Codeforces Global Round 12 D. Rating Compression(思维)

最新推荐文章于 2022-10-28 14:15:45 发布

Chastelovee

最新推荐文章于 2022-10-28 14:15:45 发布

阅读量444

点赞数 1

分类专栏： # 思维题

本文链接：https://blog.youkuaiyun.com/Chstelove/article/details/110912562

版权

数组压缩双端队列排列算法编程竞赛

关键词由优快云通过智能技术生成

思维题专栏收录该内容

1 篇文章

订阅专栏

本文介绍了一种解决数组压缩问题的方法，即判断数组经过k次压缩后是否形成排列。通过特判k=1和n的情况，再使用双端队列维护较小元素的位置，确保每个元素出现次数不超过1，且等于n-k+1的元素至少出现1次，从而确定输出结果。代码中展示了具体实现过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目链接：http://codeforces.com/contest/1450/problem/D

D. Rating Compression

On the competitive programming platform CodeCook, every person has a rating graph described by an array of integers a of length n. You are now updating the infrastructure, so you’ve created a program to compress these graphs.

The program works as follows. Given an integer parameter k, the program takes the minimum of each contiguous subarray of length k in a.

More formally, for an array a of length n and an integer k, define the k-compression array of a as an array b of length n−k+1, such that
bj=minj≤i≤j+k−1ai
For example, the 3-compression array of [1,3,4,5,2] is [min{1,3,4},min{3,4,5},min{4,5,2}]=[1,3,2].
A permutation of length m is an array consisting of m distinct integers from 1 to m in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (m=3 but there is 4 in the array).

A k-compression array will make CodeCook users happy if it will be a permutation. Given an array a, determine for all 1≤k≤n if CodeCook users will be happy after a k-compression of this array or not.

Input
The first line contains a single integer t (1≤t≤104) — the number of test cases.

The first line of the description of each test case contains a single integer n (1≤n≤3⋅105) — the length of the array.

The second line of the description of each test case contains n integers a1,…,an (1≤ai≤n) — the elements of the array.

It is guaranteed, that the sum of n for all test cases does not exceed 3⋅105.

Output
For each test case, print a binary string of length n.

The k-th character of the string should be 1 if CodeCook users will be happy after a k-compression of the array a, and 0 otherwise.

Example
input

5

5

1 5 3 4 2

4
1 3 2 1
5
1 3 3 3 2
10
1 2 3 4 5 6 7 8 9 10
3
3 3 2

output

10111
0001
00111
1111111111
000

Note
In the first test case, a=[1,5,3,4,2].

The 1-compression of a is [1,5,3,4,2] and it is a permutation.
The 2-compression of a is [1,3,3,2] and it is not a permutation,since 3 appears twice.
The 3-compression of a is [1,3,2] and it is a permutation.
The 4-compression of a is [1,2] and it is a permutation.
The 5-compression of a is [1] and it is a permutation.

题意大意：

给你一个长度为n的数组a，一个定义bj = min(ai) (j≤i≤j+k−1)，k表示将数组a分为连续的长度为k的小段，k的值可以从1到n。当k=1的时， b1 = a1，b2 = a2……k=2的时， b1 = min(a1,a2),b2 = min(a2,a3)……判断b数组若不是由1到n-k+1的数字组成的排列，那么就输出0，否则就输出1。

思路：

首先特判一下当k=1和n时的答案，当k=1时，显然只有1到n的每个数字都出现一次才能输出1，否则输出0。当k=n时，显然只有当存在数字1才能输出1，否则输出0。
接下来我们进行假设，如果将最小的数放在序列的中间位置，必然会使一些段的最小值连续出现这个最小的数，显然应该把此最小的数放在序列的两端。所以可以维护一个双端队列进行判断，从1到n进行遍历，将小的数字依次放在双端队列的左右两端，并且小于n-k+1的数字的个数必须都为1，否则会出现上述使一些段的最小值连续出现某一最小值的现象，而等于n-k+1的数字的个数只要大于0即可。满足上述的则输出1，否则输出0。

代码：

#include<bits/stdc++.h> 
using namespace std;
const int maxn=5e5+10;

int ans[maxn],a[maxn],cnt[maxn];

int main()
{
    int t,n;
    scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
	    memset(ans,0,sizeof(ans));
	    memset(cnt,0,sizeof(cnt));
		for(int i=1;i<=n;i++){
			scanf("%d",&a[i]);
			cnt[a[i]]++;
		}
		bool flag=1;
		for(int i=1;i<=n;i++){
			if(cnt[i]>1) flag=0;
		}
		if(flag) ans[1]=1;
		if(cnt[1]) ans[n]=1;
		int l=1,r=n,cur=1;
	    for(int i=n-1;i>=2;i--){
			if(cnt[cur+1]&&cnt[cur]==1&&(a[l]==cur||a[r]==cur)){
				ans[i]=1;
				if(a[l]==cur)
					l++;
				else
					r--;
				++cur;
			}
			else break;
		}
	    for(int i=1;i<=n;i++) printf("%d",ans[i]);
		puts("");
	}
}