Almost Sorted Array(o(nlgn)求解LIS)

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原文链接：http://www.cnblogs.com/bianjunting/p/11586531.html

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#java

本文深入探讨了AlmostSortedArray问题的高效解决方案，该问题旨在判断一个数组是否通过移除一个元素即可变为有序。文章提供了使用LIS算法进行优化的代码示例，并详细解释了其背后的逻辑，对于理解算法优化及LIS的应用具有较高的参考价值。

Almost Sorted Array

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 10562 Accepted Submission(s): 2449

Problem Description

We are all familiar with sorting algorithms: quick sort, merge sort, heap sort, insertion sort, selection sort, bubble sort, etc. But sometimes it is an overkill to use these algorithms for an almost sorted array.

We say an array is sorted if its elements are in non-decreasing order or non-increasing order. We say an array is almost sorted if we can remove exactly one element from it, and the remaining array is sorted. Now you are given an array

Input

The first line contains an integer

Output

For each test case, please output "`YES`" if it is almost sorted. Otherwise, output "`NO`" (both without quotes).

Sample Input

2 1 7

3 2 1

3 1 4 1 5

Sample Output

YES

Source

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

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hujie


onlgn求解LIS的基本思想是，用dp[i]保存长度为i的最长子序列的最大值的最小值。
遍历数组，如果a >= dp[len]，接到后面，否则，在dp中寻找第一个大于这a的数，把他替换掉，原因很好想，要想使得序列足够长，那么dp[i]越小越好。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 100000 + 5;

int n, val[maxn];
int cnt1, cnt2, dp[maxn];

int main() {
   int t;
   int a, temp;
   scanf("%d", &t);
   while(t --) {
      cnt1 = 0, cnt2 = 0;
      memset(dp, 0, sizeof dp);
      scanf("%d", &n);
      for(int i = 0; i < n; i ++) scanf("%d", &val[i]);
      for(int i = 0; i < n; i ++) {
         if(val[i] >= dp[cnt1]) dp[++ cnt1] = val[i];
         else {
            temp = upper_bound(dp + 1, dp + 1 + cnt1, val[i]) - dp;
            dp[temp] = val[i];
         }
      }
      memset(dp, 0, sizeof dp);
      for(int i = n - 1; i >= 0; i --) {
         if(val[i] >= dp[cnt2]) dp[++ cnt2] = val[i];
         else {
            temp = upper_bound(dp + 1, dp + 1 + cnt2, val[i]) - dp;
            dp[temp] = val[i];
         }
      }
      if(cnt1 >= n - 1 || cnt2 >= n - 1) printf("YES\n");
      else printf("NO\n");
   }
   return 0;
}