題目描述

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.

農民約翰的某N(1 < N < 80000)頭奶牛正在過亂頭發節！由于每頭牛都意識到自己凌亂不堪 的發型，約翰希望統計出能夠看到其他牛的頭發的牛的數量.

每一頭牛i有一個高度所有N頭牛面向東方排成一排，牛N在最前面，而 牛1在最后面.第i頭牛可以看到她前面的那些牛的頭，只要那些牛的高度嚴格小于她的高度，而且 中間沒有比hi高或相等的奶牛阻隔.

讓N表示第i頭牛可以看到發型的牛的數量；請輸出Ci的總和

輸入輸出格式

輸入格式：

?

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.

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輸出格式：

?

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

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輸入輸出樣例

輸入樣例#1：?復制 6 10 3 7 4 12 2 輸出樣例#1：?復制 5


只要能看出是單調棧
這題就做完了


#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; }

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與50位技術專家面對面20年技術見證，附贈技術全景圖

總結

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