BestCoder Round #29 1003 （hdu 5172） GTY's gay friends [线段树 判不同 预处理 好题]

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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): 264    Accepted Submission(s): 57

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 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.

 

Input

Multi test cases (about 3) . The first line contains two integers n and m ( 1≤n,m≤100000

), 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.

 

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

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 ;
 }