WHU-1551-Pairs(莫队算法+分块实现)

Description

Give you a sequence consisted of n numbers. You are required to answer how many pairs of numbers (ai, aj) satisfy that | ai - aj | <= 2 and L <= i < j <= R.

Input

The input consists of one or more test cases.



For each case, the first line contains two numbers n and m, which represent the length of the number sequence and the number of queries. ( 1 <= n <= 10^5 and 1 <= m <= 10^5 )

The second line consists of n numbers separated by n - 1 spaces.( 0 <= ai <= 10^5 )

Then the m lines followed each contain two values Li and Ri.

Output

For each case, first output “Case #: “ in a single line, where # will be replaced by case number.

Then just output the answers in m lines.

Sample Input

10 10

5 5 1 3 6 3 5 7 1 7

3 4

6 8

8 9

2 8

5 7

6 7

1 9

3 10

3 10

5 6

Sample Output

Case 1:

1

2

0

13

2

1

22

14

14

0

思路：把n个数分成 sqrt(n)块。用pos数组保存i所在的块。将全部查询按(pos(l),r)双keyword排序(这样子是为了降低后面的移动次数),然后模拟l向左向右移,r向左向右移,求出全部区间的结果。详见代码。

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;

struct Q{
int l,r,index,ans;
}query[100001];

int pos[100001],f[100001],w[100001];

bool cmplr(struct Q a,struct Q b)
{
    if(pos[a.l]==pos[b.l]) return a.r<b.r;//先按l所在的块排。假设相等就按r排
    else return pos[a.l]<pos[b.l];
}

bool cmpid(struct Q a,struct Q b)
{
    return a.index<b.index;
}

int update(int p,int ans,int add)//每移一步就更新一次结果
{
    int i;

    if(add==1)//1代表区间长度加1
    {
        for(i=w[p]-2;i<=w[p]+2;i++)
        {
            if(i>=0) ans+=f[i];
        }
        f[w[p]]++;//f[i]表示当前区间里面包括了f[i]个i
    }
    else//-1代表区间长度减1
    {
        f[w[p]]--;//f[i]表示当前区间里面包括了f[i]个i
        for(i=w[p]-2;i<=w[p]+2;i++)
        {
            if(i>=0) ans-=f[i];
        }
    }

    return ans;
}

int main()
{
    int n,m,cnt,i,ans,l,r,casenum=1;

    while(~scanf("%d%d",&n,&m))
    {
        memset(f,0,sizeof(f));//f[i]表示当前区间里面包括了f[i]个i

        cnt=(int)sqrt(n);//分成sqrt(n)块

        for(i=1;i<=n;i++)
        {
            scanf("%d",&w[i]);
            pos[i]=i/cnt;
        }

        for(i=1;i<=m;i++)
        {
            scanf("%d%d",&query[i].l,&query[i].r);
            query[i].index=i;
        }

        sort(query+1,query+m+1,cmplr);

        ans=0;
        l=1,r=0;
        for(i=1;i<=m;i++)//每一次都使l和r指向查询区间的l和r，这样子区间长度添加时就不用把自己加进去了，区间长度减小时要把自己减去
        {
            if(query[i].l==query[i].r)
            {
                query[i].ans=0;
                continue;
            }

            if(r<query[i].r)
            {
                for(r=r+1;r<=query[i].r;r++)
                {
                    ans=update(r,ans,1);
                }
                r--;
            }
            if(r>query[i].r)
            {
                for(;r>query[i].r;r--)
                {
                    ans=update(r,ans,-1);
                }
            }
            if(l<query[i].l)
            {
                for(;l<query[i].l;l++)
                {
                    ans=update(l,ans,-1);
                }
            }
            if(l>query[i].l)
            {
                for(l=l-1;l>=query[i].l;l--)
                {
                    ans=update(l,ans,1);
                }
                l++;
            }

            query[i].ans=ans;
            //printf("index=%d,l=%d,r=%d,ans=%d\n",query[i].index,query[i].l,query[i].r,query[i].ans);
        }

        sort(query+1,query+m+1,cmpid);

        printf("Case %d:\n",casenum++);

        for(i=1;i<=m;i++)
        {
            printf("%d\n",query[i].ans);
        }
    }

}