給用戶一個隨機生成的4x4二維數組,其中一個元素肯定是0.考慮到0是空位置,用戶必須重複交換剩餘的15個元素,直到他們以升序獲得數組,0作爲最後一個元素。 在這一點上,他們被允許與0交換任何元素。 但是,如何修改此代碼以確保只能交換那些與它相鄰的元素(上方,下方或旁邊) ?與將二維數組按升序排序相關的謎題
#include<iostream>
#include<stdlib.h>
#include<time.h>
using namespace std;
int check_asc(int a[][4])
{
int i, j, previous = a[0][0];
for (i = 0; i < 4; i++)
{
for (j = 0; j < 4; j++)
{
if(i == 3 && j == 3)
{
if (a[i][j] == 0)
return 1;
}
else if (a[i][j] < previous)
{
return 0;
}
previous = a[i][j];
}
}
return 1;
}
void swap(int a[][4], int &xpos, int &ypos)
{
int arr, temp;
cout << "\n\nEnter number to be swapped with 0: ";
cin >> arr;
for (int i = 0; i < 4; i++)
{
for (int j = 0; j < 4; j++)
{
if (a[i][j] == arr)
{
temp = a[xpos][ypos];
a[xpos][ypos] = a[i][j];
a[i][j] = temp;
xpos = i;
ypos = j;
return;
}
}
}
}
int check_rep(int a[][4], int assign)
{
for (int i = 0; i < 4; i++)
{
for (int j = 0; j < 4; j++)
{
if (assign == a[i][j])
return 0;
}
}
return 1;
}
void main()
{
int a[4][4], assign, xpos = 0, ypos = 0, asc_result, rep_result;
srand((unsigned)time(NULL));
for (int i = 0; i < 4; i++)
for (int j = 0; j < 4; j++)
{
if (i == 0 && j == 0)
a[i][j] = 0;
else
{
do {
assign = rand() % 50;
rep_result = check_rep(a, assign);
} while (rep_result == 0);
a[i][j] = assign;
}
}
cout << "\n\nArrange the 4x4 matrix into ascending order. (Consider 0 as a blank space)" << endl;
for (int i = 0; i < 4; i++)
{
cout << endl;
for (int j = 0; j < 4; j++)
cout << a[i][j] << '\t';
}
do {
swap(a, xpos, ypos);
system("cls");
for (int i = 0; i < 4; i++)
{
cout << endl;
for (int j = 0; j < 4; j++)
cout << a[i][j] << '\t';
}
asc_result = check_asc(a);
} while (asc_result == 0);
cout << "\n\tYou win"<<endl;
system("pause");
}
它類似於十五謎題,只是數字不需要是1..15(和重新可能是重複的)。 – CiaPan
是的,這裏的數字可以是任何東西,直到50,沒有重複。 – Abhay
沒錯,沒有重複。順便說一句,你知道你的'check_rep'函數測試整個數組的重複部分,包括尚未初始化的部分? – CiaPan