HDU 3410【单调栈】

思路：
单调栈。
鄙人的记忆：按当前为最大值的两边延伸就是维护单调递减栈。

//#include <bits/stdc++.h>
#include <iostream>
#include <stack>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N=50000+10;
struct asd{
    int id;
    int left,right,w;
};
int n,a[N];
int ans[N][2];
stack<asd>q;
int main()
{
    while(!q.empty())
        q.pop();
    int T,cas=1;
    scanf("%d",&T);
    while(T--)
    {
        asd now,nex;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        now.id=1;
        now.left=now.right=1;
        now.w=a[1];
        q.push(now);
        for(int i=2;i<=n;i++)
        {
            nex.id=i;
            nex.left=nex.right=i;
            nex.w=a[i];
            while(!q.empty()&&q.top().w<a[i])
            {
                now=q.top();q.pop();
                ans[now.id][0]=now.left!=now.id?now.left:0;
                ans[now.id][1]=now.right!=now.id?now.right:0;
                if(!q.empty())
                    q.top().right=now.id;
                nex.left=now.id;
            }
            q.push(nex);
        }
        while(!q.empty())
        {
            now=q.top();q.pop();
            ans[now.id][0]=now.left!=now.id?now.left:0;
            ans[now.id][1]=now.right!=now.id?now.right:0;
            if(!q.empty())
                q.top().right=now.id;
        }
        printf("Case %d:\n",cas++);
        for(int i=1;i<=n;i++)
            printf("%d %d\n",ans[i][0],ans[i][1]);
    }
    return 0;
}