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  • Almost Sorted Array(o(nlgn)求解LIS)

    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 a1,a2,,an, is it almost sorted?
     

     

    Input
    The first line contains an integer T indicating the total number of test cases. Each test case starts with an integer n in one line, then one line with n integers a1,a2,,an.

    1T2000
    2n105
    1ai105
    There are at most 20 test cases with n>1000.
     
    Output
    For each test case, please output "`YES`" if it is almost sorted. Otherwise, output "`NO`" (both without quotes).
     

     

    Sample Input
    3
    3
    2 1 7
    3
    3 2 1
    5
    3 1 4 1 5
     

     

    Sample Output
    YES
    YES
    NO
     

     

    Source
     

     

    Recommend
    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
    ");
          else printf("NO
    ");
       }
       return 0;
    }
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  • 原文地址:https://www.cnblogs.com/bianjunting/p/11586531.html
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