（第七场）A Minimum Cost Perfect Matching 【位运算】

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

A、Minimum Cost Perfect Matching

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 x^y  (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 ≤ 5e5).

输出描述：

Output n space-separated integers, where the i-th integer denotes p i (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

3

输出

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.

题意概括：

求0~N-1的一个排列 p， 使得p[ i ] ^ i 的总和最小。

官方题解：

• 最优解的和一定是0。
• 当n是2的次幂的时候，非常简单p[i]=n-1-i即可。
• 否则，设b为<=n最大的2的次幂，
• 对于b <= x < n，交换x和x-b。
• 分别求两边的最优解。

举例
Minimum Cost Perfect Matching
• n = 11，初始为 0, 1, 2 ,3, 4, 5, 6, 7, 8, 9, 10
• 取b为8，开始交换 8, 9, 10, 3, 4, 5, 6, 7 / 0, 1, 2
• 取b为2，开始交换 8, 9, 10, 3, 4, 5, 6, 7 / 2, 1 / 0
• 翻转每段得到
• 7, 6, 5, 4, 3, 10, 9, 8 / 1, 2 / 0

解题思路：

题解给的是常识和构造。

最优和为 0 毋庸置疑。结合这道题思考，有按位与运算，我们可以从大到小把 i 取反（则相与的结果肯定为0），然后求该数 i 的最高位，保证取反的情况下不超过N的范围。

AC code：

 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 #include <bitset>
 #define INF 0x3f3f3f3f
 using namespace std;

 const int MAXN = 5e5+;
 int num[MAXN], a;
 int N;
 int hb(int k)
 {
     int n = ;
     while(k > )
     {
         n++;
         k>>=;
     }
     return n;
 }
 int main()
 {
     scanf("%d", &N);
     bitset<> b;
     memset(num, -, sizeof(num));
     for(int i = N-; i >= ; i--)
         {
             if(num[i] == -)
             {
                 b = ~i;
                 int len = hb(i);
                 int t = ;
                 int sum = ;
                 for(int k = ; k < len; k++)
                 {
                     sum+=b[k]*t;
                     t*=;
                 }
                 num[i] = sum;
                 num[sum] = i;
             }
         }
     for(int i = ; i < N; i++)
         {
             printf("%d", num[i]);
             if(i < N-) printf(" ");
         }
     puts("");
     return ;
 }