poj 3624 (背包入门)
入门题,最近又一次想不起来背包的dp方法,无奈又得找道水题找找感觉...
Charm Bracelet
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 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 |
#include <stdio.h> #include <string.h> #include <iostream> using namespace std; #define N 13000 int dp[N]; int main() { int n,m; memset(dp,0,sizeof(dp)); scanf("%d%d",&n,&m); for(int i=0;i<n;i++) { int tmp,w; scanf("%d%d",&w,&tmp); for(int j=m;j>=w;j--) { dp[j]=max(dp[j],dp[j-w]+tmp); } } printf("%d\n",dp[m]); return 0; }