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): 24129    Accepted Submission(s): 9269

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

 

Source

kicc

 

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题意就是要修路，给出代表城市数目，又给出城市的邻接矩阵，有些城市之间已经有道路了，这些道路连接的两个城市之间的距离直接置零就可以了，然后就是求最小生成树，可以用prim算法或者kruskal算法，因为给出的是邻接矩阵，所以用prim算法比较好，用kruskal算法可以提前将有道路的两个顶点合并就可以了，prim算法适用于稠密图O(N^2)，kruskal适合稀疏图O(eloge).

代码:

#include<iostream>
#include<cstring>
#include<cstdio>

using namespace std;
const int N = 1000 + 5;
const int INF = 0x3f3f3f3f;
int mat[N][N],lowc[N];
bool visit[N];

int prim(int n){
    int i,k,cnt,ret=0;
    memset(lowc,INF,sizeof(lowc));
    memset(visit,0,sizeof(visit));
    for(lowc[1]=cnt=0;cnt < n;cnt++){
        for(k=-1,i=1;i<=n;i++)
            if(!visit[i]&&(k==-1||lowc[i] < lowc[k]))
            k = i;
        for(visit[k]=true,ret+=lowc[k],i=1;i<=n;i++)
            if(!visit[i]&&mat[k][i] < lowc[i])
            lowc[i] = mat[k][i];
    }
    return ret;
}
int main(){
    int n,m,u,v;
    while(scanf("%d",&n)==1){
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++) scanf("%d",&mat[i][j]);
        scanf("%d",&m);
        for(int i=1;i<=m;i++){
            scanf("%d %d",&u,&v);
            mat[u][v] = mat[v][u] = 0;
        }
        printf("%d
",prim(n));
    }
}