P1259. 黑白棋子的移动

考点

  • 模拟

题解

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#include <bits/stdc++.h>

using namespace std;

int N;

char str[550];

void init()
{
int i = 1;
for (; i <= N; ++i)
str[i] = 'o';
for (; i <= 2 * N; ++i)
str[i] = '*';
str[i++] = '-', str[i] = '-';
}

void print()
{
for (int i = 1; i <= 2 * N + 2; ++i)
cout << str[i];
cout << endl;
}

void deal()
{
int blank_idx = 2 * N + 1;
print();
for (int i = N; i > 4; --i)
{
swap(str[i], str[blank_idx]), swap(str[i + 1], str[blank_idx + 1]);
print();
swap(str[i], str[blank_idx - 2]), swap(str[i + 1], str[blank_idx - 1]);
print();
blank_idx -= 2;
}
// 单独处理只剩4个的时候
swap(str[4], str[blank_idx]), swap(str[5], str[blank_idx + 1]);
--blank_idx;
print();
swap(str[4], str[blank_idx]), swap(str[5], str[blank_idx + 1]);
print();
swap(str[2], str[blank_idx]), swap(str[3], str[blank_idx + 1]);
--blank_idx;
print();
swap(str[2], str[blank_idx]), swap(str[3], str[blank_idx + 1]);
print();
swap(str[1], str[blank_idx]), swap(str[2], str[blank_idx + 1]);
print();
}

int main()
{
cin >> N;
init();
deal();
return 0;
}

思路

纯粹的模拟题

每次两个空白格子和一对相邻的白黑棋子交换位置,然后两个空白格子的指针向前移,再换两个黑子即可,循环往复

只剩4个的时候单独处理即可