Charm Bracelet (POJ 3624）
Charm Bracelet
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 40141 | Accepted: 17439 |
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 6 1 4 2 6 3 12 2 7
Sample Output
23
用二维动态肯定不行，这就需要优化成一维的了。
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 using namespace std; 5 const int N = 3410; 6 int w[N],d[N]; 7 int dp[12881]; 8 int main(){ 9 int n,m; 10 cin>>n>>m; 11 for(int i = 0; i < n; i ++)scanf("%d %d",&w[i],&d[i]); 12 for(int i = 0; i < n; i ++){ 13 for(int j = m; j >= w[i]; j--){ 14 dp[j] = max(dp[j],dp[j-w[i]]+d[i]); 15 } 16 } 17 printf("%d\n",dp[m]); 18 return 0; 19 }
