HDU-1102-Constructing Roads（最小生成树）

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

3

0 990 692

990 0 179

692 179 0

1

1 2

 

 

Sample Output

179

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思路：克鲁斯卡尔算法呗。已经连的就merge（）先处理一下。

坑点：

1.题没说多组测试用例，但其实是多组...一般地（可能），输出一行就是多组，输出一个数是一组。

2.剪枝的时候cnt初始值要注意一下，本题也可以不用剪枝。

---

#include<cstdio>
#include<algorithm>
using namespace std;
int N,Q;
int pre[105];

struct node{
    int a,b,len;    
}l[5000];

int find(int x){
    while(pre[x]!=x){
        int r=pre[x];
        pre[x]=pre[r];
        x=r;
    }
    return x;
}

void merge(int x,int y){
    int fx=find(x);
    int fy=find(y);
    if(fx!=fy) pre[fx]=fy;
}

int cmp(node x,node y){
    return x.len<y.len;
}

int main(){
    while(~scanf("%d",&N)){
    
        int cnt=0;
        int val;
        for(int i=1;i<=N;i++)
            pre[i]=i;
    
        for(int i=1;i<=N;i++){
            for(int j=1;j<=N;j++){
                scanf("%d",&val);
                if(j<=i) continue;
            
                l[cnt].a=i;
                   l[cnt].b=j;
                l[cnt].len=val;
                cnt++;            
            }
        }
        sort(l,l+cnt,cmp);
        scanf("%d",&Q);
        cnt=0;
        for(int i=0;i<Q;i++){
            int a,b;
            scanf("%d%d",&a,&b); 
            if(find(a)!=find(b)) {//这个要注意一下，因为a，b可能已经有路，也有可能被别的城市间接连起来了。//为了后边剪枝 
            cnt++;  
            merge(a,b);
            }
        }

        int sum=0;        
        for(int i=0;i<N*(N-1)/2;i++){
            if(cnt==N-1) break;//这个剪枝可以不要，也能AC 
            if(find(l[i].a)==find(l[i].b)) continue;
            sum+=l[i].len;
            merge(l[i].a,l[i].b);
            cnt++;
        }
         printf("%d\n",sum);
    }
    return 0;
}