现得到城镇道路统计表。表中列出了随意两城镇间修建道路的费用,以及该道路是否已经修通的状态。现请你编敲代码。计算出全省畅通须要的最低成本。
3 1 0
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这道题是用的最小生成树做的,和dijstra非常像
代码:
#include<stdio.h>
#include<string.h>
#define INF 1 << 30
int map[1001][1001] ;
int dis[1001] ;
int used[1001] ;
void Prim(int N)
{?
?int i = 0 ,j = 0 ;
?int c = 0 ;?
?int sum = 0 ;//用来记录最后所须要的花费
?dis[1] = 0 ;
??? for( i = 1 ; i <= N ; i++)
?{
??int min = INF ;
??for( j = 1 ; j <= N ; j++)
??{
??????????? if(!used[j] && dis[j] < min)
???{
????min = dis[j] ;
????c = j ;
???}
??}
??used[c] = 1 ;
??for(j = 1 ; j <= N ; j++)
??{
???if(!used[j] && dis[j] > map[c][j])
????dis[j] = map[c][j] ;
??}
?}
??? for(i = 1 ; i <= N ; i++)
??sum += dis[i] ;
?printf("%d\n",sum);
}
int main()
{
?int N = 0 ;
?while(~scanf("%d",&N))
?{
??if(N == 0)
???break ;
???? int a = 0 , b = 0 , c = 0 , d = 0 ;
??int i = 0 , j = 0 ;
??for(i = 1 ; i <= N ; i++)
??{
???for(j = 1 ; j <= N ; j++)
????map[i][j] = INF ;
????? dis[i] = INF ;
????? used[i] = 0 ;
??}
??int m = 0 ;
??m = N*(N-1)/2 ;
??for( i = 0 ; i < m ; i++)
??{
???scanf("%d%d%d%d" , &a , &b , &c , &d );
???//当d = 1时,对于此时两点间的花费能够不用计入当中。由于已经修建了。则花费可变为0
???if( d == 1 )
????map[a][b] = map[b][a] = 0 ;
???else
???{
????//推断是否会有重边
????if(map[a][b] > c)
??????? map[a][b] = map[b][a] = c ;?
???}
??}
??Prim( N ) ;
?}
?return 0 ;
}
?
原文:https://www.cnblogs.com/ldxsuanfa/p/10856451.html