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 7Sample Output
23 題意:經典0—1背包問題,有n個物品,編號為i的物品的重量為w[i],價值為v[i],現在要從這些物品中選一些物品裝到一個容量為m的背包中,使得背包內物體在總重量不超過m的前提下價值盡量大. #include <stdio.h>
#include <string.h>int max(int a ,int b)
{
return a>b?a:b;
}int main()
{
int n,m,w[40000],d[40000],i,j;
int dp[40000];
while(~scanf("%d%d",&n,&m))
{
memset(dp,0,sizeof(dp));
for(i = 0;i<n;i++) scanf("%d%d",&w[i],&d[i]);
for(i = 0;i<n;i++)
{
for(j = m;j>=w[i];j--)
{
dp[j] = max(dp[j],dp[j-w[i]]+d[i]);
}
}
printf("%d\n",dp[m]);
} return 0;
}
