Constructing Roads 【poj-2421】【最小生成树】

Constructing Roads

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 25040 Accepted: 10897

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：要连接所有的村庄，其中已经有一些路是已存在的，需要再修尽可能短的路使所有村庄相连。

题解：最小生成树（Prim算法）。将已存在的路更新为0，这样在后续的计算时，就不用再修建这些路了。

代码如下：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int maxn=105;const int INF=0x3f3f3f3f;int n,m;int cost[maxn][maxn];int mincost[maxn];bool used[maxn];int prim(){for(int i=1;i<=n;i++){mincost[i]=INF;used[i]=false;}mincost[1]=0;int res=0;while(true){int v=-1;for(int u=1;u<=n;u++){if(!used[u]&&(v==-1||mincost[u]<mincost[v])) v=u;}if(v==-1) break;used[v]=true;res+=mincost[v];for(int u=1;u<=n;u++){mincost[u]=min(mincost[u],cost[v][u]);}}return res;}int main(){while(~scanf("%d",&n)){for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){scanf("%d",&cost[i][j]);}}scanf("%d",&m);int a,b;for(int i=0;i<m;i++){scanf("%d%d",&a,&b);cost[a][b]=cost[b][a]=0;}printf("%d\n",prim());}return 0;}