Sicily 1782. Knapsack

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背包问题中最简单的01背包问题，属于动态规划问题中的一种，相关状态转移方程是：capacity[i]=max(capacity[i],capacity[i-item[i]]+item[i]),理解为在i容量的背包中可以容纳的最大量为：不放入当前物品（即在i容量背包中的最大量）和放入当前所占容量为item[i]的物品（即在i-item[i]容量背包中的最大量加上当前物品的所占容量item[i]），这两者中的最大值，循环迭代后capacity[m]就是所求。时间耗时是0.16s,应该有更好的算法，目前未知。

Run Time: 0.16sec

Run Memory: 304KB

Code length: 648Bytes

SubmitTime: 2012-02-04 12:25:26

// Problem#: 1782// Submission#: 1205523// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include <cstdio>using namespace std;int max( int a, int b ) { return ( a > b ? a: b ); }int main(){    int T, n, m;    int item[ 1001 ], capacity[ 10001 ];    int i, j;    scanf( "%d", &T );    while ( T-- ) {        scanf( "%d%d", &n, &m );        for ( i = 1; i <= n; i++ )            scanf( "%d", &item[ i ] );        for ( i = 0; i <= m; i++ )            capacity[ i ] = 0;        for ( i = 1; i <= n; i++ ) {            for ( j = m; j >= item[ i ]; j-- )                capacity[ j ] = max( capacity[ j ], capacity[ j - item[ i ] ] + item[ i ] );        }        printf( "%d\n", capacity[ m ] );    }    return 0;}

在评论中见到alipu07的回复，研究了一下其贴出的代码，发现了更高效率的算法。基本思路就是用布尔类型的capacity[i]表示能否容纳进i重量的物品，类似于上面的方法，但在处理方法上更为高效，因为简略了capacity[i]=max(capacity[i],capacity[i-item[i]]+item[i])中的计算和比较，取而代之为直接的放入物品。

Run Time: 0.05sec

Run Memory: 288KB

Code length: 751Bytes

SubmitTime: 2013-03-01 21:53:50

// Problem#: 1782// Submission#: 1932564// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include <cstdio>using namespace std;int main(){    int T, n, m;    int item[ 1000 ];    bool capacity[ 10000 ];    int i, j;    scanf( "%d", &T );    while ( T-- ) {        scanf( "%d%d", &n, &m );        for ( i = 0; i < n; i++ )            scanf( "%d", &item[ i ] );        for ( i = 0; i <= m; i++ )            capacity[ i ] = false;        for ( i = 0; i < n; i++ ) {            for ( j = m; j > item[ i ]; j-- ) {                if ( capacity[ j - item[ i ] ] )                    capacity[ j ] = true;            }            capacity[ item[ i ] ] = true;        }        for ( i = m; i > 0; i-- ) {            if ( capacity[ i ] )                break;        }        printf( "%d\n", i );    }    return 0;}