2016多校联合训练10&&HDU5857 Median

最新推荐文章于 2020-08-06 10:44:35 发布

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本文介绍了一种在已排序序列中快速找出特定组合元素中位数的算法。该算法适用于处理多个查询，每个查询涉及从两个指定区间内抽取元素形成新序列，并要求计算新序列的中位数。

Problem Description

There is a sorted sequence A of length n. Give you m queries, each one contains four integers, l1, r1, l2, r2. You should use the elements A[l1], A[l1+1] ... A[r1-1], A[r1] and A[l2], A[l2+1] ... A[r2-1], A[r2] to form a new sequence, and you need to find the median of the new sequence.

Input

First line contains a integer T, means the number of test cases. Each case begin with two integers n, m, means the length of the sequence and the number of queries. Each query contains two lines, first two integers l1, r1, next line two integers l2, r2, l1<=r1 and l2<=r2.
T is about 200.
For 90% of the data, n, m <= 100
For 10% of the data, n, m <= 100000
A[i] fits signed 32-bits int.

Output

For each query, output one line, the median of the query sequence, the answer should be accurate to one decimal point.

Sample Input

Sample Output

2.0
1.5

【解题方法】由于序列是有序的，按照l1<l2的顺序排好之后，分3种情况讨论即可，水题。具体看代码。

【AC 代码】

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
double a[200010];
double queryans(int l1,int r1,int l2,int r2,int x)
{
    if(r1<l2){//分离
        if(x<=r1-l1+1) return a[l1+x-1];
        else return a[l2+(x-(r1-l1+1))-1];
    }
    else if(r1>=r2){//完全重合
        if(x<=l2-l1) return a[l1+x-1];
        else if(x>r2-l1+1+r2-l2+1) return a[r2+x-(r2-l1+1+r2-l2+1)];
        else return a[l2-1+(x-(l2-l1)+1)/2];
    }
    else{//部分重合
        if(x<=l2-l1) return a[l1+x-1];
        else if(x>r1-l1+1+r1-l2+1) return a[r1+x-(r1-l1+1+r1-l2+1)];
        else return a[l2-1+(x-(l2-l1)+1)/2];
    }
}
int main()
{
    int T,n,m;
    int l1,r1,l2,r2;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1; i<=n; i++){
            scanf("%lf",&a[i]);
        }
        for(int i=1; i<=m; i++){
            scanf("%d%d%d%d",&l1,&r1,&l2,&r2);
            if(l1>l2) swap(l1,l2),swap(r1,r2);
            int len = (r1-l1+1)+(r2-l2+1);
            if(len%2==1){
                printf("%.1f\n",queryans(l1,r1,l2,r2,len/2+1));
            }else{
                printf("%.1f\n",(queryans(l1,r1,l2,r2,len/2)+queryans(l1,r1,l2,r2,len/2+1))/2.0);
            }
        }
    }
    return 0;
}