HDU5172 - GTY's gay friends - 哈希

最新推荐文章于 2020-12-11 16:01:01 发布

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本文介绍了一道算法题目，需要判断给定区间内的数列是否构成特定排列。通过使用哈希技巧，实现O(1)时间复杂度的回答。代码采用C++实现。

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1.题目描述：

GTY's gay friends

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1687 Accepted Submission(s): 429

Problem Description

GTY has

n gay friends. To manage them conveniently, every morning he ordered all his gay friends to stand in a line. Every gay friend has a characteristic value

ai , to express how manly or how girlish he is. You, as GTY's assistant, have to answer GTY's queries. In each of GTY's queries, GTY will give you a range

[l,r] . Because of GTY's strange hobbies, he wants there is a permutation

[1..r−l+1] in

[l,r] . You need to let him know if there is such a permutation or not.

Input

Multi test cases (about 3) . The first line contains two integers n and m (

1≤n,m≤1000000 ), indicating the number of GTY's gay friends and the number of GTY's queries. the second line contains n numbers seperated by spaces. The

ith number

ai (

1≤ai≤n ) indicates GTY's

ith gay friend's characteristic value. The next m lines describe GTY's queries. In each line there are two numbers l and r seperated by spaces (

1≤l≤r≤n ), indicating the query range.

Output

For each query, if there is a permutation

[1..r−l+1] in

[l,r] , print 'YES', else print 'NO'.

Sample Input

  
   8 5
2 1 3 4 5 2 3 1
1 3
1 1
2 2
4 8
1 5
3 2
1 1 1
1 1
1 2

Sample Output

  
   YES
NO
YES
YES
YES
YES
NO

Source

BestCoder Round #29

Recommend

hujie

2.题意概述：

给出一个数组a[n]（1<=a[i]<=n）,可能会有重复，然后m组询问每次询问两个数：l,r 在区间[l,r]内是否构成一个1,2,..,r-l+1的排列；

3.解题思路：

对于

#include <bits/stdc++.h>
#define INF 0x7fffffff
#define maxn 1001000
#define eps 1e-6
#define pi acos(-1.0)
#define e 2.718281828459
#define mod (int)1e9 + 7;
using namespace std;
typedef long long ll;
int n, m;
ll sum[maxn], hs[maxn], val[maxn];
inline ll randval()
{
  ll one = 1;
  ll RAND = (rand() + rand()) * (one << 44) + (rand() + rand()) * (one << 33) + (rand() + rand()) * (one << 22) + (rand() + rand());
  return RAND;
}
int main()
{
#ifndef ONLINE_JUDGE
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
  long _begin_time = clock();
#endif
  srand(time(NULL));
  hs[0] = 0;
  for (int i = 1; i < maxn; i++)
  {
    val[i] = randval();
    hs[i] = hs[i - 1] ^ val[i];
  }
  while (scanf("%d%d", &n, &m) != EOF)
  {
    sum[0] = 0;
    for (int i = 1; i <= n; i++)
    {
      int x;
      scanf("%d", &x);
      sum[i] = sum[i - 1] ^ val[x];
    }
    for (int i = 0; i < m; i++)
    {
      int l, r;
      scanf("%d%d", &l, &r);
      if (hs[r - l + 1] == (sum[r] ^ sum[l - 1]))
        puts("YES");
      else
        puts("NO");
    }
  }
#ifndef ONLINE_JUDGE
  long _end_time = clock();
  printf("time = %ld ms\n", _end_time - _begin_time);
#endif
  return 0;
}