AcWing 400 Greedy + Euler Circuit

Title

Portal AcWing 400 Taiko Master

answer

Length is $K$ of$01$ string total$2^K$ types, need to move clockwise on the ring as much as possible to traverse different$01$ string, and return to the initial string. Will$K$ long$The\; 01$ string is regarded as an edge, and the weight is$0$ The last digit of the string of$1$ will be the01 01represented by the side$01$ string of$K-1$ long prefix is ​​considered as a point. Each point has$With\; two$ outgoing edges and incoming edges, there are Euler circuits of the directed graph. Take the point$0$ is the starting point to solve the Euler circuit. In order to ensure the minimum lexicographic order, recursively start from the edge with the smaller weight. Total time complexity$O(2^K)$。

#include <bits/stdc++.h>
using namespace std;
const int maxk = 12, maxn = 1 << maxk, maxm = maxn << 1;
int K, nr, res[maxn];
int tot, head[maxn], to[maxm], nxt[maxm], cost[maxm];
bool vs[maxn];

inline void add(int x, int y, int z)
{
    
    
    to[++tot] = y, cost[tot] = z, nxt[tot] = head[x], head[x] = tot;
}

void euler(int x)
{
    
    
    for (int &i = head[x], y, z; i;)
    {
    
    
        y = to[i], z = cost[i], i = nxt[i];
        euler(y);
        res[++nr] = z;
    }
}

int main()
{
    
    
    scanf("%d", &K);
    int t = 1 << (K - 1);
    for (int i = 0; i < t; ++i)
    {
    
    
        int j = (i << 1) & (t - 1);
        add(i, j | 1, 1), add(i, j, 0);
    }
    euler(0);
    printf("%d ", 1 << K);
    for (int i = 1; i < K; ++i)
        putchar('0');
    while (nr >= K)
        putchar(res[nr--] + '0');
    putchar('\n');
    return 0;
}