POJ 3624 0-1背包问题

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 45540		Accepted: 19429

Description

Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight W_i (1 ≤ W_i ≤ 400), a 'desirability' factor D_i (1 ≤ D_i ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).

Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: W_i and D_i

Output

* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

Sample Input

Sample Output

Source

USACO 2007 December Silver

题目意思：有N件物品放入容积为M的背包，每个物品所占的体系为W[i],价值为D[i],使得价值最大

动态规划，用F[i][j]表示放入前i个商品，使得容积为j的价值最大的最大价值

边界条件为

if w[1]<j

F[1][j]=d[1];

else F[1][j]=0;

F[i][j]=max[F[i-1][j],F[i-1][j-w[i]]+D[i];

没有优化的二维代码如下

#include <stdio.h>
#include <string.h>
#include <string>
#include <iostream>
using namespace std;
#define max(a,b) a>b?a:b;
//数组要设的比给的范围稍大一些
int dp[3410][12900];
int path[3410][12900];
int w[3410];
int d[3410];
int totalN;
int totalW;
int main()
{
    int i,j;
    scanf("%d",&totalN);
    scanf("%d",&totalW);

    for(i=1;i<=totalN;i++)
    {
        scanf("%d",&w[i]);
        scanf("%d",&d[i]);
    }

    memset(dp,0,sizeof(dp));
    for(i=1;i<=totalN;i++){
        for(j=1;j<=totalW;j++){
            dp[i][j] = dp[i-1][j];
            if(j>=w[i] && dp[i][j]<dp[i-1][j-w[i]]+d[i]){
                dp[i][j] =dp[i-1][j-w[i]]+d[i];
                //path[i][j] = 1;
            }
        }
    }
    printf("%d\n",dp[totalN][totalW]);
    return 0;
}

优化逆序计算容积j，还可以用一维数组优化

POJ 3624 0-1背包问题

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