洛谷P2866 [USACO06NOV]糟糕的一天Bad Hair Day(单调栈)

题目描述

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.

农民约翰的某N(1 < N < 80000)头奶牛正在过乱头发节！由于每头牛都意识到自己凌乱不堪的发型，约翰希望统计出能够看到其他牛的头发的牛的数量.

每一头牛i有一个高度所有N头牛面向东方排成一排，牛N在最前面，而牛1在最后面.第i头牛可以看到她前面的那些牛的头，只要那些牛的高度严格小于她的高度，而且中间没有比hi高或相等的奶牛阻隔.

让N表示第i头牛可以看到发型的牛的数量；请输出Ci的总和

输入输出格式

输入格式：

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.

输出格式：

Line 1: A single integer that is the sum of c1 through cN.

输入输出样例

输入样例#1：复制

输出样例#1：复制

5

只要能看出是单调栈
这题就做完了

#include<cstdio>

#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)

#define LL long long

char buf[<<],*p1=buf,*p2=buf;

const int MAXN = ;

inline LL read() {

    char c = getchar(); LL x = , f = ;

    while(c < '' || c > '') {if(c == '-') f = -;c = getchar();}

    while(c >= '' && c <= '') {x = x *  + c - '';c = getchar();}

    return x * f;

}

LL a[MAXN];

int  Ans[MAXN], s[MAXN];

int N, top = ;

int main() {

    N = read();

    for(int i = ; i <= N; i++) a[i] = read();

    s[] = N + ;

    LL ans = ;

    for(int i = N; i >= ; i--)    {

        while(top >  && a[i] > a[ s[top] ]) top--;

        ans += (s[top] - i - );

        s[++top] = i;

    }

    printf("%lld", ans);

    return ;

}