poj 3624 Charm Bracelet

题目：

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

思路：

经典01背包问题…..
有n个物品,编号为i的物品的重量为w[i],价值为v[i],现在要从这些物品中选一些物品装到一个容量为m的背包中,使得背包内物体在总重量不超过m的前提下价值尽量大.

代码：

#include<iostream>#include<string.h>#include<stdio.h>#include<stdlib.h>#include<algorithm>using namespace std;int dp[12888];int Di[3407];int Wi[3407];int main(){    int N, M;    scanf("%d%d", &N, &M);        memset(dp,0,sizeof(dp));        memset(Di,0,sizeof(Di));        memset(Wi,0,sizeof(Wi));        for (int i = 0; i < N; i++){            scanf("%d%d",&Wi[i],&Di[i]);        }        for (int i = 0; i < N; i++){            for (int j = M; j>=Wi[i]; j--){                dp[j] = max(dp[j],dp[j-Wi[i]]+Di[i]);            }        }        printf("%d\n",dp[M]);    return 0;}