HDU2602 Bone Collector 01背包DP 模板题

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 2³¹).

 

Sample Input

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

 

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;}