来源:https://www.cnblogs.com/jinglecjy/p/5674796.html

题目链接:http://bailian.openjudge.cn/practice/4131/

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 Wi (1 ≤ Wi ≤ 400), a 'desirability' factor Di (1 ≤ Di ≤ 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: Wi and Di

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、题目大意

有N个物品,分别有不同的重量Wi和价值Di,Bessie只能带走重量不超过M的物品,要是总价值最大,并输出总价值。

二、解题思路

典型的动态规划题目,用一个数组记录背包各个重量的最优解,不断地更新直到穷尽所有可能性。

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

 // 背包问题(动态规划)
#include <iostream>
#include <cstdio>
#include <cstring>
#define MAXN 3402
#define MAXM 12880
using namespace std; int main(){
int N, M, W[MAXN+], D[MAXN+], dp[MAXM+];
while(scanf("%d%d", &N, &M) != EOF){
for(int i=; i<N; i++){
scanf("%d%d", &W[i], &D[i]);
}
memset(dp, , sizeof(dp));
for(int i=; 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 ;
}

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