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 weightWi (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 inte
gers: 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
典型的动态规划问题————0-1背包
dp[i] 表示表示前i件物品放入一个容量为v的背包可以获得的最大价值
因此 状态转移方程可得：
dp[j] = dp[j] > (dp[j-d[i]] + w[i]) ? dp[j] : (dp[j-d[i]] + w[i]);
1 #include<stdio.h> 2 #include<string.h> 3 #define N 12888 4 5 int n, m, d[N],w[N], dp[N]; 6 int main() 7 { 8 int i, j; 9 scanf("%d %d", &n, &m); 10 for(i = 0; i < n; i++) 11 scanf("%d %d", &d[i], &w[i]); 12 13 memset(dp, 0, sizeof(dp)); 14 for(i = 0; i < n; i++) 15 for(j = m; j >= d[i]; j--) 16 dp[j] = dp[j] > (dp[j-d[i]] + w[i]) ? dp[j] : (dp[j-d[i]] + w[i]); 17 18 printf("%d ", dp[m]); 19 }