POJ 3624 Charm Bracelet

Charm Bracelet

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 31436 Accepted: 13982

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

题意：经典0—1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.

#include <stdio.h>#include <string.h>int dp[13000];int main(){int N, M, W, D, i, j;while(scanf("%d%d", &N, &M) != EOF){memset(dp, 0, sizeof(dp));for(i = 1; i <= N; i++){scanf("%d%d", &W, &D);for(j = M; j >= W; j--)dp[j] = dp[j] > dp[j - W] + D ? dp[j] : dp[j - W] + D;}printf("%d\n", dp[M]);}return 0;}