zoj 3034 - The Bridges of Kolsberg

Posted 2020-09-21 cynchanpin

题目：在河两端有两排server，如今要把河两边同样的品牌型号的机器连起来。每一个电脑有个值，

           每一个机器仅仅能与还有一台机器链接。而且不同的链接不交叉，如今要求链接的电脑总之最大。

分析：dp，最大公共子序列，字符串。还要加一个字符串处理。

说明：（2011-09-19 11:08）。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define max( a, b ) ((a)>(b)?(a):(b))

char LeftOS[ 1001 ][ 12 ];
char RightOS[ 1001 ][ 12 ];
int  LeftID[ 1001 ];
int  LeftV[ 1001 ];
int  RightID[ 1001 ];
int  RightV[ 1001 ];
char OSList[ 1001 ][ 12 ]; 
int  Match[ 1001 ][ 1001 ];
int  Count[ 1001 ][ 1001 ];
int  Number = 0;

int ID( char * Data ) 
{
    for ( int i = 0 ; i < Number ; ++ i )
        if ( !strcmp( OSList[ i ], Data ) )
            return i;
    strcpy( OSList[ Number ], Data );
    return Number ++;
}

int main()
{
    int  t,n,m;
    char City[ 12 ];
    while ( ~scanf("%d",&t) ) 
    while ( t -- ) {
        scanf("%d",&n);
        for ( int i = 1 ; i <= n ; ++ i )
            scanf("%s %s %d",City,LeftOS[ i ],&LeftV[ i ]);
        scanf("%d",&m);
        for ( int i = 1 ; i <= m ; ++ i )
            scanf("%s %s %d",City,RightOS[ i ],&RightV[ i ]);
        
        Number = 0;
        for ( int i = 1 ; i <= n ; ++ i )
            LeftID[ i ] = ID( LeftOS[ i ] );
        for ( int i = 1 ; i <= m ; ++ i )
            RightID[ i ] = ID( RightOS[ i ] );
        
        memset( Match, 0, sizeof( Match ) );
        memset( Count, 0, sizeof( Count ) );
        
        for ( int i = 1 ; i <= n ; ++ i )
        for ( int j = 1 ; j <= m ; ++ j ) {
            if ( Match[ i ][ j ] < Match[ i-1 ][ j ] ) {
                Match[ i ][ j ] = Match[ i-1 ][ j ];
                Count[ i ][ j ] = Count[ i-1 ][ j ];
            }
            if ( Match[ i ][ j ] < Match[ i ][ j-1 ] ) {
                Match[ i ][ j ] = Match[ i ][ j-1 ];
                Count[ i ][ j ] = Count[ i ][ j-1 ];
            }
            if ( LeftID[ i ] == RightID[ j ] && Match[ i ][ j ] < Match[ i-1 ][ j-1 ] + LeftV[ i ] + RightV[ j ] ) {
                Match[ i ][ j ] = Match[ i-1 ][ j-1 ] + LeftV[ i ] + RightV[ j ];
                Count[ i ][ j ] = Count[ i-1 ][ j-1 ] + 1;
            }
        }

        printf("%d %d\n",Match[ n ][ m ],Count[ n ][ m ]);
    }
    return 0;
}