HDU2602 Bone Collector 01背包DP 模板题

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本文介绍了一个经典的背包问题变种——骨收集者问题，通过给定若干骨头的重量和价值，求解在限定体积内所能获得的最大总价值。文章详细阐述了使用动态规划算法解决该问题的方法，并提供了一个具体的实现案例。

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Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 46602 Accepted Submission(s): 19405

Problem Description

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?

Input

The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

Output

One integer per line representing the maximum of the total value (this number will be less than 2 ³¹).

Sample Input

Sample Output

Author

Teddy

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

给出若干物体的重量和价值，问在一定重量内能得的最大价值。复杂度O（V*W）。

int dp[1009][1009];//从第i个物品开始，挑选重量不大于j时的最大值

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <iomanip>
#include <time.h>
#include <set>
#include <map>
#include <stack>
using namespace std;
typedef long long LL;
const int INF=0x7fffffff;
const int MAX_N=10000;

int T,N,W;
int value[1009];
int weight[1009];
int dp[1009][1009];//从第i个物品开始，挑选重量不大于j时的最大值

void solve(){
    memset(dp,0,sizeof(dp));
    for(int i=N;i>=1;i--){
        for(int j=0;j<=W;j++){
            if(j<weight[i]){
                dp[i][j]=dp[i+1][j];
            }
            else{
                dp[i][j]=max(dp[i+1][j],dp[i+1][j-weight[i]]+value[i]);
            }
        }
    }
}
int main(){
    cin>>T;
    while(T--){
        cin>>N>>W;

        for(int i=1;i<=N;i++){
            scanf("%d",&value[i]);
        }
        for(int i=1;i<=N;i++){
            scanf("%d",&weight[i]);
        }

        solve();
        cout<<dp[1][W]<<endl;


    }
    return 0;
}