POJ 3624 Charm Bracelet 动态规划（01背包问题）

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15203		Accepted: 6950

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
USACO 2007 December Silver

//POJ3624 01背包问题
#include<stdio.h>
#include<string.h>
/*注意数组要开大，然后输入不需要判断EOF
Input：
4 6
1 4
2 6
3 12
2 7
动态规划表：
0  1  2  3  4  5  6
0 0  0  0  0  0  0  0
1 0  4  4  4  4  4  4
2 0  4  6 10 10 10 10
3 0  4  6 12 16 18 22
4 0  4  7 11 13 19 23
*/
int W[20000],D[20000],f[20000];
int main()
{
int N,M;
int i,j;
scanf("%d%d",&N,&M);
memset(W,0,sizeof(W));
memset(D,0,sizeof(D));
memset(f,0,sizeof(f));
for(i=1;i<=N;i++)
scanf("%d%d",&W[i],&D[i]);
for(i=0;i<=M;i++)  f[i]=0;

for(i=1;i<=N;i++)
for(j=M;j>=W[i];j--)
{
if(f[j]<(f[j-W[i]]+D[i])) f[j]=f[j-W[i]]+D[i];
}
printf("%d\n",f[M]);
return 0;
}