[poj3250]单调栈 Bad Hair Day

解题关键：将每头牛看到的牛头数总和转化为每头牛被看到的次数，然后用单调栈求解，其实做这道题的目的只是熟悉下单调栈

此题为递减栈

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<stack>
#include<iostream>
using namespace std;
typedef long long ll;
ll a[80002];
stack<ll>ss;
int main(){
    ll n,ans;
    while(cin>>n){
        ans=0;    
        for(int i=0;i<n;i++) cin>>a[i];
        for(int i=0;i<n;i++){
            while(!ss.empty()&&a[i]>=ss.top()){
                ss.pop();
            }
            ans+=ss.size();
            ss.push(a[i]);
        }
        cout<<ans<<endl;
        while(!ss.empty())    ss.pop();
    }
    return 0;
} 

数组实现：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long ll;
ll ss[80002],top;
int main(){
    ios::sync_with_stdio(0);
    ll n,t,ans=0;
    while(cin>>n){
        top=-1,ans=0;
        for(int i=0;i<n;i++){
            cin>>t;
            while(top!=-1&&ss[top]<=t) top--;
            ans+=top+1;
            ss[++top]=t;
        }
        cout<<ans<<endl;
    }
    return 0;
}