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最新推荐文章于 2024-06-26 16:53:26 发布

Jerry_wang119

最新推荐文章于 2024-06-26 16:53:26 发布

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分类专栏： dp 背包文章标签： dp 背包

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本文通过一道NOIP题目介绍了动态规划的应用方法。详细展示了如何使用C++实现动态规划算法来解决背包问题变种，并针对特定条件进行了优化。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

啧，YL 居然选了 Noip 的题来做练习？

唉，点了发布居然没有发布成功，神 TM 我写了一个下午，沃日 …… 垃圾优快云 ……

直接放代码吧 ……

#include <bits/stdc++.h>

const  int  N = 10000 + 50 , M = 1000 + 50 ;  

// using  namespace  std ;

int  dp [ N ] [ M ] , tube [ N ] [ 3 ] ;
int  x [ N ] , y [ N ] , tag [ N ] ;
int  n , m , k , p , inf , ans ;
bool  ok ;

int  main ( ) {
    scanf ( "%d%d%d" , & n , & m , & k ) ;
    for ( int  i = 0 ; i < n ; i ++ )
        scanf ( "%d%d" , & x [ i ] , & y [ i ] ) ;
    for ( int  i = 1 ; i <= n ; i ++ )
        tube [ i ] [ 0 ] = 0 , tube [ i ] [ 1 ] = m + 1 ;
    for ( int  i = 1 ; i <= k ; i ++ ) {
        scanf ( "%d" , & p ) ;
        tag [ p ] = 1 ;
        scanf ( "%d%d" , & tube [ p ] [ 0 ] , & tube [ p ] [ 1 ] ) ;
    }
    memset ( dp , 0x7f , sizeof ( dp ) ) ;
    inf = dp [ 0 ] [ 0 ] ;
    for ( int  i = 0 ; i <= m ; i ++ )
        dp [ 0 ] [ i ] = 0 ; 
    for ( int  i = 1 ; i <= n ; i ++ ) {
        ok = true ;
        for ( int  j = x [ i - 1 ] + 1 ; j <= m ; j ++ ) 
            dp [ i ] [ j ] = std :: min ( dp [ i ] [ j ] , std :: min ( dp [ i - 1 ] [ j - x [ i - 1 ] ] + 1 , dp [ i ] [ j - x [ i - 1 ] ] + 1 ) ) ;
        for ( int  j = 1 ; j <= m - y [ i - 1 ] ; j ++ )
            dp [ i ] [ j ] = std :: min ( dp [ i ] [ j ] , dp [ i - 1 ] [ j + y [ i - 1 ] ] ) ;
        for ( int  j = m - x [ i - 1 ] ; j <= m ; j ++ )
            dp [ i ] [ m ] = std :: min ( dp [ i ] [ m ] , std :: min ( dp [ i ] [ j ] + 1 , dp [ i - 1 ] [ j ] + 1 ) ) ;
        for ( int  j = 1 ; j <= tube [ i ] [ 0 ] ; j ++ ) 
            dp [ i ] [ j ] = inf ;
        for ( int  j = tube [ i ] [ 1 ] ; j <= m ; j ++ )
            dp [ i ] [ j ] = inf ;
        for ( int  j = 1 ; j <= m ; j ++ )
            if ( dp [ i ] [ j ] < inf ) {
                ok = false ;
                break ;
            }
        if ( ! ok ) {
            if ( tag [ i ] )
                ans ++ ;
        }
        else  break ; 
    }
    if ( ok )
        printf ( "0\n%d" , ans ) ;
    else {
        int  cnt = inf ;
        for ( int  i = 1 ; i <= m ; i ++ )
            cnt = std :: min ( cnt , dp [ n ] [ i ] ) ;
        printf ( "1\n%d" , cnt ) ;
    }
    return  0 ;
}