POJ 3579 Median （二分搜索）

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本文介绍了一种高效的方法来解决中位数差异问题，包括输入格式、输出要求及AC代码实现，适用于大规模数据集。

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Median

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4379		Accepted: 1340

Description

Given N numbers, X₁, X₂, ... , X_N, let us calculate the difference of every pair of numbers: ∣X_i - X_j∣ (1 ≤ i ＜ j ≤ N). We can get C(N,2) differences through this work, and now your task is to find the median of the differences as quickly as you can!

Note in this problem, the median is defined as the (m/2)-th smallest number if m,the amount of the differences, is even. For example, you have to find the third smallest one in the case of m = 6.

Input

The input consists of several test cases.
In each test case, N will be given in the first line. Then N numbers are given, representing X₁, X₂, ... , X_N, ( X_i≤ 1,000,000,000 3 ≤ N ≤ 1,00,000 )

Output

For each test case, output the median in a separate line.

Sample Input

Sample Output

1
8

经典的二分套二分题目每次二分a[i]+x的位置来确定有多少比它大的然后确定它是不是中位数

AC代码如下：

//
//  Created by TaoSama on 2015-04-28
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int N = 1e5 + 10;

int n, a[N];
long long half;

bool check(int x) {
	int cnt = 0;
	for(int i = 1; i <= n; ++i)
		cnt += a + n - lower_bound(a + 1, a + 1 + n, a[i] + x) + 1;
	return cnt > half;
}

int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
//  freopen("out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    while(scanf("%d", &n) == 1) {
        for(int i = 1; i <= n; ++i) scanf("%d", a + i);
        sort(a + 1, a + 1 + n);

		half = n * (n - 1LL) >> 2;
        int l = 0, r = 1e9;
        while(l + 1 < r) {
            int mid = l + r >> 1;
            if(check(mid)) l = mid;
            else r = mid;
        }
        printf("%d\n", l);
    }
    return 0;
}