LightOJ 1188 Fast Queries(离线树状数组)

描述

Given an array of N integers indexed from 1 to N, and q queries, each in the form i j, you have to find the number of distinct integers from index i to j(inclusive).

输入

Input starts with an integer T (≤ 5), denoting the number of test cases.

The first line of a case is a blank line. The next line contains two integers N (1 ≤ N ≤ 105), q (1 ≤ q ≤ 50000). The next line contains N space separated integers forming the array. There integers
range in [0, 105].

Each of the next q lines will contain a query which is in the form i j (1 ≤ i ≤ j ≤ N).

输出

For each test case, print the case number in a single line. Then for each query you have to print a line containing number of distinct integers from index i to j.

样例输入

1

 

8 5

1 1 1 2 3 5 1 2

1 8

2 3

3 6

4 5

4 8

样例输出

Case 1:

4

1

4

2

4

这题的题意就是给你一个数组，q次询问，每次询问的区间内有几个不同的数。

这题的话就是离线处理每次的操作，先把询问按照r的大小排序，然后每次找的话有重复的就把前面的更新为0，后面的更新为1，然后求个区间和就好了，比如样例1 的第三组数据，这时候的值为 0 0 1 1 1 1，所以3 6 为4.

/* ***********************************************
Author        :yzkAccepted
Created Time  :2016/2/28 10:16:32
TASK		  :ggfly.cpp
LANG          :C++
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <stack>
using namespace std;
const int maxn=1e6+5;
struct node{
int l,r;
int idx;
}p[maxn];
int b[maxn],n,q,ans[maxn],vis[maxn],c[maxn];
bool cmp(node a,node b)
{
return a.r<b.r;
}
void add(int i,int v)
{
while(i<=n)
{
b[i]+=v;
i+=i&(-i);
}
}
int sum(int i)
{
int res=0;
while(i>0)
{
res+=b[i];
i-=i&(-i);
}
return res;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int t,cas=1;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&q);
for(int i=1;i<=n;i++)
scanf("%d",&c[i]);
memset(vis,0,sizeof(vis));
memset(b,0,sizeof(b));
for(int i=1;i<=q;i++)
{	scanf("%d %d",&p[i].l,&p[i].r);p[i].idx=i;}
sort(p+1,p+q+1,cmp);
int cnt=1;
for(int i=1;i<=n;i++)
{
add(i,1);
if(vis[c[i]]) add(vis[c[i]],-1);
vis[c[i]]=i;
while(cnt<=q && p[cnt].r==i)
{
ans[p[cnt].idx]=sum(p[cnt].r)-sum(p[cnt].l-1);
cnt++;
}
if(cnt>q)
break;
}
printf("Case %d:\n",cas++);
for(int i=1;i<=q;i++)
{
printf("%d\n",ans[i]);
}
}
return 0;
}