0826 openJ#7113 Charm Bracelet

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M_H5211

最新推荐文章于 2022-05-04 00:00:39 发布

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分类专栏： DP

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摘要

-01背包类型使用状态压缩为一维数组节省空间进行dp

原题目摘要

-Charm Bracelet

描述

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 iin 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.

输入

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

输出

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

样例输入

样例输出

来源

USACO 2007 December Silver

题目理解

-01背包模型不超过m的条件下的最大结果。

最开始直接使用二维数组进行递归设置状态dp[i][m]:不超过m的情况下的最大值，dp[i][m]=max(dp[i-1][m],dp[-1][m-w[i]]+d[i]); 可是m远远大于i造成极度的浪费后思考使用一维数组进行递推。

dp[j] 为i情况选的最大值；dp[j]>=w[i]的情况下可以再增加反之则只能是不选择i使用上一次的数据；

  for(int i=1;i<n;i++){
        for(int j=m;j>=w[i];j--){
            dp[j] =max(dp[j],dp[j-w[i]]+d[i]) ;
        }
    }

注意

- 因为更新数据需要用到上一轮前面的数据所以j的方向是从m到w[i];

日期

-2017 08 26

附加

代码

#include <algorithm>
#include <iostream>
#include <memory>
#include <cstdio>

using namespace std;

#define MAX 20004
int n;
int d[MAX],w[MAX];
int dp[MAX];
int m;

void solve(){
    //edge condition
    for(int j=0;j<=m;j++){
        if(j>=w[0]) dp[j]=d[0];
    }
    //i个之前的最大值
    for(int i=1;i<n;i++){
        for(int j=m;j>=w[i];j--){
            dp[j] =max(dp[j],dp[j-w[i]]+d[i]) ;
        }
    }
}

int main(){
    scanf("%d%d",&n,&m);
    memset(dp,0,sizeof(dp));

    for(int i=0;i<n;i++){
            scanf("%d%d",&w[i],&d[i]);
    }
    solve();
    printf("%d\n",dp[m]);
    return 0;
}