Kruskal次小生成树 :The Unique MST

The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 14402		Accepted: 4981

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!

题意是问有没有耗费相同的不同的最小生成树

此题的数据都是连通的，先最小生成树一次，把树里每条边标记，然后依次删除来看新生成的最小树的耗费是否与原来相同，相同则不唯一

唯一要注意的是删边之后要注意判断这次图还连不连通，如果不连通那这次得到的结果就不能拿来比较

#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

const long long INF = 1 << 31 -1;
int p[11000];
long long mm;
long long tmp;
int n, m;
int fflag;
int del;

struct edge{
    int u;
    int v;
    long long w;
    int flag;
}e[10001000];

int find(int x){
    return x == p[x] ? x : find(p[x]);
}

int cmp(edge a, edge b){
    return a.w < b.w;
}

int Kruskal(){
    long long ans = 0;
    int i;
    for(i = 1; i <= n; i ++)
        p[i] = i;
    sort(e + 1, e + m + 1, cmp);
    for(i = 1; i <= m; i ++){
        int x = find(e[i].u);
        int y = find(e[i].v);
        if(x != y){
            e[i].flag = 1;
            ans += e[i].w;
            p[x] = y;
        }
    }
    return ans;
}

long long Kruskalm(){
    long long ans = 0;
    int i;
    for(i = 1; i <= n; i ++)
        p[i] = i;
    sort(e + 1, e + m + 1, cmp);
    int j = 0;
    fflag = 0;
    for(i = 1; i <= m; i ++){
        if(i == del)
            continue;
        int x = find(e[i].u);
        int y = find(e[i].v);
        if(x != y){
            ans += e[i].w;
            p[x] = y;
            j ++;
        }
    }
    if(j < n - 1) //如果最后生成树的边数小于n-1，显然就是因为删边之后不连通了，所以这次的结果直接赋为无穷大，相当于不行
        ans = INF;
    return ans;
}

int main(){
    int t, i, tt;
    cin >> t;
    while(t --){
        int ff = 0;
        scanf("%d %d", &n ,&m);
        for(i = 1; i <= m; i ++){
            scanf("%d %d %d", &e[i].u, &e[i].v, &e[i].w);
            e[i].flag = 0;
        }
        mm = Kruskal();
        for(i = 1; i <= m; i ++){
            if(e[i].flag){
                //tt = e[i].w;
                del = i;
                tmp = Kruskalm();
                if(tmp == mm){
                    ff = 1;
                    break;
                }
            }
        }
            if(ff){
                printf("Not Unique!\n");
            }
            else
                printf("%I64d\n", mm);
    }
    return 0;
}