【HDU】-1102-Constructing Roads（最小生成树）

Constructing Roads

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20691    Accepted Submission(s): 7903

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

 

题意：给了一个矩阵，可以看成 x，y 轴。如题目(1,1),(2,2),(3,3)就是自己本身，长度为0；以此类推......

#include<cstdio>#include<cstring>#include<queue>#include<stack>#include<cmath>#include<vector>#include<algorithm>using namespace std;#define CLR(a,b)  memset(a,b,sizeof(a))struct node{int from,to,va;}a[5010];bool cmp(node x,node y){return x.va<y.va;}int f[110];int find(int x){if(x!=f[x])f[x]=find(f[x]);return f[x];}bool join(int x,int y){int fx=find(x);int fy=find(y);if(fx!=fy){f[fx]=fy;return true;} return false;}int main(){int n;while(~scanf("%d",&n)){for(int i=1;i<=n;i++)f[i]=i;int t,num=1;for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){if(j>i)//只看一半，减少时间 {scanf("%d",&t);a[num].from=i;a[num].to=j;a[num].va=t;num++;}elsescanf("%d",&t);}}sort(a+1,a+num,cmp);int m;scanf("%d",&m);for(int i=1;i<=m;i++){int a,b;scanf("%d %d",&a,&b);//a,b 已经修建，直接连接 join(a,b); } int ant=0;num=0;for(int i=1;i<=n*(n-1)/2;i++){if(join(a[i].from,a[i].to)){ant+=a[i].va;num++;}if(num==n-1)break;}printf("%d\n",ant);}return 0;}