POJ3250——Bad Hair Day（栈的应用）

Bad Hair Day

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 17352 Accepted: 5846

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 h_i (1 ≤ h_i ≤ 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 c_i denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c₁ through c_N.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 c₁ through c_N.

Sample Input

610374122

Sample Output

5

题意：每头牛能看到它右边的比他矮的牛的发型，问每头牛能看到的牛的总和。

思路：相当于是每头牛能没前面的牛看到的数量相加。栈顶保存最高的牛的编号。

#include <cmath>#include <cstring>#include <cstdio>#include <vector>#include <string>#include <algorithm>#include <string>#include <map>#include <queue>#include <set>#include <stack>using namespace std;#define MAXN 80010#define LEN 200010#define INF 1e9+7#define MODE 1000000#define pi acos(-1)#define g 9.8typedef long long ll;int n;long long h[MAXN];int r[MAXN];stack <int> ans;int main(){        scanf("%d",&n);        for(int i=0;i<n;i++)            scanf("%I64d",h+i);        while(!ans.empty())            ans.pop();        unsigned long long res=0;        for(int i=0;i<n;i++)        {            while(!ans.empty()&&h[ans.top()]<=h[i])            {                ans.pop();            }            res+=ans.size();            ans.push(i);        }        printf("%I64d\n",res);}