212 lines
5.6 KiB
Markdown
212 lines
5.6 KiB
Markdown
##[$AcWing$ $896$. 最长上升子序列 II](https://www.acwing.com/problem/content/description/898/)
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### 一、题目描述
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给定一个长度为 $N$ 的数列，**求数值严格单调递增的子序列的长度最长** 是多少。
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**输入格式**
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第一行包含整数 $N$。
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第二行包含 $N$ 个整数，表示完整序列。
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**输出格式**
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输出一个整数，表示最大长度。
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**数据范围**
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$1≤N≤100000$，
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$−10^9≤数列中的数≤10^9$
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**输入样例：**
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```cpp {.line-numbers}
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7
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3 1 2 1 8 5 6
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```
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**输出样例**：
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```cpp {.line-numbers}
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4
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```
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<!-- 让表格居中显示的风格 -->
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<style>
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.center
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{
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width: auto;
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display: table;
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margin-left: auto;
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margin-right: auto;
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}
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</style>
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### 二、贪心+二分优化
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#### 1、朴素算法
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> **状态表示**：
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> - 集合:$f[i]$表示以 $a[i]$<font color='red'><b> 结尾 </b></font>的最长的上升序列长度
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> - 属性：$max$
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前一版本：$N≤1000$，本题要求：$N≤100000$
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$N$的数据范围大了$100$倍,前一版本动态规划代码的时间复杂度是$O(N^2)$,$1000^2=1000000$，是$1e6$，是可以$1$秒过的，但如果是$100000^2=10000000000$,是$1e10$，超时，需要优化~
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#### 2、贪心+二分算法
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><font color='red' size=4><b>思想 : 对于同样长度的子串，希望它的末端越小越好，这样后面更易扩展它,使数列更长。</b></font>
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> **状态表示**：
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> - 集合：$f[i]$表示长度为$i$的递增子序列中，末尾元素最小的是$f[i]$
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> - 属性：这个挺特殊，不是$max,min$,而是最终数列的长度$fl$
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**算法步骤**
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> $f[fl]$:$f$中的最后一个数字,$fl$是数组中的游标。
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扫描每个原序列中的数字$a[i]$:
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① 如果$a[i]$大于$f[fl]$,$f[++fl]=a[i]$,把$a[i]$ **接** 在$f[]$数组的最后面
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② 如果$a[i]$小于$f[fl]$,在$f$中查找并替换第一个大于等于它元素
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> **解释**：这样操作就可以使得长度一样的上升子序列的每个位置是最小的数字！
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**举栗子**
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<div class="center">
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| $a$ | $3$ | $1$ | $2$ | $1$ | $8$ | $5$ | $6$ |
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| --- | --- | --- | --- | --- | --- | --- | --- |
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</div>
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开始时$f[]$为空，数字$3$进入序列
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<div class="center">
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| $a$ | <font color='red'>$3$</font> | $1$ | $2$ | $1$ | $8$ | $5$ | $6$ |
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| --- | ---------------------------- | --- | --- | --- | --- | --- | --- |
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| $f$ | <font color='red' size=3><b>$3$</b></font> |
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| --- | ------------------------------------------ |
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</div>
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$1$ 比 $3$ 小, $3$出序列 ，$1$入序列
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<div class="center">
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| $a$ | $3$ | <font color='red'> $1$ </font> | $2$ | $1$ | $8$ | $5$ | $6$ |
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| --- | --- | ------------------------------ | --- | --- | --- | --- | --- |
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| $f$ | <font color='red' size=4><b>$1$</b></font> |
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| --- | ------------------------------------------ |
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</div>
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$2$ 比 $1$ 大,$2$入序列
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<div class="center">
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| $a$ | $3$ | $1$ | <font color='red'>$2$ </font> | $1$ | $8$ | $5$ | $6$ |
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| --- | --- | --- | ----------------------------- | --- | --- | --- | --- |
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| $f$ | $1$ | <font color='red' size=4><b>$2$</b></font> |
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| --- | --- | ------------------------------------------ |
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</div>
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$1$ 比 $2$ 小,在$f$中找到第一个大于等于$1$的位置，并替换掉原来的数字
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<div class="center">
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| $a$ | $3$ | $1$ | $2$ | <font color='red'>$1$</font> | $8$ | $5$ | $6$ |
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| --- | --- | --- | --- | ---------------------------- | --- | --- | --- |
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| $f$ | <font color='red' size=4><b>$1$</b></font> | $2$ |
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| --- | ------------------------------------------ | --- |
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</div>
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$8$ 比 $2$ 大
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<div class="center">
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| $a$ | $3$ | $1$ | $2$ | $1$ | <font color='red'>$8$</font> | $5$ | $6$ |
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| --- | --- | --- | --- | --- | ---------------------------- | --- | --- |
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| $f$ | $1$ | $2$ | <font color='red' size=4><b>$8$</b></font> |
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| --- | --- | --- | ------------------------------------------ |
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</div>
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$5$ 比 $8$ 小,在$f$中找到第一个大于等于$5$的数字，并替换掉原来的数字
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<div class="center">
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| $a$ | $3$ | $1$ | $2$ | $1$ | $8$ | <font color='red'>$5$</font> | $6$ |
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| --- | --- | --- | --- | --- | --- | ---------------------------- | --- |
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| $f$ | $1$ | $2$ | <font color='red' size=4><b>$5$</b></font> |
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| --- | --- | --- | ------------------------------------------ |
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</div>
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$6$ 比 $5$ 大
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<div class="center">
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| $a$ | $3$ | $1$ | $2$ | $1$ | $8$ | $5$ | <font color='red' size=4><b>$6$ </b></font> |
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| --- | --- | --- | --- | --- | --- | --- | ------------------------------------------- |
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| $f$ | $1$ | $2$ | $5$ | <font color='red' size=4><b>$6$</b></font> |
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| --- | --- | --- | --- | ------------------------------------------ |
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</div>
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<font color='blue' size=6><b><center>$f$ 的长度$fl$就是最长递增子序列的长度</center></b></font>
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#### 3、时间复杂度
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$$\large O(N*logN)$$
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### 四、实现代码
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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 = 100010;
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int n, a[N];
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// 数组模拟
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int f[N], fl;
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int main() {
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cin >> n;
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for (int i = 0; i < n; i++) cin >> a[i];
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// 1、首个元素
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f[0] = a[0];
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// 2、后续元素开始计算
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for (int i = 1; i < n; i++) {
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if (a[i] > f[fl])
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f[++fl] = a[i];
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else
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// 利用STL的二分，在f中查找第一个大于等于a[i]的值，并完成替换
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*lower_bound(f, f + fl, a[i]) = a[i];
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}
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// 栈内元素数量就是答案
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printf("%d\n", fl + 1);
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return 0;
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}
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``` |