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:
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= - = Cows facing right -->
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= - = = =
= = = = = =
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.
输出样例#1： 5
模拟题。可以搞一下
………………万脸懵逼。。。。
woc，我要举报翻译，翻译的和屎一样啊!!!
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想想咋做呢。
可以用一个栈保持一个单调的序列。
然后每次把栈的长度加到Ans中。保证这个序列单调即保证我们前面的牛都是可以看到后面的牛的。
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#include <stack> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define MAXN 80007 using namespace std; int n, h[MAXN], c[MAXN]; long long Ans; stack<int> S; int main() { scanf("%d", &n); for(int i=1; i<=n; i++) { scanf("%d", &h[i]); while(!S.empty() && S.top()<=h[i]) { S.pop(); } Ans += S.size(); S.push(h[i]); } printf("%lld", Ans); }
转载于:https://www.cnblogs.com/bljfy/p/9062445.html
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