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POJ 3624
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

题意：在容量M下，能获得的最大Di，，Wi是各自对应的容量。

这题我之前用贪心做的，给的数据对了，但是WA，才发现这是01背包

#include <stdio.h>#include <string.h>int a[22222],b[22222],c[52222]={0};int _max(int a,int b){    return a>b?a:b;}int main(){    int i,j,k,n,m;    scanf("%d %d",&n,&m);    for(i=1; i<=n; i++)        scanf("%d %d",&a[i],&b[i]);        for(i=1;i<=n;i++)        for(j=m;j>=0;j--)        {            if(j>=a[i])            c[j]=_max(c[j],c[j-a[i]]+b[i]);        }    printf("%d\n",c[m]);    return 0;}