多重背包练习-计数法-POJ-1276-Cash Machine

Cash Machine

Time Limit:1000MS		Memory Limit:10000K
Total Submissions:22644		Accepted:7943

Description

A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The machine uses exactly N distinct bill denominations, say Dk, k=1,N, and for each denomination Dk the machine has a supply of nk bills. For example,

N=3, n1=10, D1=100, n2=4, D2=50, n3=5, D3=10

means the machine has a supply of 10 bills of @100 each, 4 bills of @50 each, and 5 bills of @10 each.

Call cash the requested amount of cash the machine should deliver and write a program that computes the maximum amount of cash less than or equal to cash that can be effectively delivered according to the available bill supply of the machine.

Notes:
@ is the symbol of the currency delivered by the machine. For instance, @ may stand for dollar, euro, pound etc.

Input

The program input is from standard input. Each data set in the input stands for a particular transaction and has the format:

cash N n1 D1 n2 D2 ... nN DN

where 0 <= cash <= 100000 is the amount of cash requested, 0 <=N <= 10 is the number of bill denominations and 0 <= nk <= 1000 is the number of available bills for the Dk denomination, 1 <= Dk <= 1000, k=1,N. White spaces can occur freely between the numbers in the input. The input data are correct.

Output

For each set of data the program prints the result to the standard output on a separate line as shown in the examples below.

Sample Input

735 3  4 125  6 5  3 350
633 4  500 30  6 100  1 5  0 1
735 0
0 3  10 100  10 50  10 10

Sample Output

735
630
0
0

解题报告：多重背包题，为体积与价值相等的特殊情况

1、可以采有计数的方法代替单调队列将时间复杂度降到o(vn)

注：计数的这种方法解决多重背包时有限制，只能解决体积与价值相等的情况

#include <iostream>
#include <string.h>
using namespace std;
int main() {
	int cash,n,v[11],nums[11];
	int opt[10001];
	int cnt[10001];
	while(cin >> cash){
		cin >> n;
		for(int i=0; i<n; i++)
			cin  >> nums[i]>> v[i] ;
		for(int i=0; i<=cash; i++)
			opt[i] = 0;
		for(int i=0; i<n; i++){
			memset(cnt, 0 ,sizeof(int)*(cash+1)); //计数器清0
			for(int j=v[i];  j<=cash ; j++){
				if(opt[j] < opt[j-v[i]]+v[i]  && cnt[ j-v[i] ] < nums[i]){
					cnt[j] = cnt[ j-v[i] ]+ 1; //计数器加1
					opt[j] =  opt[j-v[i]]+v[i];
				}
			}
		}
		cout << opt[cash] << endl;
	}
	return 0;
}