HDU 5172 GTY's gay friends（线段树）

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 52 1 3 4 5 2 3 11 31 12 24 81 53 21 1 11 11 2

 

Sample Output

YESNOYESYESYESYESNO

 

题意：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。

大小很容易我们通过记录前缀和很容易求出来，但是关键是去重。

考虑线段树做法，我们记录每个点的靠左最近的相同元素的位置，然后求

整个区间的最大值（即最大的前驱）如果小于l,即满足条件，输出YES。

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<string>#include<iostream>#include<queue>#include<cmath>#include<map>#include<stack>#include<bitset>using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )typedef long long LL;typedef pair<int,int>pil;const int MOD = 10000007;const int INF=0x3f3f3f3f;const int maxn=1000000+10;int mp[maxn];int ans[maxn],pre[maxn];LL s[maxn];int num[maxn],sum[maxn<<2];int n,m;void pushup(int rs){    sum[rs]=max(sum[rs<<1],sum[rs<<1|1]);}void build(int rs,int l,int r){    sum[rs]=0;    if(l==r)  return ;    int mid=(l+r)>>1;    build(rs<<1,l,mid);    build(rs<<1|1,mid+1,r);}void update(int x,int c,int l,int r,int rs){    if(l==r)    {        sum[rs]=c;        return ;    }    int mid=(l+r)>>1;    if(x<=mid)  update(x,c,l,mid,rs<<1);    else   update(x,c,mid+1,r,rs<<1|1);    pushup(rs);}int query(int x,int y,int l,int r,int rs){    if(l>=x&&r<=y)        return sum[rs];    int mid=(l+r)>>1;    int res=0;    if(x<=mid) res=max(res,query(x,y,l,mid,rs<<1));    if(y>mid)  res=max(res,query(x,y,mid+1,r,rs<<1|1));    pushup(rs);    return res;}int main(){    int x,y;    while(~scanf("%d%d",&n,&m))    {        s[0]=0;        CLEAR(mp,0);        CLEAR(ans,0);        REPF(i,1,n)        {            scanf("%d",&num[i]);            s[i]=s[i-1]+num[i];            if(!mp[num[i]])            {                pre[i]=0;                mp[num[i]]=i;            }            else            {                pre[i]=mp[num[i]];                mp[num[i]]=i;            }            update(i,pre[i],1,n,1);        }        while(m--)//离线做法超时了        {            scanf("%d%d",&x,&y);            LL temp=(1+y-x+1)*(y-x+1)/2;            if(s[y]-s[x-1]==temp&&query(x,y,1,n,1)<x)  puts("YES");            else  puts("NO");        }    }    return 0;}