poj1787（记录路径+多重背包）（可以用完全背包来模拟）

地址：http://poj.org/problem?id=1787

Charlie's Change

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 2689		Accepted: 721

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. 

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.

题意：面值为1、5、10、25的四种硬币分别有c1、c2、c3、c4枚，问买一杯P元的咖啡最多用多少硬币。

思路：多重：因为只有四种硬币，所以用了十分暴力的记录路径的方法。

           完全：模拟出来的，详细见代码。

代码：

多重背包：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int dp[10010][5];  //多开一维用来记录路径
int main()
{
int v,c[5],i;
int cent[5]= {0,1,5,10,25};
while(scanf("%d%d%d%d%d",&v,&c[1],&c[2],&c[3],&c[4])>0)
{
if(!v&&!c[1]&&!c[2]&&!c[3]&&!c[4]) break;
memset(dp,0,sizeof(dp));
for(int s=1; s<5; s++)
{
if(c[s]*cent[s]>=v)
{
for(i=cent[s]; i<=v; i++)
{
if(!(i==cent[s]||dp[i-cent[s]][0])) continue;
int k=max(dp[i][0],dp[i-cent[s]][0]+1);
if(k>dp[i][0])
{
dp[i][0]=k;
for(int j=1;j<=s;j++)
{
if(j==s) dp[i][j]=dp[i-cent[s]][j]+1;
else dp[i][j]=dp[i-cent[s]][j];
}
}
}
}
else
{
int m=1;
while(c[s])
{
c[s]-=m;
for(i=v; i>=m*cent[s]; i--)
if(i==m*cent[s]||dp[i-m*cent[s]][0])
{
int k=max(dp[i-m*cent[s]][0]+m,dp[i][0]);
if(k>dp[i][0])
{
dp[i][0]=k;
for(int j=1;j<=s;j++)
{
if(j==s) dp[i][j]=dp[i-m*cent[s]][j]+m;
else dp[i][j]=dp[i-m*cent[s]][j];
}
}
}
m=m<<1;
if(m>c[s]) m=c[s];
}
}
}
if(dp[v][0]) printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n",dp[v][1],dp[v][2],dp[v][3],dp[v][4]);
else printf("Charlie cannot buy coffee.\n");
}
return 0;
}  //因为是求最多的硬币数，所以dp[i][0]表示总硬币数，当总硬币数没变大时，就不用更新方案。（与01背包思想有点相似，都是保留相同体积的最优方案）

完全背包:

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int dp[10010][2],used[10010];
int main()
{
int v,c[50],i,j;
int cent[5]= {0,1,5,10,25};
while(scanf("%d",&v)>0,v)
{
memset(dp,0,sizeof(dp));
for(i=1; i<=4; i++) scanf("%d",&c[i]);
for(i=1; i<=4; i++)
{
memset(used,0,sizeof(used));
for(j=cent[i]; j<=v; j++)
{
if(dp[j-cent[i]][0]+1>dp[j][0]&&(dp[j-cent[i]][0]||j==cent[i])&&used[j-cent[i]]+1<=c[i])
{
dp[j][0]=dp[j-cent[i]][0]+1;  //根据总硬币数来更新最优结果
dp[j][1]=j-cent[i];  //记录路径
used[j]=used[j-cent[i]]+1;  //判断硬币有没有用多
}
}
}
if(!dp[v][0]) printf("Charlie cannot buy coffee.\n");
else
{
c[1]=c[5]=c[10]=c[25]=0;
i=v;
while(i)
{
c[i-dp[i][1]]++;  //根据路径来得出结果
i=dp[i][1];
}
printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n",c[1],c[5],c[10],c[25]);
}
}
return 0;
}