Lucky - HDU 5213 莫队算法

Lucky

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 419 Accepted Submission(s): 143



Problem Description

WLD is always very lucky.His secret is a lucky number K.k is
a fixed odd number. Now he meets a stranger with N numbers:a1,a2,...,aN.The
stranger asks him M questions.Each
question is like this:Given two ranges [Li,Ri] and [Ui,Vi],you
can choose two numbers X and Y to
make aX+aY=K.The X you
can choose is between Li and Ri and
the Y you
can choose is between Ui and Vi.How
many pairs of numbers(X,Y) you
can choose?

If WLD can answer all the questions correctly,he'll be the luckiest man in the world.Can you help him?



Input

There are multiple cases.(At MOST 5)

For each case:

The first line contains an integer N(1≤N≤30000).

The following line contains an integer K(2≤K≤2∗N),WLD's
lucky number.K is odd.

The following line contains N integers a1,a2,...,aN(1≤ai≤N).

The following line contains an integer M(1≤M≤30000),the
sum of the questions WLD has to answer.

The following M lines,the
i-th line contains 4 numbers Li,Ri,Ui,Vi(1≤Li≤Ri<Ui≤Vi≤N),describing
the i-th question the stranger asks.



Output

For each case:

Print the total of pairs WLD can choose for each question.



Sample Input

5
3 
1 2 1 2 3
1 
1 2 3 5

Sample Output

2

Hinta1+a4=a2+a3=3=K.
So we have two pairs of numbers (1,4) and (2,3).
Good luck!

题意：给出n个数个m个查询，对于每个查询找到ax(l<=x<=r)和ay(u<=y<=v)中有多少对使得ax+ay=K。

思路：这道题需要一些莫队算法的知识

定义记号f(A,B)表示询问区间A，B时的答案

用记号＋表示集合的并

利用莫队算法我们可以计算出任意f(A,A)的值

不妨假设A=[l1,r1],B=[l2,r2],C=[r1+1,l2−1]

容易知道f(A,B)=f(A+B+C,A+B+C)+f(C,C)−f(A+C,A+C)−f(C+B,C+B)

因此一个询问被拆成四个可以用莫队算法做的询问

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
struct node
{
    int l,r,ans,id,type;
}arr[120010];
int n,m,M,K,pos[30010],c[30010],num[60010],ans,f[30010];
bool cmp(node a,node b)
{
    return pos[a.l]<pos[b.l] ||(pos[a.l]==pos[b.l] && a.r<b.r);
}
void update(int p,int add)
{
    ans+=num[K-c[p]]*add;
    num[c[p]]+=add;
}
void solve()
{
    int i,j,k,l,r;
    l=1;r=0;
    ans=0;
    memset(num,0,sizeof(num));
    for(i=1;i<=M;i++)
    {
        for(;r<arr[i].r;r++)
           update(r+1,1);
        for(;r>arr[i].r;r--)
           update(r,-1);
        for(;l<arr[i].l;l++)
           update(l,-1);
        for(;l>arr[i].l;l--)
           update(l-1,1);
        arr[i].ans=ans;
    }
}
int main()
{
    int i,j,k,l,r,u,v;
    while(~scanf("%d",&n))
    {
        scanf("%d",&K);
        for(i=1;i<=n;i++)
           scanf("%d",&c[i]);
        k=(int)sqrt(n);
        for(i=1;i<=n;i++)
           pos[i]=(i-1)/k+1;
        scanf("%d",&m);
        M=m*4;
        for(i=1;i<=m;i++)
        {
            scanf("%d%d%d%d",&l,&r,&u,&v);
            arr[i*4-3].l=l;arr[i*4-3].r=v;arr[i*4-3].id=i;arr[i*4-3].type=1;
            arr[i*4-2].l=l;arr[i*4-2].r=u-1;arr[i*4-2].id=i;arr[i*4-2].type=-1;
            arr[i*4-1].l=r+1;arr[i*4-1].r=v;arr[i*4-1].id=i;arr[i*4-1].type=-1;
            arr[i*4  ].l=r+1;arr[i*4  ].r=u-1;arr[i*4  ].id=i;arr[i*4  ].type=1;
        }
        sort(arr+1,arr+1+M,cmp);
        solve();
        memset(f,0,sizeof(f));
        for(i=1;i<=M;i++)
           f[arr[i].id]+=arr[i].type*arr[i].ans;
        for(i=1;i<=m;i++)
           printf("%d\n",f[i]);
    }
}