POJ1679: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

23 31 2 12 3 23 1 34 41 2 22 3 23 4 24 1 2

Sample Output

3Not Unique!

题意：判断最小生成树是否唯一

思路：首先求出最小生成树，记录现在这个最小生成树上所有的边，然后通过取消其中一条边，找到这两点上其他的边形成一棵新的生成树，求其权值，通过枚举所有可能，通过这些权值看与原最小生成树的权值比较看其是否唯一

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;struct node{    int x,y,dis;    int flag;} a[10005];int cmp(node x,node y){    return x.dis<y.dis;}int father[105],n;int kruskal(int num,int m){    int i,j,k;    int ans = 0,cnt = 1;    for(i = 0; i<m; i++)    {        if(i == num)//除去这条边之后再求一次最小生成树            continue;        int s1 = father[a[i].x];        int s2 = father[a[i].y];        if(s1!=s2)        {            ans+=a[i].dis;            cnt++;            father[s2] = s1;            for(j = 0; j<=n; j++)                if(father[j] == s2)                    father[j] = s1;        }    }    if(cnt!=n)        return -1;    else        return ans;}int main(){    int m,i,j,t,sum,ans,cnt;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        for(i = 0; i<=n; i++)            father[i] = i;        for(i = 0; i<m; i++)        {            scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].dis);            a[i].flag = 0;        }        sort(a,a+m,cmp);        cnt = 1;        ans = 0;        for(i = 0; i<m; i++)        {            int s1 = father[a[i].x];            int s2 = father[a[i].y];            if(s1!=s2)            {                a[i].flag = 1;                ans+=a[i].dis;                cnt++;                father[s2] = s1;                for(j = 0; j<=n; j++)                    if(father[j] == s2)                        father[j] = s1;            }        }        int flag = 0;        for(i = 0; i<m; i++)//枚举所有原最小生成树上的边        {            if(a[i].flag==0)                continue;            sum = 0;            for(j = 0; j<=n; j++)//初始化                father[j] = j;            sum = kruskal(i,m);            if(sum == ans)//与之前的最小生成树比较，如果相等，那么肯定不是唯一的            {                flag = 1;                break;            }        }        if(flag)            printf("Not Unique!\n");        else            printf("%d\n",ans);    }    return 0;}