POJ 1784 Charlie's Change 记录路径的多重背包
Description
Charlie is a driver of Advanced Cargo Movement, Ltd. Charlie drives a lot and so he often buys coffee at coffee vending machines at motorests. Charlie hates change. That is basically the setup of your next task.
Your program will be given numbers and types of coins Charlie has and the coffee price. The coffee vending machines accept coins of values 1, 5, 10, and 25 cents. The program should output which coins Charlie has to use paying the coffee so that he uses as many coins as possible. Because Charlie really does not want any change back he wants to pay the price exactly.
Your program will be given numbers and types of coins Charlie has and the coffee price. The coffee vending machines accept coins of values 1, 5, 10, and 25 cents. The program should output which coins Charlie has to use paying the coffee so that he uses as many coins as possible. Because Charlie really does not want any change back he wants to pay the price exactly.
Input
Each line of the input contains five integer numbers separated by a single space describing one situation to solve. The first integer on the line P, 1 <= P <= 10 000, is the coffee price in cents. Next four integers, C1, C2, C3, C4, 0 <= Ci <= 10 000, are the numbers of cents, nickels (5 cents), dimes (10 cents), and quarters (25 cents) in Charlie's valet. The last line of the input contains five zeros and no output should be generated for it.
Output
For each situation, your program should output one line containing the string "Throw in T1 cents, T2 nickels, T3 dimes, and T4 quarters.", where T1, T2, T3, T4 are the numbers of coins of appropriate values Charlie should use to pay the coffee while using as many coins as possible. In the case Charlie does not possess enough change to pay the price of the coffee exactly, your program should output "Charlie cannot buy coffee.".
Sample Input
12 5 3 1 2 16 0 0 0 1 0 0 0 0 0
Sample Output
Throw in 2 cents, 2 nickels, 0 dimes, and 0 quarters. Charlie cannot buy coffee.
Source
CTU Open 2003
网上有不少题解了，但是都没有明确的注释。对于初学背包问题的人，本题还是较难理解的。在此我贴上自己的代码。本题讲了一种很巧妙地计入路径的方法，大家可以借鉴，要说的内容包含在注释中。
#include <iostream>
#include <sstream>
#include <ios>
#include <iomanip>
#include <functional>
#include <algorithm>
#include <vector>
#include <string>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#include <cctype>
using namespace std;
#define XINF INT_MAX
#define INF 0x3F3F3F3F
#define MP(X,Y) make_pair(X,Y)
#define PB(X) push_back(X)
#define REP(X,N) for(int X=0;X<N;X++)
#define REP2(X,L,R) for(int X=L;X<=R;X++)
#define DEP(X,R,L) for(int X=R;X>=L;X--)
#define CLR(A,X) memset(A,X,sizeof(A))
#define IT iterator
typedef long long ll;
typedef pair<int,int> PII;
typedef vector<PII> VII;
typedef vector<int> VI;
//const int MAXN = 10010;
//#define INF 0x3FFFFFFF
/*************************************
**************头文件*****************
*************************************/
int dp[10010];//dp[i]表示当费用为i时最多的硬币个数
int path[10010];//巧妙的记录路径
int used[10010];//某种货币已经使用的次数
int w[4] = {1,5,10,25};//4种货币
int ans[28];//答案，根据代码应该大于25
int main()
{
int P,a[5];
while(cin>>P)
{
REP(i,4)
cin>>a[i];
if(P==0&&a[0]==0&&a[1]==0&&a[2]==0&&a[3]==0)break;
//memset(dp,-INF,sizeof(dp));
for(int i = 0;i<=P;i++) dp[i] = -INF;//随意的设置为任何小于0的数，初始化
memset(path,0,sizeof(path));//初始化路径
dp[0] = 0;//当金额为0时，花费硬币个数为0
REP(i,4)
{
memset(used,0,sizeof(used));//每一次大循环都要更新来表示每一次钱币用到的个数；
//used[j]表示到那个货币为j时用得到的i种货币的数量
for(int j = w[i];j<=P;j++)//这一重循环在学背包问题的时候大家都已经接触过
{
if(dp[j]<dp[j-w[i]]+1&&dp[j-w[i]]>=0&&used[j-w[i]]<a[i])
{
//////要注意的是dp[j]的上一个状态是dp[j-w[i]]不是dp[i-1]
dp[j] = dp[j-w[i]]+1;
used[j] = used[j-w[i]]+1;//
path[j] = j-w[i];//具体的路径是怎么标记得最好做一遍单步调试，会更容易理解,近似于模拟指针
} //w[i] = j-path[j]这是导出下面ans的原理
}
}
memset(ans,0,sizeof(ans));//给答案初始化，实际上只用到ans[1],ans[5],ans[10],ans[25]
if(dp[P] <0)//如果dp[P]依旧小于零就说明上面遍历时没有用到，就不能表示出来
{
cout<<"Charlie cannot buy coffee."<<endl;
continue;
}
else //下面的东西只要模拟一边或者但不调试一下很容易明白
{
int i = P;
while(1)
{
if(i==0) break;
ans[i-path[i]]++;
i = path[i];
}
/*REP(i,P)
{
cout<<dp[i]<<" ";//大家可以把它加到代码里有助于理解
}
cout<<endl;
REP(i,P)
{
cout<<path[i]<<" ";
}*/
cout<<"Throw in "<<ans[1]<<" cents, "<<ans[5];
cout<<" nickels, "<<ans[10]<<" dimes, and ";
cout<<ans[25]<<" quarters."<<endl;
}
//0012341234567这是样例一中path的结果
}
return 0;
}
这篇文章的代码是根据对网上大神们的代码的理解写出来的。如有问题请留言。