标签：des style blog http io ar color os sp
http://poj.org/problem?id=3624
一维数组的01背包，v一定要逆序枚举，不然会多次放同一种物品。
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 24456 | Accepted: 11031 |
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
#include<iostream> #include<cstring> #include<cstdio> using namespace std; int c,n; int v[3412],w[3412],f[12888]; void backpack() { memset(f,0,sizeof(f)); int i,j; for(i=1;i<=n;i++) for(j=c;j>=0;j--) if(j>=v[i]) f[j]=max(f[j],f[j-v[i]]+w[i]); printf("%d\n",f[c]); } int main() { int i; scanf("%d%d",&n,&c); for(i=1;i<=n;i++) scanf("%d%d",&v[i],&w[i]); backpack(); return 0; }
标签：des style blog http io ar color os sp
原文地址：http://www.cnblogs.com/cancangood/p/4165930.html
