POJ1679 The Unique MST(Kruskal)（最小生成树的唯一性）

最新推荐文章于 2021-04-04 11:09:28 发布

转载最新推荐文章于 2021-04-04 11:09:28 发布 · 97 阅读

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原文链接：http://www.cnblogs.com/jianrenfang/p/5736204.html

本文详细介绍了如何判断一个无向图的最小生成树是否唯一的方法，并提供了完整的C++实现代码。通过定义生成树和最小生成树的概念，文章进一步解释了输入输出要求及示例，最后给出了具体的算法实现步骤。

The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 27141		Accepted: 9712

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

Sample Output

3
Not Unique!
【分析】最小生成树的唯一性，思路是先判断每条边是否有重边，有的话eq=1，否则0.然后第一次求出最小生成树，将结果记录下来，
 然后依次去掉第一次使用过的且含有重边的边，再求一次最小生成树，若结果与第一次结果一样，则不唯一。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include<functional>
#define mod 1000000007
#define inf 0x3f3f3f3f
#define pi acos(-1.0)
using namespace std;
typedef long long ll;
const int N=11000;
const int M=15005;
int n,m,cnt;
int parent[N];
bool flag;
struct man {
    int u,v,w;
    int eq,used,del;
} edg[N];
bool cmp(man g,man h) {
return g.w<h.w;
}
void init() {
    for(int i=0; i<=10005; i++) {
        parent[i]=i;
    }
}
int Find(int x) {
    if(parent[x] != x) parent[x] = Find(parent[x]);
    return parent[x];
}//查找并返回节点x所属集合的根节点
void Union(int x,int y) {
    x = Find(x);
    y = Find(y);
    if(x == y) return;
    parent[y] = x;
}//将两个不同集合的元素进行合并
int Kruskal() {
    init();
   int sum=0;
   int num=0;
   for(int i=0;i<m;i++){
    if(edg[i].del==1)continue;
    int u=edg[i].u;int v=edg[i].v;int w=edg[i].w;

    if(Find(u)!=Find(v)){
        sum+=w;
        if(!flag)edg[i].used=1;
        num++;
        Union(u,v);
    }
    if(num>=n-1)break;
   }
   return sum;
}
int main() {
    int t,d;
    cin>>t;
    while(t--) {
        cnt=0;
        cin>>n>>m;
        for(int i=0; i<m; i++) {
            cin>>edg[i].u>>edg[i].v>>edg[i].w;
            edg[i].del=0;
            edg[i].used=0;
            edg[i].eq=0;//一开始这个地方eq没有初始化，WA了好几发，操
        }
        for(int i=0;i<m;i++){
            for(int j=0;j<m;j++){
                if(i==j)continue;
                if(edg[i].w==edg[j].w)edg[i].eq=1;
            }
        }
        sort(edg,edg+m,cmp);
        flag=false;
        cnt=Kruskal();
        flag=true;
        bool gg=false;
        for(int i=0;i<m;i++){
            if(edg[i].used==1&&edg[i].eq==1){
                edg[i].del=1;
                int s=Kruskal();//printf("%d %d\n",i,s);
                if(s==cnt){
                    gg=true;
                    printf("Not Unique!\n");
                    break;
                }
                edg[i].del=0;
            }
        }
        if(!gg)cout<<cnt<<endl;
    }
    return 0;
}

View Code

转载于:https://www.cnblogs.com/jianrenfang/p/5736204.html