Bone Collector（01背包）

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

5 10 1 2 3 4 5 5 4 3 2 1 
       
    

Sample Output

14

代码如下：

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<deque>
using namespace std;
typedef long long ll;
#define MAX 0x3f3f3f3f
#define N 1005
struct bag
{
    int c;
    int val;
}bag;
int main()
{
    int t;
    struct bag a[1111];
    cin>>t;
    while(t--)
    {
    	int n,v;
        cin>>n>>v;
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i].val);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i].c);
		int f[1111][1111]={0};
        for(int i=1;i<=n;i++)
            for(int j=0;j<=v;j++)
                if(j-a[i].c>=0&&f[i-1][j]<f[i-1][j-a[i].c]+a[i].val)
                    f[i][j]=f[i-1][j-a[i].c]+a[i].val;
                else
                    f[i][j]=f[i-1][j];
        printf("%d\n",f[n][v]);
    }
    return 0;
}

附某大神代码：

#include<stdio.h>
#include<iostream>
#include<queue>
#include<cstring>
#include<cmath>
using namespace std;
#define MAXN 1005
int N,V;
int vol[MAXN],val[MAXN];
int dp[MAXN];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
memset(dp,0,sizeof(dp));
scanf("%d %d",&N,&V);
for(int i=0 ; i<N ; i++)
{
scanf("%d",&val[i]);
}
for(int i=0 ; i<N ; i++)
{
scanf("%d",&vol[i]);
}
for(int i=0 ; i<N ; i++)
{
for(int j=V ; j>=vol[i] ; j--)
{
dp[j] = max(dp[j],dp[j-vol[i]]+val[i]);
}
}
printf("%d\n",dp[V]);
}
return 0;
}

扫描二维码关注公众号，回复： 1480512 查看本文章

Bone Collector（01背包）

猜你喜欢