POJ 1276 Cash Machine 多重背包+二进制优化

来源：http://poj.org/problem?id=1276

题意：就是有一些一定数量和一定面值的钱，还有一个总钱数，让你在所给的纸币中选出一定数目的钱，使这些钱的和小于等于总钱数，并且使这个和尽可能的大。

思路：将总钱数看作是背包的体积，将钱币的面值看作是物体的体积，物体的价值和体积相等。即转化为多重背包问题，由于数据比较大，所以需要用到二进制优化。

代码：

#include <iostream>
#include <cstdio>
#include <string.h>
using namespace std;

#define CLR(arr,val) memset(arr,val,sizeof(arr))
struct cash{
	int num,val,weight;
}cc[15];
int totalweight,dp[100010];
int max(int a,int b){
	return a>b?a:b;
}
void full_bag(int x){
	int k = 1;
	while(k < cc[x].num){
		for(int j = k*cc[x].weight;j <= totalweight;++j){
		   dp[j] = max(dp[j],dp[j - k*cc[x].weight] + k*cc[x].val);
		}
		cc[x].num = cc[x].num - k;
		k = k*2;
	}
	for(int j = cc[x].num * cc[x].weight;j <= totalweight;++j)
		dp[j] = max(dp[j],dp[j - cc[x].num * cc[x].weight] + cc[x].num * cc[x].val);
}
void one_zero_bag(int x){
	int k = 1;
	while(k < cc[x].num){
		for(int j = totalweight;j >= k*cc[x].weight;--j)
			dp[j] = max(dp[j],dp[j - k*cc[x].weight] + k*cc[x].val);
		cc[x].num = cc[x].num - k;
		k = k*2;
	}
	for(int j = totalweight;j >= cc[x].num * cc[x].weight;--j)
		dp[j] = max(dp[j],dp[j - cc[x].num * cc[x].weight] + cc[x].num*cc[x].val);
}
int main(){
	//freopen("1.txt","r",stdin);
	while(scanf("%d",&totalweight) != EOF){
      int n;
	  scanf("%d",&n);
	  for(int i = 0;i < n;++i){
		  scanf("%d%d",&cc[i].num,&cc[i].val);
		  cc[i].weight = cc[i].val;
	  }
	  CLR(dp,0);
	  for(int i = 0;i < n;++i){
		  if(cc[i].num * cc[i].weight >= totalweight){
		    full_bag(i);
		  }
		  else
			  one_zero_bag(i);
	  }
	  printf("%d\n",dp[totalweight]);
	}
	return 0;
}