HDU 2602: Bone Collector ← 0-1背包问题

【题目来源】
http://acm.hdu.edu.cn/showproblem.php?pid=2602

【题目描述】
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 ?

【数据输入】
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.

【数据输出】
One integer per line representing the maximum of the total value (this number will be less than 2^31).

【算法代码】

#include <bits/stdc++.h>
using namespace std;
 
const int maxn=1005;
int c[maxn][maxn];
int val[maxn],vol[maxn];
 
int main(){
    int t,N,V;
    cin>>t;
    while(t--){
		cin>>N>>V;
    	for(int i=1;i<=N;i++)
    		cin>>val[i];
    	for(int i=1;i<=N;i++)
    		cin>>vol[i];
    	for(int i=0;i<=N;i++)
    		c[i][0]=0;
    	for(int j=0;j<=V;j++)
    		c[0][j]=0;
	    for(int i=1;i<=N;i++)
	        for(int j=0;j<=V;j++)
	            if(j<vol[i]) c[i][j]=c[i-1][j];
	            else c[i][j]=max(c[i-1][j],c[i-1][j-vol[i]]+val[i]);
    	cout<<c[N][V]<<endl;
	}
    return 0;
}
 
 
 
/*
in:
1
5 10
1 2 3 4 5
5 4 3 2 1

out:
14
*/

【参考文献】
https://blog.csdn.net/hnjzsyjyj/article/details/126067759
https://blog.csdn.net/hnjzsyjyj/article/details/125944790