算法---动态规划0/1背包与找零钱

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关于动态规划法，我怎么觉得就是蛮力法，有时候比蛮力法看起来还复杂

1.0/1背包问题

目前存在的问题，可能可以忽视？

#include<iostream>
#include<algorithm>
using namespace std;
int KnapSack(int w[], int v[], int n, int c);
int main()
{
 int w[6] = { 0,2,2,6,5,4 };
 int v[6] = { 0,6,3,5,4,6 };
 int n = 5, c = 10;
 cout << KnapSack(w, v, n, c) << endl;
}

int KnapSack(int w[], int v[], int n, int c) {
 int i, j;int V[6][11] = { {0} };
 int x[6];
 for (i = 0;i <= n;i++) {
  V[i][0] = 0;
 }
 for (j = 0;j <= c;j++) {
  V[0][j] = 0;
 }
 for (i = 1;i <= n;i++) {
  for (j = 1;j <= c;j++) {
   if (j < w[i]) {
    V[i][j] = V[i - 1][j];
   }
   else {
    V[i][j] = max(V[i - 1][j], V[i - 1][j - w[i]] + v[i]);
   }
  }
 }
  for (j = c, i = n;i > 0;i--) {
   if (V[i][j] > V[i-1][j]) {
    x[i] = 1;j = j - w[i];
   }
   else x[i] = 0;
  }

 return V[n][c];
}

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找零钱问题

建立了一个三维数组存取每个硬币的数量

但是在    int V[6][99] = { {0} }, L[6][99][6] = { { { 0 } } };

不定义V，L的输出就会变很奇怪，希望取得答案。

#include<iostream>
using namespace std;
int* find(int change, int Change[], int n);
int main() {
    int i, Collection, Price;
    cin >> Collection >> Price;
    int change = Collection - Price;
    int Change[6] = { 0,1,2,5,7,10 };
    int  count = 0, n = 5;
    int* L;L = find(change, Change, n);
    cout << "应找金额：" << endl;
    for (i = 1;i <= 5;i++)
        cout << Change[i] << "元硬币的个数：" << L[i] << endl;

    system("pause");
}
int* find(int change, int Change[], int n) {
    int i, j;
    int V[6][99] = { {0} }, L[6][99][6] = { { { 0 } } };


    for (i = 1;i <= n;i++) {
        for (j = 1;j <= change;j++) {
            L[i][j][i] = j / Change[i];int m = i;
            int x;int dchange = j;
            for (x = i - 1;x > 0;x--) {
                int k = Change[x + 1]* L[i][dchange][x + 1];

                dchange -= k;m -= 1;
    //            if (m == 0)L[i][j][x] = change;
                L[i][j][x] = L[m][dchange][x];
            }

        }
    }return L[n][change];
}

 2019年11月21日

错误原因：

如果不设置足够大的V，会释放函数内申请的内存；

所以由函数传入一个数组，然后存储再传出

#include<iostream>
using namespace std;
void find(int change, int Change[], int n, int res[]);
int main() {
    int i, Collection, Price;
    cin >> Collection >> Price;
    int change = Collection - Price;
    int Change[6] = { 0,1,2,5,7,10 };
    int count = 0, n = 5;
    int L[6]; 
    find(change, Change, n, L);
    cout << "应找金额：" << endl;
    for (i = 1; i <= 5; i++)
        cout << Change[i] << "元硬币的个数：" << L[i] << endl;

}
void find(int change, int Change[], int n, int res[]) {
    int i, j;
    int V[6][99] = { {0} }, L[6][99][6] = { { { 0 } } };


    for (i = 1; i <= n; i++) {
        for (j = 1; j <= change; j++) {
            L[i][j][i] = j / Change[i]; int m = i;
            int x; int dchange = j;
            for (x = i - 1; x > 0; x--) {
                int k = Change[x + 1] * L[i][dchange][x + 1];

                dchange -= k; m -= 1;
                //            if (m == 0)L[i][j][x] = change;
                L[i][j][x] = L[m][dchange][x];
            }

        }
    }
    for (int i = 1; i <= n; i++) {
        res[i] = L[n][change][i];
    }
}