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@ -107,44 +107,49 @@ int res;
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int main() {
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cin >> n;
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for (int i = 1; i <= n; i++) cin >> a[i];
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for (int i = 0; i < n; i++) cin >> a[i];
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// 正向
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f[++fl] = a[1];
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p1[1] = 1;
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f[0] = a[0];
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p1[0] = 1;
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for (int i = 2; i <= n; i++)
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for (int i = 0; i < n; i++)
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if (a[i] > f[fl]) {
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f[++fl] = a[i];
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p1[i] = fl;
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} else {
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int t = lower_bound(f + 1, f + fl + 1, a[i]) - f;
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int t = lower_bound(f, f + fl, a[i]) - f;
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f[t] = a[i];
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p1[i] = t;
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}
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// 反向
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g[++gl] = a[n];
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p2[n] = 1;
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g[0] = a[n - 1];
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p2[n - 1] = 1;
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for (int i = n - 1; i >= 1; i--)
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for (int i = n - 1; i >= 0; i--)
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if (a[i] > g[gl]) {
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g[++gl] = a[i];
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p2[i] = gl;
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} else {
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int t = lower_bound(g + 1, g + gl + 1, a[i]) - g;
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int t = lower_bound(g, g + gl, a[i]) - g;
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g[t] = a[i];
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p2[i] = t;
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}
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for (int i = 1; i <= n; i++) res = max(res, p2[i] + p1[i] - 1);
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for (int i = 0; i < n; i++) res = max(res, p2[i] + p1[i] + 2 - 1);
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printf("%d\n", res);
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return 0;
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}
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```
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### 六、状态机分析法
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>**总结**:
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① 朴素版本性能较差$O(N^2)$,但有一个先天的优势:可以知道每个数字作为最高点时,左边有多少个,右边有多少个,对于求左右最长这样的题目不用再拐弯了。
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② 贪心+二分版本的性能好$O(LogN*N)$,但有一个缺点,就是只能获取到最长上升序列的长度,不知道 **以每个数字为最高点时,左边有多少个,右边有多少个**,如果非得用这个算法不可的话,那么就需要对贪心+二分版本的代码进行改造:用数组记录第几个数字在上升序列中应该是排在第几位的,显得麻烦一些。
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### 六、状态机分析法【选读$O(N^2)$】
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这是我做过 $AcWing$第二场热身赛的$C$题——[$AcWing$ $3549$. 最长非递减子序列](https://www.acwing.com/problem/content/3552/) 后总结出来的一类模型
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那就是,**利用状态机模型$DP$解决最长$xxx$子序列模型** 的方法
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