poj1276 多重揹包問題（二進位制解決方案）

life4711發表於2014-10-22

http://poj.org/problem?id=1276

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

Hint

The first data set designates a transaction where the amount of cash requested is @735. The machine contains 3 bill denominations: 4 bills of @125, 6 bills of @5, and 3 bills of @350. The machine can deliver the exact amount of requested cash. 

In the second case the bill supply of the machine does not fit the exact amount of cash requested. The maximum cash that can be delivered is @630. Notice that there can be several possibilities to combine the bills in the machine for matching the delivered cash. 

In the third case the machine is empty and no cash is delivered. In the fourth case the amount of cash requested is @0 and, therefore, the machine delivers no cash.
題目大意：給定揹包的容量V,和n中物品，每種物品的數量、重量、價值分別為num，w，v。求揹包可以獲得的最大價值。

解題思路：一道多重揹包的問題，一開始採用了以前用過的方法，結果超時了，趁機重讀《揹包九講》，豁然開朗，去年怎麼也看不懂的二進位制做法，而今看來是這麼的容易==詳見《揹包九講》

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=200100;
int v,n,T,c[maxn],dp[maxn];
int main()
{
    while(~scanf("%d",&v))
    {
        scanf("%d",&T);
        n=0;
        memset(c,0,sizeof(c));
        while(T--)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            int k=0;
            while(x-(1<<(k+1))+1>0)k++;
            for(int i=0;i<k;i++)
            {
                c[++n]=y*(1<<i);
            }
            c[++n]=y*(x+1-(1<<k));
        }
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            for(int j=v;j>=c[i];j--)
            {
                dp[j]=max(dp[j],dp[j-c[i]]+c[i]);
            }
        }
        printf("%d\n",dp[v]);
    }
    return 0;
}


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