Codeforces round # ff (255) C. dzy loves sequences (LIS upgrade)

Source: Internet
Author: User

Title: C. dzy loves sequences (LIS upgrade)


Of the N numbers, a maximum of one number is changed, and the maximum length of the subsequence (continuous) that can be reached is obtained.


Consider the number of I, whether it can be changed, and then splice the two strings before and after, and maintain the current maximum value


Left [I]: indicates the maximum length of the strictly ascending subsequence ending with I

Right [I]: indicates the maximum length of the strictly ascending subsequence starting from I.

DP [I]: indicates the length of the string before and after splicing after changing the number of I

Transfer equation:

      dp[i] = max{left[i-1] + right[i+1] + 1 | a[i-1] + 1 < a[i+1]};


for(i = 1; i<=n; i++){    if(a[i-1] >= a[i])        ans = max(ans, right[i] + 1);    if(a[i+1] <= a[i])        ans = max(ans, left[i] + 1);    if(a[i-1] + 1 < a[i+1])        ans = max(ans, left[i-1] + right[i+1] + 1);}


#include <stdio.h>#include <iostream>#include <math.h>#include <algorithm>#include <string.h>#include <string>#include <queue>#include <stack>#include <map>#include <vector>#include <time.h>using namespace std;int a[100000+10];int L[100000+10];int R[100000+10];int main(){//freopen("a.txt", "r", stdin);int n, i, j;while(~scanf("%d", &n)){for(i = 1; i<=n; i++){scanf("%d", &a[i]);}memset(L, 0, sizeof(L));for(i = 1; i<=n; i++){L[i] = 1;if(i>1 && a[i] > a[i-1]){L[i] = max(L[i], L[i-1]+1);}}memset(R, 0, sizeof(R));int ans = 0;for(i = n; i>=1; i--){R[i] = 1;if(i<n && a[i] < a[i+1]){R[i] = max(R[i], R[i+1]+1);}ans = max(ans, R[i]);}for(i = 1; i<=n; i++){if(i>1 && a[i-1] >= a[i])ans = max(ans, L[i-1] + 1);if(i<n && a[i] >= a[i+1])ans = max(ans, R[i+1] + 1);if(i>1 && i<n && a[i-1] + 1 < a[i+1])ans = max(ans, L[i-1] + R[i+1] + 1);}printf("%d\n", ans);}return 0;}

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