POJ - 3624 《Charm Bracelet》 【0-1背包】

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

【实现代码】

#include<cstdio>
#include<cstring>
int W[3404];
int D[3404];
int dp[12881];
int max(int a,int b){return a>b?a:b;}
int main()
{
/*
char STD[102400]="";
if(freopen("in.txt","r",stdin)==NULL)freopen("CON","r",stdin);
else
{
fread(STD,1,1024,stdin);
printf("===input===\n%s\n===output===\n",STD);
freopen("in.txt","r",stdin);
}
*/
int N,M;
scanf("%d%d",&N,&M);
for(int i=0;i<N;i++)scanf("%d%d",&W[i],&D[i]);
memset(dp,0,sizeof(dp));
for(int i=0;i<N;i++)
for(int j=M;j>=W[i];j--) dp[j]=max(dp[j],dp[j-W[i]]+D[i]);
printf("%d",dp[M]);
}