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Problem Description
You are given an array of nn positive integers a1,a2,…,ana1,a2,…,an.
In one operation you do the following:
Choose any integer xx. For all ii such that ai=xai=x, do ai:=0ai:=0 (assign 00 to aiai). Find the minimum number of operations required to sort the array in non-decreasing order.
Input
Each test contains multiple test cases. The first line contains the number of test cases tt (1≤t≤1041≤t≤104). Description of the test cases follows.
The first line of each test case contains a single integer nn (1≤n≤1051≤n≤105).
The second line of each test case contains nn positive integers a1,a2,…,ana1,a2,…,an (1≤ai≤n1≤ai≤n).
It is guaranteed that the sum of nn over all test cases does not exceed 105105.
Output
For each test case print one integer — the minimum number of operations required to sort the array in non-decreasing order.
Example
input
5
3
3 3 2
4
1 3 1 3
5
4 1 5 3 2
4
2 4 1 2
1
1
output
1
2
4
3
0
Note
In the first test case, you can choose x=3x=3 for the operation, the resulting array is [0,0,2][0,0,2].
In the second test case, you can choose x=1x=1 for the first operation and x=3x=3 for the second operation, the resulting array is [0,0,0,0][0,0,0,0].
题意：
给出一个数组，你可以进行若干次操作，每次操作可以选择数组中某个数字x，令所有的x变成0，如果想得到一个不减序列的最少操作次数是多少。
分析：
操作次数满足二分性质，所以可以尝试一下二分，在check(x)函数中需要判断能否通过至多x次操作得到不减序列，那具体选哪几个数变成0呢？贪心地想，选择最靠前的非0数字最优，这是因为假如选择的是中间的非0数字，那么为了序列整体不减，前面的非0数字也一定要变成0，这样至少也要操作两次，但是直接选择最靠前的非0数字一定不会比这种方案差，毕竟它可以之后再选择中间的非0数字，甚至还会更优，因为这样是至少操作一次。所以贪心策略就是操作机会都用在最靠前的非0数字上，最后检查一下序列是否不减，最终复杂度为O(nlogn)。
具体代码如下：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <unordered_map>
using namespace std;
int a[100005], n, cpy[100005];
bool check(int x){
unordered_map<int, bool> mp;
for(int i = 1; i <= n; i++){
if(mp.count(cpy[i]))
cpy[i] = 0;
else if(mp.size() < x){
mp[cpy[i]] = true;
cpy[i] = 0;
}
}
for(int i = 2; i <= n; i++)
if(cpy[i]-cpy[i-1] < 0) return false;
return true;
}
signed main()
{
int T;
cin >> T;
while(T--){
scanf("%d", &n);
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]);
int l = 0, r = n, ans = -1;
while(l <= r){
for(int i = 1; i <= n; i++)
cpy[i] = a[i];
int mid = l+r>>1;
if(check(mid)){
ans = mid;
r = mid-1;
}
else l = mid+1;
}
printf("%d\n", ans);
}
return 0;
}
