79 lines
1.9 KiB
Markdown
79 lines
1.9 KiB
Markdown
## [$AcWing$ $420$. 火星人](https://www.acwing.com/problem/content/description/422/)
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### 1.算法：$STL$
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```cpp {.line-numbers}
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 10010;
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int n, m;
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int a[N];
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/*
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5
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3
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1 2 3 4 5
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答案
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1 2 4 5 3
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变化过程
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(1) 1 2 3 5 4
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(2) 1 2 4 3 5
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(3) 1 2 4 5 3
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*/
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int main() {
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cin >> n >> m;
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for (int i = 1; i <= n; i++) cin >> a[i];
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while (m--) next_permutation(a + 1, a + 1 + n); // 输入数据保证这个结果不会超出火星人手指能表示的范围。
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/*
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int a[4] = {1, 2, 3, 4};
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sort(a, a + 4);
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do {
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for (int i = 0; i < 4; i++) cout << a[i] << " ";
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cout << endl;
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} while (next_permutation(a, a + 4));
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*/
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for (int i = 1; i <= n; i++) printf("%d ", a[i]);
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return 0;
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}
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```
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### 2 、算法：利用贪心法解决字典序最小相关问题：
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```cpp {.line-numbers}
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 10010;
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int q[N];
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int n, m;
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int main() {
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cin >> n >> m;
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for (int i = 0; i <= n; i++) cin >> q[i];
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while (m--) { // 执行 m次
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int k = n - 1; // 从后向前来
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/*
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Q:123654321 问下一个是啥？
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算法步骤：
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1、从向向前找第一个下降点，本例就是6
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*/
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while (q[k - 1] > q[k]) k--;
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// 2、 k--：找到3
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int t = k--;
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// 3、从3开始，向后找第一个大于3的数字，就是4
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while (t + 1 < n && q[t + 1] > q[k]) t++;
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// 4、交换3和4
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swap(q[t], q[k]);
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// 5、对于折点的后半部进行翻转，即 124 653321 =>翻转=> 124 123356
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reverse(q + k + 1, q + n);
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}
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for (int i = 0; i < n; i++) printf("%d ", q[i]);
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return 0;
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}
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``` |