Bad Hair Day(POJ 3250) 单调栈

来自《挑战程序设计竞赛》

单调栈的应用

这道题可以说是单调站的入门题。。。

单调栈，顾名思义，栈中的元素是单调递增或者单调递减的。

1.题目原文

http://poj.org/problem?id=3250

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17768		Accepted: 5998

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  c ₁ through  c_N.

Sample Input

6
10
3
7
4
12
2

Sample Output

5

Source

USACO 2006 November Silver

2.解题思路

题目要求每头牛能看到的牛的数目之和，换个角度思考，题目即是求每头牛可以被其他牛看到的数目之和。

可以声明一个栈，每加入一个牛时，就把栈中的比该元素小的牛删除，这样栈中的牛都能看到这头牛，栈中元素的个数就是能看到这头牛的个数。每次都把这头牛加入栈中，维护单调栈，可以得知栈中元素总是单调递减的。(栈底的元素最大)时间复杂度为O(n)。

3.AC代码

#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<string>
#include<set>
#include<vector>
#include<cmath>
#include<bitset>
#include<stack>
#include<sstream>
#include<deque>
using namespace std;
#define INF 0x7fffffff
const int maxn=1005;

typedef long long ll;
int main()
{
    int n,h,t;
    ll sum=0;
    scanf("%d",&n);
    stack<int> s;
    scanf("%d",&h);
    s.push(h);
    for(int i=1;i<n;i++){
        scanf("%d",&t);
        while(!s.empty()&&t>=s.top()){
            s.pop();
        }
        sum+=s.size();
        s.push(t);
    }
    printf("%lld\n",sum);
    return 0;
}