1. 问题描述

找零问题，在贪心算法讲过。但是贪心不一定能得出最优解。假设有几种不同币值的硬币v1，v2，.……vn（单位是元）。如果要支付w元，求最少需要多少个硬币。比如，有3种不同的硬币，1元、3元、5元，我们要支付9元，最少需要3个硬币（3个3元的硬币）。

2. 问题分析

2.1 回溯法求解

/*** @description: 找零钱，需要张数最少，回溯法* @author: michael ming* @date: 2019/7/20 22:50* @modified by:*/
#include <iostream>
#define N 3
int rmb[N] = {1,9,10};//钞票面额
int amount[N];
int minAmount[N];
using namespace std;
void exchange(const int &targetMoney, int curMoney, int &minPiece, int piece)
{if(curMoney > targetMoney)//超过目标，返回return;if(curMoney == targetMoney)//达到目标金额{if(piece < minPiece){minPiece = piece;//更新最小张数for(int i = 0; i < N; ++i)minAmount[i] = amount[i];//获取每张钞票的张数}return;}for(int i = 0; i < N; ++i){//递归调用,拿取每张面额的钞票amount[i]++;exchange(targetMoney,curMoney+rmb[i],minPiece,piece+1);amount[i]--;//恢复上次的状态}
}
int main()
{int minPiece = 65535, piece = 0,targetMoney = 18, curMoney = 0;exchange(targetMoney,curMoney,minPiece,piece);cout << "凑成" << targetMoney << "元，最少需要：" << minPiece << "张(枚)。" << endl;int i = 0;while(i < N){if(minAmount[i] != 0)cout << minAmount[i] << "个" << rmb[i] << " ";i++;}cout << endl;cout << "----------------------" << endl;
}

2.2 DP状态转移方程法

由于上面的钞票面额可能不止3种，递归树是多叉树，所以状态转移表法画起回溯的递归图比较麻烦，我们采用状态转移方程法。

状态转移方程如下：

minPiece(targetMoney) = 1 + min{minPiece(targetMoney-rmb[0]), ... , minPiece(targetMoney-rmb[N-1])}

targetMoney = 18;//目标金额
rmb[N] = {1,9,10};//钞票面额
对于题目的情况，代入具体数值，状态转移方程如下

minPiece(18) = 1 + min{minPiece(18-1), minPiece(18-9) , minPiece(18-10)}= 1 + min{minPiece(17),minPiece(9),minPiece(8)}

DP（递归+备忘录）代码如下：

/*** @description: 找零钱，需要张数最少* @author: michael ming* @date: 2019/7/20 18:35* @modified by: */
#include <iostream>
#include <algorithm>
#include <memory.h>#define N 3
const int targetMoney = 18;//目标金额
int rmb[N] = {1,9,10};//钞票面额
int mem[targetMoney+1];//备忘录，存放最小张数
using namespace std;
int minP(int Money)
{if(Money < 0)//超过目标，返回很大的张数，表示不可能凑成return 65535;if(Money == 0)//达到目标金额return 0;if(mem[Money] > 0)//计算过了，直接读取备忘录return mem[Money];int minAmount[N];memset(minAmount,65535,N*sizeof(int));for(int i = 0; i < N; ++i){//递归调用,拿取每张面额的钞票minAmount[i] = minP(Money-rmb[i]);}sort(minAmount,minAmount+N);mem[Money] = minAmount[0]+1;//记录最小的张数return mem[Money];
}
int main()
{cout << "凑成" << targetMoney << "元，最少需要："<< minP(targetMoney) << "张(枚)。" << endl;//如何打印出选取钞票的面额和张数？？？
}

2.3 DP状态转移表法

/*** @description: 找零钱，需要张数最少,dp状态表法* @author: michael ming* @date: 2019/7/21 20:01* @modified by: */
#include <iostream>
#include <algorithm>
#include <memory.h>#define N 3
const int targetMoney = 18;//目标金额
int rmb[N] = {1,9,10};//钞票面额,从小到大
using namespace std;
void exchange(int Money)
{int maxPiece = targetMoney/rmb[0];//最大张数int i, j, k;int (*states)[targetMoney+1] = new int [maxPiece][targetMoney+1];//memset(states,65535,maxPiece*(targetMoney+1)*sizeof(int));//上面错误！！！memset一般只付0或极大值for(i = 0; i < maxPiece; ++i)for(j = 0; j <= targetMoney; ++j)states[i][j] = 65535;//初始化for(k = 0, j = 0; j <= targetMoney; ++j){if(k < N && j == rmb[k]){//初始化第一行数据states[0][j] = 1;//一张rmbk++;}}for(i = 1; i < maxPiece; ++i)//动态规划{for(j = 0; j <= targetMoney; ++j)//上面一行的数据考下来states[i][j] = states[i-1][j];for(j = 0; j <= targetMoney; ++j){if(states[i-1][j] != 65535){for(k = 0; k < N; ++k){if(j+rmb[k] <= targetMoney && states[i-1][j+rmb[k]] > states[i-1][j]+1)states[i][j+rmb[k]] = states[i-1][j]+1;}}}}cout << "凑成" << targetMoney << "元，最少需要："<< states[maxPiece-1][targetMoney] << "张(枚)。" << endl;//------------打印选择的信息---------------------------for(i = maxPiece-1; i >= 1 && states[i][targetMoney] == states[i-1][targetMoney]; --i);//此时i等于最早出现的答案处的行for(j = targetMoney; j > 0; ){if(i != 0){for(k = 0; k < N; ++k){if(states[i-1][j-rmb[k]] == states[i][j]-1){cout << "1张" << rmb[k] << " ";j = j-rmb[k];i--;break;}}}else{cout << "1张" << j << " ";break;}}delete [] states;//释放资源
}
int main()
{exchange(targetMoney);return 0;
}