WHU 1551 Pairs（分块暴力）

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

Hint

题意：给你N个数，M次询问，每次询问求区间[L,R]的符合|ai-aj|<=2&&i<j的个数

分析：我们考虑加入一个数u对个数的影响，加入一个u后，那么增加的个数就是cnt[u]+cnt[u-1]+cnt[u-2]

+cnt[u+1]+cnt[u+2]，删除一个u后，减少的个数就是cnt[u]+cnt[u-1]+cnt[u-2]+cnt[u+1]+cnt[u+2].

每次操作将cnt[u]的个数更新。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int INF=0x3f3f3f3f;
typedef long long LL;
const int maxn=1e5+10;
int n,m,sum;
int c[maxn],pos[maxn];
int num[maxn],ans[maxn];
struct node{
    int l,r;
    int id;
}q[maxn];
int cmp(node l1,node l2)
{
    if(pos[l1.l]==pos[l2.l])
        return l1.r<l2.r;
    return l1.l<l2.l;
}
void Add(int i)
{
    int val=c[i];
    int temp=num[val-1]+num[val-2]+num[val+1]+num[val+2];
    temp+=num[val];
    sum+=temp;  num[val]++;
}
void Sub(int i)
{
    int val=c[i];
    int temp=num[val-1]+num[val-2]+num[val+1]+num[val+2];
    temp+=num[val];
    sum-=temp-1;  num[val]--;
}
int main()
{
    int l,r;
    int cas=1;
    while(~scanf("%d%d",&n,&m))
    {
        int block=(int)sqrt(n*1.0+0.5);
        CLEAR(num,0);
        REPF(i,1,n)
        {
            scanf("%d",&c[i]);
            c[i]+=1;
            pos[i]=(i-1)/block+1;
        }
        REP(i,m)
        {
            scanf("%d%d",&l,&r);
            q[i].l=l;q[i].r=r;
            q[i].id=i;
        }
        sort(q,q+m,cmp);
        sum=0;
        int L=1,R=0;
        for(int i=0;i<m;i++)
        {
            while(L>q[i].l) Add(--L);
            while(R<q[i].r) Add(++R);
            while(L<q[i].l) Sub(L++);
            while(R>q[i].r) Sub(R--);

            ans[q[i].id]=sum;
        }
        printf("Case %d:\n",cas++);
        for(int i=0;i<m;i++)
            printf("%d\n",ans[i]);
    }
    return 0;
}

/*
10 1
5 5 1 3 6 3 5 7 1 7
3 4
*/