Bone Collector(01背包模板)

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Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 66915 Accepted Submission(s): 27941

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

一个袋子，容量为V,N个骨头，各自的体积价值不同；所能获得的最大价值；

#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>using namespace std;int main(){int T;cin >> T;while(T--){int N, V;cin >> N >> V;int v[1005], w[1005];for(int i=1; i<=N; i++){cin >> w[i];}for(int i=1; i<=N; i++){cin >> v[i];}int dp[1005];memset(dp, 0, sizeof(dp));for(int i=1; i<=N; i++){for(int j=V; j>=v[i]; j--)dp[j]=max(dp[j-v[i]]+w[i], dp[j]);}cout << dp[V] << endl;}return 0;
}

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