HDU 3410【单调栈】

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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;}

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