1188 - Fast Queries

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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).

Input

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).

Output

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.

Note

Dataset is huge. Use faster I/O methods.

题目链接：LighOJ 1188

莫队大法好，看到n的范围是1e5也就直接上莫队了，用vis记录数字出现次数，显然对于区间的增减进行更新vis，若缩小区间导致vis[val]为0则说明少了一个数，反之为1则就是增加一个数，700MS还可以，另外一种不同的做法是用树状数组更新统计。

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<bitset>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=100010;
const int M=50010;
struct info
{
    int l,r;
    int id,b;
    bool operator<(const info &t)const
    {
        if(b==t.b)
            return r<t.r;
        return b<t.b;
    }
};
info Q[M];
int arr[N];
int vis[N];
int ans[M];
int main(void)
{
    int tcase,i,j;
    int n,m,l,r;
    scanf("%d",&tcase);
    for (int q=1; q<=tcase; ++q)
    {
        scanf("%d%d",&n,&m);
        for (i=1; i<=n; ++i)
            scanf("%d",&arr[i]);
        int unit=(int)sqrt(n);
        for (i=1; i<=m; ++i)
        {
            scanf("%d%d",&Q[i].l,&Q[i].r);
            Q[i].id=i;
            Q[i].b=Q[i].l/unit;
        }
        sort(Q+1,Q+1+m);
        CLR(vis,0);
        int L=1,R=0;
        int temp=0;
        for (i=1; i<=m; ++i)
        {
            while (L>Q[i].l)
            {
                --L;
                ++vis[arr[L]];
                if(vis[arr[L]]==1)
                    ++temp;
            }
            while (L<Q[i].l)
            {
                --vis[arr[L]];
                if(vis[arr[L]]==0)
                    --temp;
                ++L;
            }
            while (R>Q[i].r)
            {
                --vis[arr[R]];
                if(vis[arr[R]]==0)
                    --temp;
                --R;
            }
            while (R<Q[i].r)
            {
                ++R;
                ++vis[arr[R]];
                if(vis[arr[R]]==1)
                    ++temp;
            }
            ans[Q[i].id]=temp;
        }
        printf("Case %d:\n",q);
        for (i=1; i<=m; ++i)
            printf("%d\n",ans[i]);

    }
    return 0;
}