[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

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<cstdlib>
 5 #include<stack>
 6 #include<iostream>
 7 using namespace std;
 8 typedef long long ll;
 9 ll a[80002];
10 stack<ll>ss;
11 int main(){
12     ll n,ans;
13     while(cin>>n){
14         ans=0;
15         for(int i=0;i<n;i++) cin>>a[i];
16         for(int i=0;i<n;i++){
17             while(!ss.empty()&&a[i]>=ss.top()){
18                 ss.pop();
19             }
20             ans+=ss.size();
21             ss.push(a[i]);
22         }
23         cout<<ans<<endl;
24         while(!ss.empty())    ss.pop();
25     }
26     return 0;
27 } 

数组实现：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<cstdlib>
 4 #include<algorithm>
 5 #include<iostream>
 6 using namespace std;
 7 typedef long long ll;
 8 ll ss[80002],top;
 9 int main(){
10     ios::sync_with_stdio(0);
11     ll n,t,ans=0;
12     while(cin>>n){
13         top=-1,ans=0;
14         for(int i=0;i<n;i++){
15             cin>>t;
16             while(top!=-1&&ss[top]<=t) top--;
17             ans+=top+1;
18             ss[++top]=t;
19         }
20         cout<<ans<<endl;
21     }
22     return 0;
23 }