poj 3250 Bad Hair Day (单调栈)

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

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 11473		Accepted: 3871

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

6
10
3
7
4
12
2

Sample Output

5

Source

USACO 2006 November Silver

 

 

思路：

暴力N*N肯定会超时，所以就想怎样通过一次遍历就求出结果呢

这里用到单调栈

struct Nod
{
    __int64 height;
    __int64 startPos;
}node;

定义节点中包含高度height，以及所在的位置

对于 输入的 数据：10  3  7  4  12  2

初始化栈： 先在栈中push进一个节点（inf，-1）以方便后面节点的统一处理，inf为一个无限大的数

 

 i=0时 10<inf  故压入栈  (输入值跟栈顶元素比较)

(inf,-1)  (10,0) 

 

 i=1时 3<10  压入栈

(inf,-1)  (10,0)  (3,1)

 

i=2时  7>3  出栈并计算     sum+=i-H1.startPos-1;    直到 7<10  压栈

(inf,-1)  (10,0)   (7,2)

 

i=3时 4<7  压栈

(inf,-1)  (10,0)   (7,2)  (4,3)

 

i=4时  12>4  出栈并计算     sum+=i-H3.startPos-1;

          12>7  出栈并计算    sum+=i-H2.startPos-1;

　　　 12>10  出栈并计算    sum+=i-H0.startPos-1;

          12<inf  入栈

(inf,-1)  (12,4)

 

i=5时 2<12  压栈

(inf,-1)  (12,4)   (2,5)

 

最后将栈内除了（inf，-1）的其他节点出栈并计算

（2,5）出栈   sum+=i-H5.startPos-1;

（12,4） 出栈    sum+=i-H4.startPos-1;

 

最后得到的sum即为所求。

 

代码C/C++：

#include<iostream>
#include<stdio.h>
#include<stack>

#define INF 1000000001

using namespace std;

struct Nod
{
    __int64 height;
    __int64 startPos;
}node;

int main()
{
    __int64 n;
    while(~scanf("%I64d",&n))
    {
        __int64 i,a,sum=0;
        stack <Nod> stk;
        node.height = INF;
        node.startPos = -1;
        stk.push(node);
        for(i=0;i<n;i++)
        {
            scanf("%I64d",&a);
            while(a>=stk.top().height)
            {
                node = stk.top();
                stk.pop();
                sum += i - node.startPos - 1;
            }
            node.height = a;
            node.startPos = i;
            stk.push(node);
        }
        while(!stk.empty())
        {
            node = stk.top();
            stk.pop();
            if(node.height!=INF)    sum += i - node.startPos - 1;
        }
        printf("%I64d
",sum);
    }
    return 0;
}