| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 29295 | Accepted: 13143 |
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 weight Wi (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
Source
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int max(int a,int b)
{
return a>b?a:b;
}
int main()
{
int n,m,i,j;
int a[],b[];
while(~scanf("%d%d",&n,&m))
{
for(i=; i<n; i++)
scanf("%d%d",&a[i],&b[i]);
int dp[]; //dp数组的大小wa了一次，要注意必须和最大重量相同，而不是最大珠子个数
memset(dp,,sizeof(dp));
for(i=; i<n; i++)
for(j=m; j>=a[i]; j--)
dp[j]=max(dp[j],dp[j-a[i]]+b[i]); //比较加上这颗珠子和不加的价值谁更大，记录大的那个
printf("%d\n",dp[m]);
}
return ;
}