GTY's gay friends
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 264 Accepted Submission(s): 57
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 a i
, 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.
), 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 i th
number a i
( 1≤a i ≤n
) indicates GTY's i th
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.
in [l,r]
, print 'YES', else print 'NO'.
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
NO
YES
YES
YES
YES
NO
转自官方题解：http://bestcoder.hdu.edu.cn/
1003 GTY's gay friends
一个区间是排列只需要区间和为len(len+1)2 (len 为区间长度),且互不相同,对于第一个问题我们用前缀和解决,对于第二个问题,预处理每个数的上次出现位置,记它为pre,互不相同即区间中pre的最大值小于左端点,使用线段树或Sparse Table即可在O(n)/O(nlogn) 的预处理后 O(logn)/O(1) 回答每个询问.不过我们还有更简单的hash做法,对于[1..n] 中的每一个数随机一个64位无符号整型作为它的hash值,一个集合的hash值为元素的异或和,预处理[1..n] 的排列的hash和原序列的前缀hash异或和,就可以做到线性预处理,O(1) 回答询问.
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<string> #define N 1000005
#define M 105
#define mod 10000007
//#define p 10000007
#define mod2 1000000000
#define ll long long
#define LL long long
#define eps 1e-6
#define inf 100000000
#define maxi(a,b) (a)>(b)? (a) : (b)
#define mini(a,b) (a)<(b)? (a) : (b) using namespace std; ll n,m;
ll a[N];
ll f[N];
ll pr[N];
ll sum[N]; ll mapre[*N]; ll build(ll i,ll l,ll r)
{
if(l==r){
mapre[i]=pr[l];
//printf(" i=%I64d l=%I64d mapre=%I64d pr=%I64d\n",i,l,mapre[i],pr[l]);
return mapre[i];
}
ll mid=(l+r)/;
ll lm=build(*i,l,mid);
ll rm=build(*i+,mid+,r);
mapre[i]=max(lm,rm);
//printf(" i=%I64d mapre=%I64d\n",i,mapre[i]);
return mapre[i];
} ll query(ll i,ll L,ll R,ll l,ll r)
{
if(l>=L && r<=R){
return mapre[i];
}
ll mid=(l+r)/;
ll lm,rm;
lm=;rm=;
if(L<=mid){
lm=query(i*,L,R,l,mid);
}
if(R>mid){
rm=query(i*+,L,R,mid+,r);
}
return max(lm,rm);
} void ini()
{
ll i;
sum[]=;
memset(f,,sizeof(f));
for(i=;i<=n;i++){
scanf("%I64d",&a[i]);
sum[i]=sum[i-]+a[i];
}
//for(i=0;i<=n;i++){
// printf(" i=%I64d sum=%I64d\n",i,sum[i]);
// }
for(i=;i<=n;i++){
pr[i]=f[ a[i] ];
f[ a[i] ]=i;
}
// for(i=0;i<=n;i++){
// printf(" i=%I64d pr=%I64d\n",i,pr[i]);
// }
build(,,n);
} void solve()
{
ll x,y;
ll ss;
ll len;
ll qu;
while(m--){
scanf("%I64d%I64d",&x,&y);
len=y-x+;
ss=sum[y]-sum[x-];
//printf(" ss=%I64d sum=%I64d\n",ss,len*(len+1)/2);
if(ss==(len*(len+)/)){
qu=query(,x,y,,n);
if(qu<x){
printf("YES\n");
}
else{
printf("NO\n");
}
}
else{
printf("NO\n");
}
}
} void out()
{ } int main()
{
//freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
//scanf("%d",&T);
//for(int ccnt=1;ccnt<=T;ccnt++)
// while(T--)
//scanf("%d%d",&n,&m);
while(scanf("%I64d%I64d",&n,&m)!=EOF)
{
ini();
solve();
out();
}
return ;
}