Bone Collector(01背包+记忆化搜索)

Bone Collector

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 21 Accepted Submission(s) : 6

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[sup]31[/sup]).

Sample Input

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

Sample Output

题解：也就01背包；

代码：

 1 #include<stdio.h>
 2 #include<string.h>
 3 #define max(a,b) (a>b?a:b)
 4 int f[2000];
 5 int val[2000];
 6 int cos[2000];
 7 int main()
 8 {
 9     int T;
10     scanf("%d",&T);
11     while(T--)
12     {
13         memset(f,0,sizeof(f));
14         memset(val,0,sizeof(val));
15         memset(cos,0,sizeof(cos));
16         int n,v;
17         scanf("%d%d",&n,&v);
18         int i,j;
19         for(i=1;i<=n;i++)
20           scanf("%d",&val[i]);
21         for(j=1;j<=n;j++)
22           scanf("%d",&cos[j]);
23         for(i=1;i<=n;i++)
24         {
25             for(j=v;j>=cos[i];j--)
26             {
27             //    f[j]=f[j-1];
28              // if(j>=cos[i])
29                 f[j]=max(f[j],f[j-cos[i]]+val[i]);
30             } 
31         }
32     printf("%d
",f[v]);    
33     }
34     return 0;
35 }

记忆化搜索；

ac代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define SD(x) scanf("%lf",&x)
#define P_ printf(" ")
typedef long long LL;
const int MAXN=1010;
int dp[MAXN][MAXN],w[MAXN],p[MAXN];
int dfs(int i,int v){
    if(dp[i][v])return dp[i][v];
    if(i==0||v<0)return 0;// 
    if(w[i]>v)dp[i][v]=dfs(i-1,v);
    else dp[i][v]=max(dfs(i-1,v),dfs(i-1,v-w[i])+p[i]);//
    return dp[i][v];
}
int main(){
    int T,N,M;
    SI(T);
    while(T--){
        SI(N);SI(M);
        for(int i=1;i<=N;i++)SI(p[i]);
        for(int i=1;i<=N;i++)SI(w[i]);
        mem(dp,0);
        printf("%d
",dfs(N,M));
    }
    return 0;
}