poj 3624 Charm Bracelet

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 61 42 63 122 7

Sample Output

23

【题意】有n个物品，每个物品都有自己的大小和自己的价值，现在你有一个总大小，问你在不超过这个大小的前提下最大价值是多少。

【分析】应该是最基本的0/1背包了，不解释。

【代码】

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>using namespace std;int ans[12885];int main(){    int num,sum_room,room,val;    memset(ans,0,sizeof(ans));    scanf("%d%d",&num,&sum_room);    while(num--)    {        scanf("%d%d",&room,&val);        for(int i=sum_room;i>=room;i--)            ans[i]=max(ans[i],ans[i-room]+val);    }    printf("%d\n",ans[sum_room]);    return 0;}