F. Array Partition
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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
inputCopy
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
outputCopy
YES
6 1 4
NO
YES
2 5 2
YES
4 1 2
YES
1 1 3
YES
2 1 4
题意：给定一个长度为N的数组，将其分成三段，符合max(1,x) = min(x+1,x+y) = max(x+y+1,n),即第一段和第三段区间的最大值和第二段区间最小值相等。
思路：首先考虑，
- 随着区间长度的增加，区间最大值是不变，或者增大。
- 随着区间长度的增加，区间最小值是不变，或者减小。
固定第一个区间的右端点x,通过二分第二个区间的右端点即可。需要快速求得某个区间的最小值，所有这里需要用到线段树或者RMQ，整体复杂度是O(NlogN*logN)
代码如下：
#include <bits/stdc++.h>
using namespace std;
int n;
const int MAX = 2e5+10;
int a[MAX];
int lp[MAX<<2],rp[MAX<<2],mi[MAX<<2];
int lMax[MAX],rMax[MAX];
void pushup(int i){
mi[i] = min(mi[i<<1],mi[i<<1|1]);
}
void build(int l,int r,int i){
lp[i] = l;
rp[i] = r;
if(l == r){
mi[i] = a[l];
return;
}
int mid = (l+r)>>1;
build(l,mid,i<<1);
build(mid+1,r,i<<1|1);
pushup(i);
}
int query(int l,int r,int i){
if(l <= lp[i] && rp[i] <= r){
return mi[i];
}
int mid = (lp[i]+rp[i]) >> 1;//左区间的右边界
int res = INT_MAX;
if(l <= mid){
res = min(res,query(l,r,i<<1));
}
if(mid < r){
res = min(res,query(l,r,i<<1|1));
}
return res;
}
void init(){
build(1,n,1);
lMax[1] = a[1];
for(int i = 2;i <= n;i += 1){
lMax[i] = max(lMax[i-1],a[i]);
}
rMax[n] = a[n];
for(int i = n-1;i >= 1;i -= 1){
rMax[i] = max(rMax[i+1],a[i]);
}
}
void solve(){
init();
for(int i = 1;i + 2 <= n;i += 1){
int l = i+1,r = n-1;
while(l <= r){
int mid = l + (r-l)/2;
int Min = query(i+1,mid,1);
if(lMax[i] < Min || rMax[mid+1] > lMax[i]){
l = mid + 1;
}else if(lMax[i] > Min || rMax[mid+1] < lMax[i]){
r = mid - 1;
}else{
cout << "YES" << endl;
cout << i << " " << mid-i << " " << n-mid << endl;
return;
}
}
}
cout << "NO" << endl;
}
int main(void){
int T;
cin >> T;
while(T--){
cin >> n;
for(int i=1;i<=n;i+=1){
cin >> a[i];
}
solve();
}
return 0;
}