内部赛3 F Frequent values

Frequent values

Time Limit: 1000MS Memory limit: 65536K

题目描述

You are given a
sequence of n integers a1 ,
a2 , ... ,
an in
non-decreasing order. In addition to that, you are given several
queries consisting of indices i and j (1
≤ i ≤ j ≤ n[/i]). For each query, determine the most frequent value
among the integers ai , ... ,
aj.

输入

The input consists of several test
cases. Each test case starts with a line containing two
integers n and q (1
≤ n, q ≤ 100000[/i]). The next line contains n integers a1 , ... ,
an (-100000
≤ ai ≤
100000[/i], for each i
∈ {1, ..., n}[/i]) separated by spaces. You can assume that for
each i
∈ {1, ..., n-1}: ai ≤
ai+1[/i].
The following q lines
contain one query each, consisting of two integers i and j (1
≤ i ≤ j ≤ n[/i]), which indicate the boundary indices for the
query.

The last test case is followed by a line containing a
single 0[/i].

输出

For each query, print one line with
one integer: The number of occurrences of the most frequent value
within the given range.

示例输入

10 3
-1 -1 1 1 1 1 3 10 10 10
2 3
1 10
5 10
0

示例输出

1
4
3

这个题题意简单我首先是想到的是线段树，在写代码时思路他一定要理清，不然很容易在递归中陷进去，代码有点长，不过听大宝说还有简单的方法,等学会了在贴一遍吧！

代码：

#include<stdio.h>

#define N 100005

struct node

{

 int l,r,count,num;

 int rnum,lnum;

 int lcount,rcount;

}tree[N*4];

int a
,MAX;

void maketree(int t,int l,int r)

{

 int mid,sum;

 tree[t].l=l;

 tree[t].r=r;

 if(l==r)

 {

  tree[t].count=tree[t].lcount=tree[t].rcount=1;

  tree[t].num=tree[t].lnum=tree[t].rnum=a[l];

  return;

 }

 mid=(l+r)/2;

 maketree(2*t,l,mid);

 maketree(2*t+1,mid+1,r);

 if(tree[2*t].count>tree[2*t+1].count)

 {

  tree[t].count=tree[2*t].count;

  tree[t].num=tree[2*t].num;

 }

 else

 {

  tree[t].count=tree[2*t+1].count;

  tree[t].num=tree[2*t+1].num;

 }

 tree[t].lnum=tree[2*t].lnum;

 tree[t].lcount=tree[2*t].lcount;

 tree[t].rnum=tree[2*t+1].rnum;

 tree[t].rcount=tree[2*t+1].rcount;

 if(tree[2*t].rnum==tree[2*t+1].lnum)

 {

  sum=tree[2*t].rcount+tree[2*t+1].lcount;

  if(sum>tree[t].count)

  {

   tree[t].count=sum;

   tree[t].num=tree[2*t].rnum;

  }

  if(tree[2*t].lnum==tree[2*t+1].lnum)

   tree[t].lcount+=tree[2*t+1].lcount;

  if(tree[2*t].rnum==tree[2*t+1].rnum)

   tree[t].rcount+=tree[2*t].rcount;

 }

}

void search(int t,int l,int r)

{

 int mid;

 int sum1,sum2;

 if(tree[t].l==l&&tree[t].r==r)

 {

  if(tree[t].count>MAX)

   MAX=tree[t].count;

  return;

 }

 if(r<=tree[2*t].r)

  search(2*t,l,r);

 else if(l>=tree[2*t+1].l)

  search(2*t+1,l,r);

 else

 {

  mid=(tree[t].l+tree[t].r)/2;

  search(2*t,l,mid);

  search(2*t+1,mid+1,r);

  if(tree[2*t].rnum ==
tree[2*t+1].lnum)

       
{

           
if(a[l]!=tree[2*t].rnum)

    sum1=tree[2*t].rcount;

           
else

    sum1=mid-l+1;

           
if(a[r]!=tree[2*t+1].lnum) sum2=tree[2*t+1].lcount;

           
else

    sum2=r-mid;

           
if(sum1+sum2 > MAX) MAX=sum1+sum2;

       
}

 }

}

int main()

{

 int n,m;

 int i,j;

 int x,y;

 while(scanf("%d%d",&n,&m),n)

 {

  for(i=1;i<=n;i++)

  {

   scanf("%d",&a[i]);

  }

  maketree(1,1,n);

  for(j=0;j<m;j++)

  {

   MAX=0;

   scanf("%d%d",&x,&y);

   search(1,x,y);

   printf("%d\n",MAX);

  }

 }

 return 0;

}