A、Minimum Cost Perfect Matching | 时间限制：1秒 | 内存限制：256M
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}.
The input contains a single integer n (1 ≤ n ≤ 5 * 105).
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.
3
0 2 1
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.
#include <bits/stdc++.h> using namespace std; const int maxn = 5e5+5; int main() { int n; cin>>n; int now = n-1; int ans[maxn]; while(1) { if(now<0) break; for(int i=now;i>=0;i--) { int t = i&now; if(t==0) { int cnt = (now-i+1)/2; for(int j=i,k=cnt;k>0;j++,k--) { ans[j] = j+k*2-1; } for(int p=now,k=cnt;k>0;p--,k--) { ans[p] = p-(k*2-1); } now = i-1; break; } } } for(int i=0;i<n;i++) { cout<<ans[i]<<" "; } }
Minimum Cost Perfect Matching（牛客）
原文：https://www.cnblogs.com/hao-tian/p/9448736.html