HDU 2602 01揹包

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 88316    Accepted Submission(s): 36385

 

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

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

 

 

Sample Output

14

題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=2602

簡單01揹包

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 1000 + 10;

int val[maxn], vol[maxn], dp[maxn][maxn];

int main (){
    int T, N, V;
    scanf("%d",&T);
    while (T--){
        scanf("%d%d",&N, &V);
        memset(dp,0,sizeof(dp));
        for (int i = 1; i <= N; i++){
            scanf("%d",&val[i]);
        }
        for (int i = 1; i <= N; i++){
            scanf("%d",&vol[i]);
        }
        for (int i = 1; i <= N; i++){
            for (int j = 0; j <= V; j++){
                if (j < vol[i]){
                    dp[i][j] = dp[i - 1][j] ;
                }
                else dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - vol[i]] + val[i]);
            }
        }
//        for (int i = 1; i <= N; i++){
//            for (int j = 1; j <= V; j++){
//                printf("%d ",dp[i][j]);
//            }
//            puts("");
//        }
        printf("%d\n",dp[N][V]);
    }
    return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std ;
const int Maxn = 1000 + 10 ;

int dp[Maxn] ;
int val[Maxn] ;
int vol[Maxn] ;

int main (){
    int T;
    int n, m ;
    cin >> T ;
    while (T--){
        cin >> n >> m ;
        for (int i = 1; i <= n; i++) cin >> val[i] ;
        for (int i = 1; i <= n; i++) cin >> vol[i] ;
        memset(dp, 0, sizeof(dp)) ;
        for (int i = 1; i <= n; i++){
            for (int j = m; j >= vol[i]; j--){
                dp[j] = max(dp[j], dp[j - vol[i]] + val[i]) ;
            }
        }
        cout << dp[m] << endl ;
    }
    return 0 ;
}

第一種是二維的寫法，第二種對空間上進行了優化！