POJ 2421 Constructing Roads

Constructing Roads

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

这道题刚开始一看，下意识以为是单源最短路。WA了一发之后才知道是最小生成树，可以说是很蠢了。

直接做就OK，没什么难度。不过我写的时候最小生成树已经忘得差不多了，看了一下模板才写出来。

AC代码：

#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>#include<iomanip>using namespace std;int mp[1111][1111];int n,nn;int a,b;int v[1111];int minTree()  {      int i,j,flag;      memset(v,0,sizeof(v));      v[1]=1;      int sum=0;      for(i=1;i<n;i++)      {          int min=999999;          for(j=1;j<=n;j++)          {              if(!v[j]&&mp[1][j]<min)              {                  min=mp[1][j];                  flag=j;              }          }          sum+=min;          v[flag]=1;          for(j=1;j<=n;j++)          {              if(!v[j]&&mp[1][j]>mp[flag][j])              {                  mp[1][j]=mp[flag][j];              }          }      }      return sum;  }int main(){while(cin>>n){for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){cin>>mp[i][j];}}cin>>nn;while(nn--){cin>>a>>b;mp[a][b]=0;mp[b][a]=0;}int res=minTree();cout<<res<<endl;}return 0;}