POJ 3624 Charm Bracelet 动态规划（01背包问题）

                                                                                                                                    Charm Bracelet

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 15203 Accepted: 6950

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 theN (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weightW_i (1 ≤ W_i ≤ 400), a 'desirability' factorD_i (1 ≤ D_i ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more thanM (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 61 42 63 122 7

Sample Output

23

Source

USACO 2007 December Silver

 

//POJ3624 01背包问题 #include<stdio.h>#include<string.h>/*注意数组要开大，然后输入不需要判断EOF Input： 4 61 42 63 122 7动态规划表：  0  1  2  3  4  5  60 0  0  0  0  0  0  01 0  4  4  4  4  4  42 0  4  6 10 10 10 103 0  4  6 12 16 18 22 4 0  4  7 11 13 19 23*/int W[20000],D[20000],f[20000];int main(){   int N,M;   int i,j;   scanf("%d%d",&N,&M);    memset(W,0,sizeof(W));     memset(D,0,sizeof(D));     memset(f,0,sizeof(f));     for(i=1;i<=N;i++)         scanf("%d%d",&W[i],&D[i]);     for(i=0;i<=M;i++)  f[i]=0;             for(i=1;i<=N;i++)       for(j=M;j>=W[i];j--)       {         if(f[j]<(f[j-W[i]]+D[i])) f[j]=f[j-W[i]]+D[i];        }               printf("%d\n",f[M]);   return 0; }