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POJ 3250 Bad Hair Day
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 17727 | Accepted: 5981 |
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
610374122
Sample Output
5
Source
1 #include <iostream> 2 #include <algorithm> 3 #include <map> 4 #include <vector> 5 #include <functional> 6 #include <string> 7 #include <cstring> 8 #include <queue> 9 #include <stack>10 #include <set>11 #include <cmath>12 #include <cstdio>13 using namespace std;14 #define IOS ios_base::sync_with_stdio(false)15 typedef long long LL;16 const int INF = 0x3f3f3f3f;17 const double PI=4.0*atan(1.0);18 19 const int maxn=80000;20 LL ans;21 int n,h;22 int main()23 {24 while(scanf("%d",&n)!=EOF){25 stack<int> s;26 ans=0;27 for(int i=0;i<n;i++){28 scanf("%d",&h);29 while(!s.empty()&&h>=s.top()) s.pop();30 ans+=s.size();31 s.push(h);32 }33 printf("%lld\n",ans);34 }35 }
POJ 3250 Bad Hair Day