牛客网多校赛第七场A--Minimum Cost Perfect Matching【位运算】【规律】

链接：https://www.nowcoder.com/acm/contest/145/A

来源：牛客网

时间限制：C/C++ 1秒，其他语言2秒

空间限制：C/C++ 262144K，其他语言524288K

Special Judge, 64bit IO Format: %lld

题目描述

You have a complete bipartite graph where each part contains exactly n nodes, numbered from 0 to n - 1 inclusive.

The weight of the edge connecting two vertices with numbers x and y is (bitwise AND).

Your task is to find a minimum cost perfect matching of the graph, i.e. each vertex on the left side matches with exactly one vertex on the right side and vice versa. The cost of a matching is the sum of cost of the edges in the matching.

denotes the bitwise AND operator. If you're not familiar with it, see {https://en.wikipedia.org/wiki/Bitwise_operation#AND}.

输入描述:

The input contains a single integer n (1 ≤ n ≤ 5 * 105).

输出描述:

Output n space-separated integers, where the i-th integer denotes pi (0 ≤ pi ≤ n - 1, the number of the vertex in the right part that is matched with the vertex numbered i in the left part. All pi should be distinct.

Your answer is correct if and only if it is a perfect matching of the graph with minimal cost. If there are multiple solutions, you may output any of them.

示例1

输入

复制

3

输出

复制

0 2 1

说明

For n = 3, p0 = 0, p1 = 2, p2 = 1 works. You can check that the total cost of this matching is 0, which is obviously minimal.

刚开始看题解不知道他在讲什么.......

后来和czc讨论了一下总算是明白为什么了....

显然结果会是0 题目主要是去找一个序列使得这个对应的&都是0

如果n是2的次幂 那么倒一下顺序就正好了 因为他们的0和1是对称的

如果n不是2的次幂 那么我们就找到小于n的 最大的2的次幂x

把【x~n-1】的数和【0~n-x-1】的数一一交换 因为这些数只有最高位不同

而且只有【x~n-1】的数最高位是1 要先把他们解决掉 解决的方式就是找最高位是0的 低位的解决方式就相当于n是2的次幂时

那么对于被换到后面的那些数来说 继续按照上面的方法

他们的个数是len 分成len是2的次幂和len不是2的次幂

不断递归

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#include<queue>
#define inf 0x3f3f3f3f
using namespace std;

int n;
const int maxn = 5 * 1e5 + 5;
int num[maxn];

void solve(int len, int pos)
{
    int t = len, k = 0;
    while(t){
        k++;
        t /= 2;
    }
    k--;

    int x = pow(2, k);
    if(len - x > 0){
        for(int i = 0; i < len - x; i++){
            swap(num[pos + i], num[pos + x + i]);
        }
        for(int i = 0; i < x / 2; i++){
            swap(num[pos + i], num[pos + x - i - 1]);
        }
        solve(len - x, pos + x);
    }
    else{
        for(int i = 0; i < x / 2; i++){
            swap(num[pos + i], num[pos + x - i - 1]);
        }
        return;
    }
}

int main()
{
	while(scanf("%d", &n) != EOF){
        for(int i = 0; i < n; i++){
            num[i] = i;
        }

        solve(n, 0);
        printf("%d", num[0]);
        for(int i = 1; i < n; i++){
            printf(" %d", num[i]);
        }
        printf("\n");
	}
	return 0;
}