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
题意：有N 个手镯， 允许最大重量为M， 每个手镯都有两个属性， W 和 D， 分别为重量和魅力值， 求在允许重量范围内所能达到的最大魅力值。
代码：
#include<stdio.h>
#include<string.h>
#define max(a, b)(a > b ? a : b)
#define N 21000
int main(void)
{
int dp[N];
int i, j, n, m;
int w[N], d[N];
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--)//j要倒推才能保证在推dp[j]时， max里dp[j]和dp[j-w[i]]保存的是状态dp[i-1][j] 和dp[i-1][j-w[i]]的值。
{
dp[j] = max(dp[j], dp[j-w[i]] + d[i]); //在容量为j时，i件物品所能达到的最大价值。
}
}
printf("%d ", dp[m]);
}
return 0;
}