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POJ 3624 Charm Bracelet

2024-07-14 00:13:05   214人阅读

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 61 42 63 122 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\n", dp[m]);19 }

 


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