Lucky - HDU 5213 莫队算法

最新推荐文章于 2019-03-11 09:33:42 发布

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本文介绍了一种解决特定查询问题的算法实现，该问题涉及在一串整数中寻找能够组合成固定奇数K的数对，并通过莫队算法进行高效求解。

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 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

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]);
    }
}