1029 Median （25 分）--PAT甲级

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本文介绍了一种高效算法，用于找出两个已排序整数序列的中位数，通过使用队列数据结构来避免直接合并序列导致的内存超限问题。

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1029 Median （25 分）

Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1 = { 11, 12, 13, 14 } is 12, and the median of S2 = { 9, 10, 15, 16, 17 } is 15. The median of two sequences is defined to be the median of the nondecreasing sequence which contains all the elements of both sequences. For example, the median of S1 and S2 is 13.

Given two increasing sequences of integers, you are asked to find their median.
Input Specification:

Each input file contains one test case. Each case occupies 2 lines, each gives the information of a sequence. For each sequence, the first positive integer N (≤2×105) is the size of that sequence. Then N integers follow, separated by a space. It is guaranteed that all the integers are in the range of long int.
Output Specification:

For each test case you should output the median of the two given sequences in a line.
Sample Input:

4 11 12 13 14
5 9 10 15 16 17

Sample Output:

求两个有序数组的中位数。
根据计算，long int一般为4字节即4B，数组最大为200000，此时占用内存为4Bx200000x2=1600000/1024/1024=1.526MB,超过了内存限制，因此不能直接将两个数组输入存储。可以借助队列，第二个数组输入时每输入一个数，将a、b队列中最小的一个出队，直到出队元素数量达到中位数的位置。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<stack>
#include<set>
#include<unordered_map>
using namespace std;
const int maxn=10010;
const int inf=0x3fffffff;

int main()
{
    queue<int>a,b;
    int m,n,t;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d",&t);
        a.push(t);
    }
    scanf("%d",&m);
    int mid=(m+n-1)/2;
    int cnt=0;
    for(int i=0;i<m;i++){
        scanf("%d",&t);
        b.push(t);
        if(cnt<mid){
            if(!a.empty() && a.front()<t) a.pop();
            else b.pop();
            cnt++;
        }
    }
    for(;cnt<mid;cnt++){
        if(a.empty()) b.pop();
        else if(b.empty()) a.pop();
        else {
            if(a.front()>b.front()) b.pop();
            else a.pop();
        }
    }
    if(a.empty()) printf("%d",b.front());
    else if(b.empty()) printf("%d",a.front());
    else printf("%d",min(a.front(),b.front()));
    return 0;
}