Constructing Roads
Constructing Roads
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11481 Accepted Submission(s): 4312
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.
0 990 692
990 0 179
692 179 0
1
1 2
思路:求最小生成树,prim 算法,注意将已经修过的路的权值置为0。
#include<stdio.h>
int map[][],p[],dist[];
int main()
{
int i,j,k,n,m,a,b,min,len;
while(~scanf("%d",&n))
{
for(i = ;i < n;i ++)
{
for(j = ;j < n;j ++)
scanf("%d",&map[i][j]);
}
scanf("%d",&m);
for(i = ;i < m;i ++)
{
scanf("%d%d",&a,&b);
map[a-][b-] = map[b-][a-] = ;
}
for(i = ;i < n;i ++)
{
p[i] = ;
dist[i] = map[][i];
}
len = dist[] = ;
p[] = ;
for(i = ;i < n;i ++)
{
min = ;
k = ;
for(j = ;j < n;j ++)
{
if(!p[j]&&dist[j]<min)
{
min = dist[j];
k = j;
}
}
len += min;
p[k] = ;
for(j = ;j < n;j ++)
{
if(!p[j]&&dist[j]>map[k][j])
dist[j] = map[k][j];
}
}
printf("%d\n",len);
}
return ;
}
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