Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19579 Accepted Submission(s): 7474
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.
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.
#include <stdio.h> #include <iostream> #include <string.h> using namespace std; const int N = 105; const int INF = 99999999999; int graph[N][N]; int low[N]; bool vis[N]; int prim(int n,int start){ memset(vis,false,sizeof(vis)); memset(low,false,sizeof(low)); int pos = start,cost=0; vis[pos]=true; for(int i=1;i<=n;i++){ low[i] = graph[pos][i]; } for(int i=1;i<n;i++){ int Min = INF; for(int j=1;j<=n;j++){ if(!vis[j]&&low[j]<Min){ pos = j,Min = low[j]; } } vis[pos]=true; cost+=Min; for(int j=1;j<=n;j++){ if(!vis[j]&&low[j]>graph[pos][j]) low[j] = graph[pos][j]; } } return cost; } int main(){ int n,m; while(scanf("%d",&n)!=EOF){ for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ scanf("%d",&graph[i][j]); } } scanf("%d",&m); while(m--){ int a,b; scanf("%d%d",&a,&b); graph[a][b]=graph[b][a] = 0; } printf("%d\n",prim(n,1)); } }