HDU 5172 GTY's gay friends 线段树

GTY's gay friends

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

【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】

【Sample Output】

YES

NO

YES

YES

YES

YES

NO

【题意】

给出一个数列，询问连续的从l开始到r为止的数是否刚好能够组成从1开始到r-l+1的数列。

 

【分析】

每一次询问都是一个区间询问。

对于每一个区间询问，需要判断区间内的数是否刚好可以组成1到k的连续数列，主要的判断标准有两个：

1.区间数字的总和与（1+k）*k/2相等；

2.保证区间内所有数都只出现一次。

 

第一个可以在读入数据时用前缀和解决。

第二个就要用到线段树了，读入时预处理记录下与当前数相同的数最近一次出现的位置。询问l~r的区间时，检索每个数的最近出现位置位置，若得到的所有结果都在区间左端的左边，那就说明区间中所有的数都是不重复出现的，则满足条件。这里就是用线段树判断区间最大值小于区间左端的过程了。

 

 /* ***********************************************

 MYID    : Chen Fan

 LANG    : G++

 PROG    : HDU5172

 ************************************************ */

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <algorithm>

 using namespace std;

 const int N=1e6+;

 int last[N],a,sum[N];

 typedef struct treetyp

 {

     int a,b,l,r,data;

 } treetype;

 treetype tree[*N];

 int treetail;

 void maketree(int l,int r)

 {

     treetail++;

     int now=treetail;

     tree[now].a=l;

     tree[now].b=r;

     if (l<r)

     {

         tree[now].l=treetail+;

         maketree(l,(l+r)/);

         tree[now].r=treetail+;

         maketree((l+r)/+,r);

     }

 }

 void add(int n,int i,int data)

 {

     if (tree[n].data<data) tree[n].data=data;

     if (i==tree[n].a&&i==tree[n].b) return ;

     else if (i<=(tree[n].a+tree[n].b)/) add(tree[n].l,i,data);

          else add(tree[n].r,i,data);

 }

 int res;

 void search(int n,int a,int b)

 {

     if (tree[n].a>=a&&tree[n].b<=b)

     {

         if (res<tree[n].data) res=tree[n].data;

         return ;

     }

     if (tree[n].a==tree[n].b) return ;

     if (a<=(tree[n].a+tree[n].b)/) search(tree[n].l,a,b);

     if (b>=(tree[n].a+tree[n].b)/+) search(tree[n].r,a,b);

 }

 int main()

 {

     int n,m;

     while(scanf("%d%d",&n,&m)==)

     {

         memset(sum,,sizeof(sum));

         memset(last,,sizeof(last));

         treetail=;

         maketree(,n);

         for (int i=;i<=n;i++)

         {

             scanf("%d",&a);

             sum[i]=sum[i-]+a;

             add(,i,last[a]);

             last[a]=i;

         }

         for (int i=;i<=m;i++)

         {

             int l,r;

             scanf("%d%d",&l,&r);

             if ((r-l+)*(r-l+)/==sum[r]-sum[l-])

             {

                 res=;

                 search(,l,r);

                 if (res<l) printf("YES\n");

                 else printf("NO\n");

             } else printf("NO\n");

         }

     }

     return ;

 }