hdoj 5532 Almost Sorted Array(最长上升子序列)
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?
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 a
1
,a2
,…,a
n![]()
.
1≤T≤2000
2≤n≤10^5
1≤T≤2000
2≤n≤10^5
1≤ai≤10^5
There are at most 20 test cases with n>1000 .
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
1 //给你n个数的序列,问你能否从中删除一个数,让剩下的序列为不递增或不递减。 2 #include <bits/stdc++.h> 3 #define N 200100 4 using namespace std; 5 int a[100100]; 6 int b[100100]; 7 8 //寻找最长上升子序列 9 int LIS(int a[], int n){ 10 int Stack[N]; //建栈 11 int top = -1; 12 int index; 13 Stack[++top] = a[0]; //首位元素入栈 14 for(int i = 1; i < n; i++){ 15 if(a[i] >= Stack[top]) //满足a[i] >= 栈顶元素则入栈 16 Stack[++top] = a[i]; 17 else{ 18 index = upper_bound(Stack, Stack + top + 1, a[i]) - Stack; 19 //如果不满足条件从栈中寻找第一个大于a[i]的元素并将其替换 20 Stack[index] = a[i]; 21 } 22 } 23 return top + 1; //栈中元素个数即为最长上升子序列长度 24 } 25 26 int main(){ 27 int n; 28 int t; 29 scanf("%d", &t); 30 while(t--){ 31 scanf("%d", &n); 32 for(int i = 0; i < n; i++){ 33 scanf("%d", &a[i]); //顺序储存 34 b[n - i - 1] = a[i]; //逆序储存 35 } 36 37 /*int x = LIS(a, n); 38 int y = LIS(b, n); 39 printf("%d %d\n", x, y); 40 */ 41 42 if(LIS(a, n) >= n - 1 || LIS(b, n) >= n - 1) printf("YES\n"); 43 //如果满足题意则满足最长上升子序列长度 >= n-1 44 else printf("NO\n"); 45 } 46 return 0; 47 }