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POJ 3624 Charm Bracelet

时间:2014-07-25 19:06:52      阅读:339      评论:0      收藏:0      [点我收藏+]

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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\n", dp[m]);
19 }

 



POJ 3624 Charm Bracelet,布布扣,bubuko.com

POJ 3624 Charm Bracelet

标签:des   style   blog   color   os   io   for   re   

原文地址:http://www.cnblogs.com/zzy9669/p/3868141.html

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