最小生成树 TOJ 4117 Happy tree friends

最新推荐文章于 2019-12-08 20:33:53 发布

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本文介绍了一个关于最小生成树的问题，给出了具体的输入输出样例，并提供了一段使用Kruskal算法求解该问题的C++代码实现。文章重点在于如何处理特定边必须包含在最小生成树中的情况。

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链接http://acm.tju.edu.cn/toj/showp4117.html
4117. Happy tree friends

Time Limit: 1.0 Seconds Memory Limit: 65536K
Total Runs: 164 Accepted Runs: 60

yuebai has an undirected complete graph with n vertices. He wants to know the minimum spanning tree of the graph. It's so easy, so yuebai wants to challenge himself. He will choose one edge which must be in the spanning tree.

INPUT

There are multiple test cases.
For each test case, the first line contain an integer n .
In the next n lines, there is an adjacency matrix M . Mij denotes the weight of the edge i to j .
Next line contains two dinstinct integer u and v , which denotes the edge which is from u to v with the value Muv must be in the spanning tree.
(2≤n≤100,0≤Mij≤100) . Mij=0 if and only if i=j .

OUTPUT

For each case, print the result.

Sample Input

Sample Output

Hint

The edge of the spanning tree is 2->3 and 2->1

Source: TJU Team Selection 2015 Round B

用Kruskal做

只要把题目中要求的边先合和起来，其余按照模版来，题目说了无向，所以对于每边的长度取矩阵中的最小值（真是坑，一开始以为是最小树形图）

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;

#define N 50005
int a[105][105];
struct graph{
  int x,y,wei;
}nodd[N];
int m,n,ufind[N];

int cmp(graph a1,graph a2){
  return a1.wei<a2.wei;
}
int find(int x){
  return ufind[x]==x? x : ufind[x]=find(ufind[x]);
}
int Kruskal(int a,int b){
	int ans=0;
	int i,j;
	for(i=1;i<=n;i++) ufind[i]=i;
	sort(nodd,nodd+m,cmp);
	ufind[a]=b;
	for(i=0;i<m;i++){
	  int x=find(nodd[i].x);  int y=find(nodd[i].y);
	  if(x!=y){
	    ans+=nodd[i].wei;
	    ufind[x]=y;
	  }
	}
	return ans;
}

int main(){
  int x;
  int sum;
  int temp;
  int a1,a2;
  int i,j,k;
  while(scanf("%d",&x)!=EOF){
  	 sum=0;
     temp=0;
     for(i=1;i<=x;i++)
       for(j=1;j<=x;j++) 
         scanf("%d",&a[i][j]);
     m=x*(x-1)/2;
     n=x;
     scanf("%d %d",&a1,&a2);
     sum+=a[a1][a2];
     a[a2][a1]=a[a1][a2];
     for(i=1;i<=x;i++)
       for(j=i+1;j<=x;j++){ 
	     nodd[temp].x=i;  nodd[temp].y=j;  nodd[temp].wei=min(a[i][j],a[j][i]);
	     temp++;
	   }
     //for(i=0;i<temp;i++) printf("%d %d %d\n",nodd[i].x,nodd[i].y,nodd[i].wei);
     printf("%d\n",Kruskal(a1,a2)+sum);
  }
}