此题也是最小生成树,是Prim算法的直接套用,只需要处理一下,对于村庄中已经修好的路,将两村中间的距离设为0即可。然后直接用Prim算法。
# include <stdio.h>
# include <string.h>
# include <stdlib.h>
# define INF 10000000
int N,Q;
int graph[105][105];
int visit[105],dis[105];
void Init()
{
int i,j;
memset(visit,0,sizeof(visit));
for(i = 1;i <= N;i ++)
for(j = 1;j <= N;j ++)
graph[i][j] = graph[j][i] = INF;
for(i = 1;i <= N;i ++)
graph[i][i] = 0;
}
int work()
{
int i,j,sum = 0,mincost,index;
for(i = 1;i <= N;i ++)
dis[i] = graph[1][i];
visit[1] = 1;
for(i = 1;i < N;i ++)
{
mincost = INF;
for(j = 1;j <= N;j ++)
{
if(!visit[j] && dis[j] < mincost)
{
mincost = dis[j];
index = j;
}
}
sum += mincost;
visit[index] = 1;
for(j = 1;j <= N;j ++)
{
if(!visit[j] && graph[index][j] < dis[j])
dis[j] = graph[index][j];
}
}
return sum;
}
int main()
{
int i,j,a,b;
while(~scanf("%d",&N))
{
Init();
for(i = 1;i <= N;i ++)
for(j = 1;j <= N;j ++)
scanf("%d",&graph[i][j]);
scanf("%d",&Q);
for(i = 1;i <= Q;i ++)
{
scanf("%d%d",&a,&b);
graph[a][b] = graph[b][a] = 0;
}
printf("%d\n",work());
}
return 0;
}