hdu1102 很好的最小生成树克鲁斯卡尔算法

Constructing Roads

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8074    Accepted Submission(s): 3022

Problem Description

There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.

We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.

 

Input

The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.

Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.

 

Output

You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.

 

Sample Input

30 990 692990 0 179692 179 011 2

 

Sample Output

179

/*有些点城镇已经连通好了  但是我们在做的时候是求剩下的连通起来要花多少钱我们可以依然把这些已经连通好的边放进我们将要连通的所有边中  不过花费是0这样依旧求出的是最短路   注意找的是最短路*/#include<stdio.h>#include<stdlib.h>int map[105][105];struct haha{int view1;int view2;int value;}e[30000+5];int parent[10005];int cmp(const void *a,const void *b){return (*(struct haha *)a).value-(*(struct haha *)b).value;}int get_root(int x){return x==parent[x]?x:get_root(parent[x]);}int join(int x,int y){int root1,root2;root1=get_root(x);root2=get_root(y);// printf("root1=%d root2=%d\n",root1,root2);if(root1==root2) return 0;else parent[root1]=root2;return 1;}int main(){int i,j,n,k,x,y,c,ans;while(scanf("%d",&n)!=EOF){c=1;ans=0;for(i=1;i<=n;i++)for(j=1;j<=n;j++)scanf("%d",&map[i][j]);for(i=1;i<=n;i++)for(j=1;j<=n;j++){if(map[i][j]){e[c].view1=i;e[c].view2=j;e[c].value=map[i][j];c++;}}for(i=0;i<=102;i++)parent[i]=i;scanf("%d",&k);i=c;for(c;c<i+k;c++){scanf("%d %d",&x,&y);e[c].view1=x;e[c].view2=y;e[c].value=0;}//  printf("%d",c);qsort(e+1,c-1,sizeof(e[0]),cmp);//  for(i=1;i<c;i++)//   printf("value=%d\n",e[i].value);//  printf("c=%d\n",c);for(i=1;i<c;i++){//   printf("i=%d:",i);if(join(e[i].view1,e[i].view2)){ans=ans+e[i].value;}}printf("%d\n",ans);}return 0;}