poj 3250 Bad Hair Day (单调栈)
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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.
Input
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i.
Output
Sample Input
610374122
Sample Output
5
单调栈的入门题:单调栈就是维护一个栈,栈中的元素都保持着单调递增或递减的顺序。
题目意思:有n只牛站在一排,给出队伍中每只牛的高度,每只牛只能看到它右边比它矮的牛,求所有的牛能看到的牛数之和。
当我们新加入一个高度值时,如果栈中存在元素小于新加入的高度值,那这个牛肯定看不见这个高度的牛,就把这个元素弹栈。每次加入新元素,并执行完弹出操作后,栈中元素个数便是可以看见这个牛的“牛数”。
#include <iostream>#include <cstdio>#include <cstring>#include <stack>typedef long long ll;using namespace std;int main(){ int n; ll heigh,ans; stack<ll>s; while(scanf("%d",&n)!=EOF) { ans=0; cin>>heigh; s.push(heigh);//入栈 for(int i=1;i<n;i++) { cin>>heigh; while(!s.empty()&&s.top()<=heigh) //比较栈顶元素和新加入元素的关系 { s.pop(); } ans+=s.size(); s.push(heigh); } cout<<ans<<endl; while(!s.empty()) s.pop();//清空栈 } return 0;}
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