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单调栈： 维护一个单调栈

A Famous City

3
1 2 3
3
1 2 1

Case 1: 3
Case 2: 2

Hint
The possible configurations of the samples are illustrated below:
 

 

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <stack>

using namespace std;

stack<int> stk;
int n;

int main()
{
    int cas=1;
    while(scanf("%d",&n)!=EOF)
    {
        int ans=0;
        while(!stk.empty()) stk.pop();
        for(int i=0;i<n;i++)
        {
            int x;
            scanf("%d",&x);
            if(stk.empty())
            {
                if(x!=0) stk.push(x);
                continue;
            }
            if(stk.top()==x) continue;
            else if(stk.top()<x) stk.push(x);
            else
            {
                bool flag=true;
                while(!stk.empty())
                {
                    if(stk.top()>x)
                    {
                        stk.pop(); ans++;
                    }
                    else if(stk.top()==x)
                    {
                        flag=false;
                        break;
                    }
                    else if(stk.top()<x)
                    {
                        if(x!=0) stk.push(x);
                        flag=false;
                        break;
                    }
                }
                if(flag)
                {
                    if(x!=0) stk.push(x);
                }
            }
        }
        ans+=stk.size();
        printf("Case %d: %d\n",cas++,ans);
    }
    return 0;
}

HDOJ 4252 A Famous City 单调栈

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原文地址：http://blog.csdn.net/ck_boss/article/details/40194173