Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 15922 | Accepted: 5374 |
Description
Some of Farmer John's N cows (1 ≤ N ≤ 80,000) are having a bad hair day! Since each cow is self-conscious about her messy hairstyle, FJ wants to count the number of other cows that can see the top of other cows' heads.
Each cow i has a specified height hi (1 ≤ hi ≤ 1,000,000,000) and is standing in a line of cows all facing east (to the right in our diagrams). Therefore, cow i can see the tops of the heads of cows in front of her (namely cows i+1, i+2, and so on), for as long as these cows are strictly shorter than cow i.
Consider this example:
=
= =
= - = Cows facing right -->
= = =
= - = = =
= = = = = =
1 2 3 4 5 6
Cow#1 can see the hairstyle of cows #2, 3, 4
Cow#2 can see no cow's hairstyle
Cow#3 can see the hairstyle of cow #4
Cow#4 can see no cow's hairstyle
Cow#5 can see the hairstyle of cow 6
Cow#6 can see no cows at all!
Let ci denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c1 through cN.For this example, the desired is answer 3 + 0 + 1 + 0 + 1 + 0 = 5.
Input
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i.
Output
Sample Input
6
10
3
7
4
12
2
Sample Output
5
题意:每头牛都有身高,他(a)前边站的也有牛,有比他高的有比他低的,他能看见他前边比他低的,但是当有比他高的牛(b)的时候,这个高牛b前边的比a牛低的 a牛就看不见了(符合日常生活,当前边有一个比自己高的牛时,自己的视线就被挡住了)
题解:每次让牛入栈,当入栈的牛比栈顶的牛低时,继续入栈,直到遇到比栈顶高的牛弹出栈顶,
#include<stdio.h>
#include<stack>
#include<iostream>
#include<string.h>
using namespace std;
stack<int>s;
int cow[80005];
int main()
{
int n,i;
int sum=0;
scanf("%d",&n);
for(i=1;i<=n;i++)
scanf("%d",&cow[i]);
s.push(cow[1]);
for(i=2;i<=n;i++)
{
while(!s.empty()&&s.top()<=cow[i])
s.pop();
sum+=s.size();
s.push(cow[i]);
}
printf("%u\n",sum);
return 0;
}