【POJ 3624】 + 01背包 + dp

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

很明显的01背包问题，AC代码 ：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int W[20011],D[20011],dp[20011];int main(){    int N,M,i,j;    while(scanf("%d %d",&N,&M)!=EOF)    {        memset(dp,0,sizeof(dp));        for(i = 1 ; i <= N ; i++)            scanf("%d%d",&W[i],&D[i]);        for(i = 1 ; i <= N ; i++)            for(j = M ; j >= W[i] ; j--)                 dp[j] = max(dp[j],dp[j - W[i]] + D[i]);        printf("%d\n",dp[M]);    }    return 0;}