算法设计与分析: 3-2 最少硬币问题

3-2 最少硬币问题

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问题描述

设有n种不同面值的硬币，各硬币的面值存于数组T[1:n]中。现要用这些面值的硬币来找钱。可以使用的各种面值的硬币个数存于数组Coins[1:n]中。对任意钱数0≤m≤20001，设计一个用最少硬币找钱m的方法。
对于给定的1≤n≤10，硬币面值数组T和可以使用的各种面值的硬币个数数组Coins，以及钱数m，0≤m≤20001，编程计算找钱m的最少硬币数。

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深度优先搜索遍历法（DFS）

Java

public class ZuiShaoYingBi {
    //case 1
    private static int n = 3;
    private static int[] T = {1, 2, 5};
    private static int[] coins = {3, 3, 3};

    //case 2
//    private static int n = 3;
//    private static int[] T = {19, 20, 50};
//    private static int[] coins = {3, 3, 3};

    //case 3
//    private static int n = 4;
//    private static int[] T = {1, 2, 5, 18};
//    private static int[] coins = {3, 3, 3, 1};

    //case 4
//    private static int n = 4;
//    private static int[] T = {1, 2, 5, 18};
//    private static int[] coins = {3, 3, 3, 0};

    private static int m = 18;
//    private static int m = 19;
//    private static int m = 20;
//    private static int m = 21;
//    private static int m = 22;
//    private static int m = 23;
//    private static int m = 24;
//    private static int m = 25;
//    private static int m = 0;
//    private static int m = 1;
    private static int coinsCount = Integer.MAX_VALUE;
    private static int tempCount = 0;

    public static void main(String[] args) {

        dfs(0);

        if(coinsCount == Integer.MAX_VALUE || coinsCount == 0)
            coinsCount = -1;

        System.out.println(coinsCount);
    }

    private static void dfs(int charge){
        if(charge == m && tempCount < coinsCount){
            coinsCount = tempCount;
        }else if(charge > m){
            return;
        }else{
            for(int i=0; i<n; i++){
                if(coins[i] > 0){
                    coins[i]--;
                    tempCount++;
                    dfs(charge+T[i]);
                    tempCount--;
                    coins[i]++;
                }
            }
        }
    }
}

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动态规划法

Java: version 1

public class ZuiShaoYingBi2 {
    //case 1
    private static int n = 3;
    private static int[] T = {0, 1, 2, 5};
    private static int[] coins = {0, 3, 3, 3};

    //case 2
//    private static int n = 3;
//    private static int[] T = {0, 19, 20, 50};
//    private static int[] coins = {0, 3, 3, 3};

    //case 3
//    private static int n = 4;
//    private static int[] T = {0, 1, 2, 5, 18};
//    private static int[] coins = {0, 3, 3, 3, 1};

    //case 4
//    private static int n = 4;
//    private static int[] T = {0, 1, 2, 5, 18};
//    private static int[] coins = {0, 3, 3, 3, 0};

    private static int m = 18;
    //    private static int m = 19;
//    private static int m = 20;
//    private static int m = 21;
//    private static int m = 22;
//    private static int m = 23;
//    private static int m = 24;
//    private static int m = 25;
//    private static int m = 0;
//    private static int m = 1;
    private static int[][] c = new int[n+1][m+1];

    private static int MAX = 1000000;
    private static int coinsCount = MAX;

    public static void main(String[] args){

        coinsCount = charge();

        if(coinsCount == MAX)
            coinsCount = -1;

        System.out.println(coinsCount);

    }

    private static int charge(){
        int i, j, k;

        for(i=1; i<=n; i++)
            for(j=1; j<=m; j++)
                c[i][j] = MAX;

        for(i=1; i<=n; i++)
            c[i][0] = 0;

        for(j=1; j<=m; j++)
            if(j%T[1] == 0 && j/T[1] <= coins[1])
                c[1][j] = j/T[1];

        for(i=2; i<=n; i++)
            for(j=1; j<=m; j++){
                for(k=0; k<=coins[i]; k++)
                    if(j-k*T[i]>=0 && k+c[i-1][j-k*T[i]] < c[i][j])
                        c[i][j] = k+c[i-1][j-k*T[i]];
            }

        return c[n][m];
    }

}

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Java: version 2

public class ZuiShaoYingBi4 {
    //case 1
    private static int n = 3;
    private static int[] T = {0, 1, 2, 5};
    private static int[] coins = {0, 3, 3, 3};

    //case 2
//    private static int n = 3;
//    private static int[] T = {0, 19, 20, 50};
//    private static int[] coins = {0, 3, 3, 3};

    //case 3
//    private static int n = 4;
//    private static int[] T = {0, 1, 2, 5, 18};
//    private static int[] coins = {0, 3, 3, 3, 1};

    //case 4
//    private static int n = 4;
//    private static int[] T = {0, 1, 2, 5, 18};
//    private static int[] coins = {0, 3, 3, 3, 0};

    private static int m = 18;
    //    private static int m = 19;
//    private static int m = 20;
//    private static int m = 21;
//    private static int m = 22;
//    private static int m = 23;
//    private static int m = 24;
//    private static int m = 25;
//    private static int m = 0;
//    private static int m = 1;
//    private static int[][] c = new int[n+1][m+1];
    private static int[] c = new int[m+1];

    private static int MAX = 1000000;
    private static int coinsCount = MAX;

    public static void main(String[] args){

        coinsCount = charge();

        if(coinsCount == MAX)
            coinsCount = -1;

        System.out.println(coinsCount);

    }

    private static int charge(){
        for(int j=1; j<=m; j++)
            c[j] = MAX;

        for(int i=1; i<=n; i++)
            for(int k=1; k<=coins[i]; k++)
                for(int j=m; j>=T[i]; j--)
                    c[j] = Min(c[j], c[j-T[i]]+1);

        return c[m];
    }

    private static int Min(int a, int b){
        return a > b ? b : a;
    }
}

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Reference

王晓东《计算机算法设计与分析》（第3版）P88