POJ 3250 Bad Hair Day（单调栈）
| Time Limit: 2000MS | Memory Limit: 65536K | |
|---|---|---|
| Total Submissions: 21883 | Accepted: 7462 |
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 c*1 through *cN.For this example, the desired is answer 3 + 0 + 1 + 0 + 1 + 0 = 5.
Input
Line 1: The number of cows, N.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i.
Output
Line 1: A single integer that is the sum of c*1 through *cN.
Sample Input
6
10
3
7
4
12
2Sample Output
5题意
有一群牛乖乖♂站好一排，只能从左往右看，对于每头牛只能看到比它矮的牛，求每头牛能看到牛数量的和。
解题思路
找到每头牛开始右边第一个比他高的就行了，比较裸的单调栈。
代码
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======`-.____`-.___\_____/___.-`____.-'======
`=---='
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
佛祖保佑 永无BUG
*/
#include<stdio.h>
#include<algorithm>
#include<iostream>
using namespace std;
const int maxn = 8e4+50;
int arr[maxn],r[maxn],num[maxn];
int main()
{
#ifdef DEBUG
freopen("in.txt","r",stdin);
#endif // DEBUG
int n,cnt=0;
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&arr[i]);
for(int i=n;i>=1;i--)
{
while(cnt&&arr[num[cnt]]<arr[i]) cnt--;
if(cnt) r[i]=num[cnt];
else r[i]=n+1;
num[++cnt]=i;
}
long long ans=0;
for(int i=1;i<=n;i++)
ans=ans+r[i]-i-1;
printf("%lld\n",ans);
return 0;
}