【PAT】A1029. Median (25)

最新推荐文章于 2024-05-27 19:16:19 发布

ztmajor

最新推荐文章于 2024-05-27 19:16:19 发布

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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。通过使用双队列方法，该算法避免了合并序列所带来的内存超限问题。

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Description:
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
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 (<=1000000) 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
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
13

//NKW 甲级练习题1008
#pragma warning(disable:4996)
#include <stdio.h>
#include <stdlib.h>
#include <climits>
#include <queue>
using namespace std;
queue<int> d1, d2;
int main(){
	int n, m, tnum;
	scanf("%d", &n);
	for (int i = 0; i < n; i++){
		scanf("%d", &tnum);
		d1.push(tnum);
	}
	d1.push(INT_MAX);
	scanf("%d", &m);
	int mid = (n + m - 1) / 2;
	for (int i = 0; i < m; i++){
		scanf("%d", &tnum);
		d2.push(tnum);
		if (!mid--){
			printf("%d\n", min(d1.front(), d2.front()));    //d2还未输入完就找到中位数了
			system("pause");
			return 0;
		}
		if (d1.front() < d2.front())
			d1.pop();
		else
			d2.pop();
	}
	d2.push(INT_MAX);
	while (mid--){        //d1中数普遍比d2大
		if (d1.front() < d2.front())
			d1.pop();
		else
			d2.pop();
	}
	printf("%d\n", min(d1.front(), d2.front()));
	system("pause");
	return 0;
}

本题如果合并两数列或者将两数列先存好再使用two pointers的话都会有一个测试点内存超限（PTA评测中。而NKW上好像没有这么刁钻的测试点，我一开始写的就直接过了）。