POJ3624 Charm Bracelet(典型01背包問題)

Time Limit: 1000MS          Memory Limit: 65536K          Total Submissions: 32897          Accepted: 14587

Description

Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight W_i (1 ≤ W_i ≤ 400), a 'desirability' factor D_i (1 ≤ D_i ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).

Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: W_i and D_i

Output

* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

Sample Input

4 6
1 4
2 6
3 12
2 7

Sample Output

23

一、題目大意

      有N個物品，分別有不同的重量Wi和價值Di，Bessie只能帶走重量不超過M的物品，要是總價值最大，並輸出總價值。

 

二、解題思路

     典型的動態規划題目，用一個數組記錄背包各個重量的最優解，不斷地更新直到窮盡所有可能性。

     狀態更新公式：state[weight] = max{state[weight-W[i]]+D[i], state[weight]}

 

三、具體代碼 

// 背包問題（動態規划）
#include <iostream> 
#include <cstdio>
#include <cstring>
#define MAXN 3402
#define MAXM 12880
using namespace std;

int main(){
    int N, M, W[MAXN+5], D[MAXN+5], dp[MAXM+5];
    while(scanf("%d%d", &N, &M) != EOF){
        for(int i=0; i<N; i++){
            scanf("%d%d", &W[i], &D[i]);
        }
        memset(dp, 0, sizeof(dp));
        for(int i=0; i<N; i++){
            for(int left_w=M; left_w>=W[i]; left_w--){
                dp[left_w] = max(dp[left_w-W[i]]+D[i], dp[left_w]);
            }
        }
        printf("%d\n", dp[M]);
    }
    
    return 0;
}