poj 3624 Charm Bracelet (01背包模板)

题目描述：
**

Charm Bracelet

**
Time Limit: 1000MS Memory Limit: 65536K

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

模板题
最基础的01背包，直接上代码。

#include <cstdio>#include <cstring>#define MAX 3410#define MM 12885int w[MAX];int c[MAX];int dp[MM];int max(int a, int b){    return a > b ? a : b;}int main(){    int N, M;    while(~scanf("%d %d", &N, &M))    {        memset(dp, 0, sizeof(dp));        for(int i = 0; i < N; i++)        {            scanf("%d %d", &w[i], &c[i]);        }        for(int i = 0; i < N; i++)        {            for(int j = M; j >= w[i]; j--)            {                dp[j] = max(dp[j - w[i]] + c[i], dp[j]);            }        }        printf("%d\n", dp[M]);    }    return 0;}

运行结果：