POJ 3624

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

01背包

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;#define maxn 15000int dp[maxn], w[maxn], d[maxn];int main(){    int n, m, i, j;    scanf("%d %d",&n, &m);    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]);}