【c++刷题Day2】专题3栈与队列&单调栈与单调队列T4

这是C++刷题的Day2

🕋题目描述
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After Mr. B arrived in Warsaw, he was shocked by the skyscrapers and took several photos. But now when he looks at these photos, he finds in surprise that he isn’t able to point out even the number of buildings in it. So he decides to work it out as follows:
- divide the photo into n vertical pieces from left to right. The buildings in the photo can be treated as rectangles, the lower edge of which is the horizon. One building may span several consecutive pieces, but each piece can only contain one visible building, or no buildings at all.
- measure the height of each building in that piece.
- write a program to calculate the minimum number of buildings.
Mr. B has finished the first two steps, the last comes to you.
Input
Each test case starts with a line containing an integer n (1 <= n <= 100,000). Following this is a line containing n integers - the height of building in each piece respectively. Note that zero height means there are no buildings in this piece at all. All the input numbers will be nonnegative and less than 1,000,000,000.
Output
For each test case, display a single line containing the case number and the minimum possible number of buildings in the photo.
Sample
Inputcopy Outputcopy

3
1 2 3
3
1 2 1

	

Case 1: 3
Case 2: 2
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9

Hint

The possible configurations of the samples are illustrated below:
1

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🔑思路

用对于H[i]，要看它最多能延申到哪里，可以用单调栈维护即可

💯CODE代码：

#include 
#pragma GCC optimize(2)
#define fi first
#define se second
#define min3 (x , y , z) min (x , min (y , z))
#define max3 (x , y , z) max (x , max (y , z))

using namespace std ;

typedef long long ll ;
typedef double db ;
typedef pair < int , int > PII ;
typedef pair < string , int > PSI ;

const int maxn = 1e5 + 5 ;
int n , A[maxn] ;
int stk[maxn] , top ;
int main ()
{
	int C = 0 ;
	while (++ C , ~ scanf ("%d" , & n)) {
		for (int i = 1; i <= n; i ++)
			scanf ("%d" , & A[i]) ;
		top = 0 ;
		int ans = n ;
		for (int i = 1; i <= n; i ++) {
			if (! A[i]) {
				ans -- ;
				top = 0 ;
				continue ;
			}
			while (top && A[stk[top]] >= A[i]) {
				if (A[stk[top]] == A[i])
					ans -- ;
				top -- ;
			}
			stk[++ top] = i ;
		}
		printf ("Case %d: %d\n" , C , ans) ;
	}
	return 0 ;
}
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