#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int set[124],n,t,a,b,c;
long long min;
struct map
{
int x,y,val;
}m[ 100000 ];
int cmp( const void *a,const void *b )
{
return ( ( map * ) a ) -> val - ( ( map * ) b ) -> val;
}
int find( int i )
{
return set[ i ] == i ? i :set[ i ] = find( set[ i ] );
}
void merge( int x,int y )
{
int a = find( x ),b = find( y );
if( a != b )
set[ a ] = b;
}
void K( )
{
for( int i = 1;i <= c; ++i )
{
int a = find( m[ i ].x ),b = find( m[ i ].y ),val = m[ i ].val;
if( a != b )
{
set[ a ] = b;
min += val;
}
}
}
int main( )
{
while( scanf( "%d",&n ) != EOF )
{
for( int i = 1; i <= 123 ; ++i )
set[ i ] = i;
c = 1;
for( int i = 1; i <= n; ++i )
{
for( int j = 1 ; j <= n; ++j )
{
int val;
scanf( "%d", &val );
if( j >= i )//因为矩阵是对称的所以只要存入一半就可以了
{
m[ c ].x = i;
m[ c ].y = j;
m[ c ].val = val;
++c;
}
}
}
qsort( m + 1,c,sizeof( m[ 0 ] ),cmp );
scanf( "%d",&t );
while( t-- )
{
scanf( "%d%d",&a,&b );
merge( a,b );//将以连接的合并成一个集合
}
min = 0;
K( );
printf( "%I64d\n",min );
}
return 0;
}
此题是最小生成树的基础题
转载于:https://www.cnblogs.com/Lvsi/archive/2011/03/02/1969214.html