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 }
原文:http://www.cnblogs.com/elpsycongroo/p/7257735.html