poj 1787 Charlie's Change【多重背包可行性+记录路径】

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：
有面值为1,5,10,20的四种硬币，分别给出他们的数量问它是否能够组合出面值为P的价值，若果可以输出他的方案，如果存在多种方案输出使用硬币数最多的。（仔细读题）
思路：
这个就比较挑战dp思维了，二进制分解成01背包不可取，难在记录路径且硬币数最多。记录路径就新开一个二维数组，分别代表种类和钱数，更新dp时，还要更新已用的路径（好好体会，主要是各个状态下使用的次数），更新路径时就类比于更新dp数组（代表钱数，可以画个图看路径），这样就好理解了。

#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>using namespace std;int dp[10010], path[5][10010];int v[5] = {0, 1, 5, 10, 25}, c[5];//dp[i]: 为真表示能够组成i钱数//path[p][j]: 表示组成j钱数，p种硬币被用的次数int main() {    int n, p;    while(~scanf("%d %d %d %d %d",&n, &c[1], &c[2], &c[3], &c[4])) {           if(!n && !c[1] && !c[2] && !c[3] && !c[4])  break;        memset(path, 0, sizeof(path));        memset(dp, 0, sizeof(dp));        dp[0] = 1;        for(int i = 1;i <= 4; ++i) {            for(int j = v[i]; j <= n; ++j) {                if(dp[j - v[i]] && dp[j] < dp[j - v[i]] + 1 && path[i][j - v[i]] + 1 <= c[i]) {                    dp[j] = dp[j - v[i]] + 1; //由于输出使用硬币最多，即使存在了也要更新背包                     path[i][j] = path[i][j - v[i]] + 1;                      for(int p = 1; p < i; ++p) //更新其余已用的路径                         path[p][j] = path[p][j - v[i]];                }            }        }        if(dp[n] == 0)  puts("Charlie cannot buy coffee.");        else printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n", path[1][n], path[2][n], path[3][n], path[4][n]);    }    return 0;}