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POJ3624 Charm Bracelet 【01背包】
阅读量:5876 次
发布时间:2019-06-19

本文共 2557 字,大约阅读时间需要 8 分钟。

Charm Bracelet
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 22621   Accepted: 10157

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 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 61 42 63 122 7

Sample Output

23

题意:给定物品数量n和背包容量m,n个物品的重量weight和价值val,求能获得的最大价值。

题解:状态转移方程:dp[i][j] = max(dp[i-1][j], dp[i-1][j-w[i]] + v[i]);当中状态dp[i][j]表示前i个物品放在容量为j的背包中能获得的最大价值。

二维数组能够压缩成一维以节省空间,可是内层循环须要倒序。

原始版本号:耗时360ms

#include 
#define maxn 12882int dp[maxn];int max(int a, int b){ return a > b ? a : b; }int main(){ int n, totalWeight, i, j, weight, val; scanf("%d%d", &n, &totalWeight); for(i = 1; i <= n; ++i){ scanf("%d%d", &weight, &val); for(j = totalWeight; j; --j){ if(j >= weight) dp[j] = max(dp[j], dp[j - weight] + val); } } printf("%d\n", dp[totalWeight]); return 0;}

优化后的代码:耗时219ms

#include 
#define maxn 12882int dp[maxn];int main(){ int n, totalWeight, i, j, weight, val; scanf("%d%d", &n, &totalWeight); for(i = 1; i <= n; ++i){ scanf("%d%d", &weight, &val); for(j = totalWeight; j; --j){ if(j >= weight && dp[j - weight] + val > dp[j]) dp[j] = dp[j - weight] + val; } } printf("%d\n", dp[totalWeight]); return 0;}

用二维dp数组写了一个,果断的超了内存,占用内存大概12882*3404*4/1024=171兆。题目限制是65兆

TLE:

#include 
#define maxn 12882int dp[3404][maxn];int max(int a, int b){ return a > b ? a : b; }int main(){ int n, m, weight, val, i, j; scanf("%d%d", &n, &m); for(i = 1; i <= n; ++i){ scanf("%d%d", &weight, &val); for(j = 1; j <= m; ++j) if(j >= weight) dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight] + val); else dp[i][j] = dp[i-1][j]; } printf("%d\n", dp[n][m]); return 0;}

转载地址:http://jlzix.baihongyu.com/

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