Problem 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 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.
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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 cN.

Sample Input

6

10
3
7
4
12
2

Sample Output

5

題意：n頭牛排隊，每個牛都有一個高度，每頭牛只能看見它前面的比它矮的牛，當(dāng)其前面的一頭牛的高度高于當(dāng)前的牛，那么這頭牛以及前面的牛都視為看不到，求每頭牛能看見的牛的數(shù)量和。

思路：單調(diào)棧

如果一頭牛前面的牛比它矮或等于它的高度，就加入棧中，否則就出棧，這樣不斷統(tǒng)計棧的元素即可

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #define PI acos(-1.0) #define E 1e-6 #define INF 0x3f3f3f3f #define N 1000001 #define LL long long const int MOD=998244353; const int dx[]={-1,1,0,0}; const int dy[]={0,0,-1,1}; using namespace std; int main(){int n;while(scanf("%d",&n)!=EOF&&n){LL sum=0;stack<int> S;for(int i=1;i<=n;i++){int x;scanf("%d",&x);while(!S.empty()&&S.top()<=x)S.pop();sum+=S.size();S.push(x);}printf("%lld\n",sum);}return 0; }

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