洛谷P2866 [USACO06NOV]糟糕的一天Bad Hair Day(…

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

6
10
3
7
4
12
2

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[1<<21],*p1=buf,*p2=buf;
const int MAXN = 80001;
inline LL read() {
    char c = getchar(); LL x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
    while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}
    return x * f;
}
LL a[MAXN];
int  Ans[MAXN], s[MAXN];
int N, top = 0;
int main() {
    N = read();
    for(int i = 1; i <= N; i++) a[i] = read();
    s[0] = N + 1;
    LL ans = 0;
    for(int i = N; i >= 1; i--)    {
        while(top > 0 && a[i] > a[ s[top] ]) top--;
        ans += (s[top] - i - 1);
        s[++top] = i;
    }
    printf("%lld", ans);
    return 0;
}

标签：

版权申明：本站文章部分自网络，如有侵权，请联系：west999com@outlook.com
特别注意：本站所有转载文章言论不代表本站观点，本站所提供的摄影照片，插画，设计作品，如需使用，请与原作者联系，版权归原作者所有