换之后权值不变则不唯一。（现学现卖）

#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <string.h>
#define maxn 1100
#define INF 999999999
int edge[maxn][maxn];
bool vis[maxn];
int lowcost[maxn];
bool used[maxn][maxn];
int Max[maxn][maxn];
int pre[maxn];
using namespace std;
int n,m;
int prime(int num)
{
    memset(used,0,sizeof(used));
    memset(Max,0,sizeof(Max));
    vis[num] = 1;
    pre[num] = -1;
    for(int i = 1; i <= n; i++)
    {
        if(i == num) continue;
        lowcost[i] = edge[i][num];
        pre[i] =num;
    }
    int sum = 0;
    for(int i = 1; i < n; i++)
    {
        int minn = INF;
        int temp;
        for(int j = 1;j <= n; j++)
        {
            if(vis[j] == 0 && lowcost[j] < minn)
            {
                minn = lowcost[j];
                temp = j;
            }
        }
        if(minn == INF) return -1;
        vis[temp] = 1;
        sum += minn;
        used[temp][pre[temp]] = used[pre[temp]][temp] = 1;
        for(int j = 1; j <= n; j++)
        {
            if(vis[j] == 1)
            {
                Max[j][temp] = Max[temp][j] = max(Max[j][pre[temp]],lowcost[temp]);
            }
            if(vis[j] == 0 && lowcost[j] > edge[j][temp])
            {
                lowcost[j] = edge[j][temp];
                pre[j] = temp;
            }
        }
    }
    return sum;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        memset(vis,0,sizeof(vis));
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                edge[i][j] = INF;
            }
        }
        for(int i = 0; i < m; i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            edge[u][v] = edge[v][u] = w;
        }
        int ans = prime(1);
        if(ans == -1)
        {
            printf("Not Unique!\n");
            continue;
        }
        int Min = INF;
        for(int i = 1; i <= n; i++)
        {
            for(int j =i+1; j <= n; j++)
            {
                if(edge[i][j] != INF && !used[i][j])
                {
                    Min = min(Min,ans+edge[i][j]-Max[i][j]);
                }
            }
        }
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                printf("%d ",Max[i][j]);
            }
            printf("\n");
        }
        if(Min == ans) printf("Not Unique!\n");
        else printf("%d\n",ans);
    }
}