POJ 3250 Bad Hair Day(单调栈stack)
| Time Limit: 2000MS |
|
Memory Limit: 65536K |
| Total Submissions: 15958 |
|
Accepted: 5392 |
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：一群牛面朝右排列，个高的牛能看见前面个子低的牛的的头发，个子高的牛也能遮住后面个子低的牛的视线。给出这群牛从左到右的身高，求每头牛能看见的发型个数之和。
题解：比赛的时候根本没有想到用单调栈做，暴力果断超时了。这里把问题想成一只牛的发型能被其他几只牛看见？把每只牛压入栈，当下一只将要被压入栈的牛身高比栈顶牛高时，就代表栈顶牛的视线被遮住了，看不见接下来的牛了，则出栈。栈里剩下的牛的个数就是能看见将要入栈牛发型的个数。将每次求得的个数叠加就能得到结果。
代码如下：
#include<cstdio>
#include<stack>
using namespace std;
int main()
{
int n,i;
long long h,sum;//int会错，longlong过了
while(scanf("%d",&n)!=EOF)
{
stack<int>s;//注意更新
sum=0;
scanf("%lld",&h);
s.push(h);
for(i=1;i<n;++i)
{
scanf("%lld",&h);
while(!s.empty()&&h>=s.top())//相等也代表看不见前方牛了
s.pop();
sum+=s.size();
s.push(h);
}
printf("%lld\n",sum);
}
return 0;
}