codeforce 686div3 F Array Partition单调栈

Array Partition

You are given an array a consisting of n integers.

Let min(l,r) be the minimum value among al,al+1,…,ar and max(l,r) be the maximum value among al,al+1,…,ar.

Your task is to choose three positive (greater than 0) integers x, y and z such that:

x+y+z=n;
max(1,x)=min(x+1,x+y)=max(x+y+1,n).
In other words, you have to split the array a into three consecutive non-empty parts that cover the whole array and the maximum in the first part equals the minimum in the second part and equals the maximum in the third part (or determine it is impossible to find such a partition).

Among all such triples (partitions), you can choose any.

You have to answer t independent test cases.

Input
The first line of the input contains one integer t (1≤t≤2⋅104) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (3≤n≤2⋅105) — the length of a.

The second line of the test case contains n integers a1,a2,…,an (1≤ai≤109), where ai is the i-th element of a.

It is guaranteed that the sum of n does not exceed 2⋅105 (∑n≤2⋅105).

Output
For each test case, print the answer: NO in the only line if there is no such partition of a that satisfies the conditions from the problem statement. Otherwise, print YES in the first line and three integers x, y and z (x+y+z=n) in the second line.

If there are several answers, you can print any.

Example
input
6
11
1 2 3 3 3 4 4 3 4 2 1
8
2 9 1 7 3 9 4 1
9
2 1 4 2 4 3 3 1 2
7
4 2 1 1 4 1 4
5
1 1 1 1 1
7
4 3 4 3 3 3 4
output
YES
6 1 4
NO
YES
2 5 2
YES
4 1 2
YES
1 1 3
YES
2 1 4
后悔没看这题…我是真的菜，反而去做了前面那道题，后来赛后看着题，第一反应就是单调栈，只需要枚举每个点作为中间部分的最小值的情况，那么通过单调栈求出这个数做为最小值的最大连续区间的位置（当然要考虑相等的情况，很明显只要不是这个数的第一个数或者最后一个数我们就能把他从栈里面拿走，用map标记一下就好了），这样贪心的选取，再根据左右区间的最值比较就好了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<vector>
#include<algorithm>
#include<queue>
#include<cstring>
#include<stack>
#include<map>
using namespace std;
typedef long long ll;
const int N=2e5+10;
int num[N];
int fmmax[N];
int rmax[N];
int l[N];
int r[N];
int book[N];
int state[N];
int main ()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin>>t;
    while(t--)
    {
        map<int,int>mp;
        int n;
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            cin>>num[i];
            fmmax[i]=0;
            rmax[i]=0;
            if(mp[num[i]]==0)
            {
                book[i]=1;
                mp[num[i]]=1;
            }
            else book[i]=0;
        }
        map<int,int>mmap;
        for(int i=n;i>=1;i--)
        {
            if(mmap[num[i]]==0)
            {
                state[i]=1;
                mmap[num[i]]=1;
            }
            else state[i]=0;
        }
        rmax[n+1]=0;
        for(int i=1;i<=n;i++)
        {
            fmmax[i]=max(num[i],fmmax[i-1]);
        }
        for(int i=n;i>=1;i--)
        {
            rmax[i]=max(num[i],rmax[i+1]);
        }
        stack<pair<int,int> >s;
        for(int i=1;i<=n;i++)
        {
            while(!s.empty()&&(s.top().first>num[i]||(s.top().first==num[i]&&!book[s.top().second])))
            {
                s.pop();
            }
            if(!s.empty())l[i]=s.top().second;
            else l[i]=0;
            s.push(make_pair(num[i],i));
        }
        while(!s.empty())s.pop();
        for(int i=n;i>=1;i--)
        {
            while(!s.empty()&&(s.top().first>num[i]||(s.top().first==num[i]&&!state[s.top().second])))
            {
                s.pop();
            }
            if(!s.empty())r[i]=s.top().second;
            else r[i]=n+1;
            s.push(make_pair(num[i],i));
        }
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            if(fmmax[l[i]]==num[i]&&num[i]==rmax[r[i]])
            {
                ans=1;
                cout<<"YES"<<endl;
                cout<<l[i]<<' '<<r[i]-l[i]-1<<' '<<n-r[i]+1<<endl;
                break;
            }
        }
        if(!ans)cout<<"NO"<<endl;
    }
    return 0;
}

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