C
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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
背包问题
核心是在双循环中比较加或不加第i件物品的价值
要注意c[]的下标是重量,要开的大一点
//背包问题 #include<stdio.h>int max(int x,int y){ return x>y?x:y;}int main(){ int n,m,i,j; int w[4000],d[4000]; //w[]存重量,d[]存价值 while(scanf("%d%d",&n,&m)!=EOF) { int c[20000]={0}; //赋零值用于比较,下标是重量 for(i=1;i<=n;i++) scanf("%d%d",&w[i],&d[i]); //重点预警 for(i=1;i<=n;i++) //n件物品循环判断 for(j=m;j>=w[i];j--) //循环重量,从0到m,不断比较加/不加第i件物品的价值,c[j]取较大值 c[j]=max(c[j],c[j-w[i]]+d[i]); printf("%d\n",c[m]); } return 0;}
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