Constructing Roads
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)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.
3 0 990 692 990 0 179 692 179 0 1 1 2
179
#include<stdio.h>
#include<string.h>
#define size 101
#define inf 100000000
int map[size][size];
int dist[size];
int visit[size];
int n;
int sum;
void prim(void){
int min;
int pos;
memset(visit,0,sizeof(visit));
for(int i=1;i<=n;i++)
dist[i]=map[1][i];
visit[1]=1;
for(int i=1;i<n;i++){
min=inf;
for(int j=1;j<=n;j++)
if(!visit[j]&&dist[j]<min){
min=dist[j];
pos=j;
}
if(min==inf)
break;
sum+=dist[pos];
visit[pos]=1;
for(int j=1;j<=n;j++)
if(!visit[j]&&map[pos][j]<dist[j])
dist[j]=map[pos][j];
}
}
int main(void){
int q;
int s;
int e;
while(scanf("%d",&n)!=EOF){
sum=0;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
scanf("%d",&map[i][j]);
scanf("%d",&q);
for(int i=0;i<q;i++){
scanf("%d%d",&s,&e);
map[s][e]=map[e][s]=0;
}
prim();
printf("%d\n",sum);
}
}
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