HDOJ 题目5172 GTY's gay friends（线段树）

最新推荐文章于 2021-02-26 23:06:41 发布

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最新推荐文章于 2021-02-26 23:06:41 发布

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分类专栏：线段树

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GTY's gay friends

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1215 Accepted Submission(s): 309

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

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题目大意：给出n个数，m个询问，问你[l,r]区间内是否为1到r-l+1的全排列。

思路：要是符合的话首先，肯定这个区间的和一定是（len*（len+1））/2,然后再利用线段树判重，记录下每个点上的值的上一个的位置，利用线段树求出这段区间中数中的上一个位置最大的数，要是这个数在左边界外，说明区间内没有重的

ac代码

15017215

2015-10-05 15:58:01

Accepted

5172

2418MS

49308K

1232 B

G++

Who_you?

#include<stdio.h>
#include<string.h>
#include<map>
#include<iostream>
using namespace std;
#define LL __int64
#define max(a,b) (a>b?a:b)
#define min(a,b) (a>b?b:a)
#define N 1000005
LL a[N],sum[N],pr[N],mp[N];
LL node[N*3];
void pushup(LL tr)
{
	node[tr]=max(node[tr<<1],node[tr<<1|1]);
}
void build(LL l,LL r,LL tr)
{
	if(l==r)
	{
		node[tr]=pr[l];
		return;
	}
	LL mid=(l+r)>>1;
	build(l,mid,tr<<1);
	build(mid+1,r,tr<<1|1);
	pushup(tr);
}
LL query(LL L,LL R,LL l,LL r,LL tr)
{
	if(L<=l&&r<=R)
	{
		return node[tr];
	}
	int mid=(l+r)>>1;
	LL x,y;
	x=y=0;
	if(L<=mid)
		x=query(L,R,l,mid,tr<<1);
	if(R>mid)
		y=query(L,R,mid+1,r,tr<<1|1);
	return max(x,y);
}
int main()
{
	LL n,m;
	while(scanf("%I64d%I64d",&n,&m)!=EOF)
	{
		LL i;
	//	map<LL,LL>mp;
		for(i=1;i<=n;i++)
		{
			scanf("%I64d",&a[i]);
			sum[i]=sum[i-1]+a[i];
			pr[i]=mp[a[i]];
			mp[a[i]]=i;
		}
		build(1,n,1);
		while(m--)
		{
			LL x,y;
			scanf("%I64d%I64d",&x,&y);
			LL len=y-x+1;
			LL ss=sum[y]-sum[x-1];
			if(ss==(len)*(len+1)/2)
			{
				LL q=query(x,y,1,n,1);
				if(q<x)
				{
					printf("YES\n");
				}
				else
					printf("NO\n");
			}
			else
				printf("NO\n");
		}
	}
}