#include<bits/stdc++.h> #define fi first #define se second #define pb push_back using namespace std; typedef pair<int,int> par; long long n, k, uf[100009], tab[2][100009], wyn[200009]; par invtab[200009], moves[3] = {{1, 0}, {0, 1}, {0, -1}}; int find(int u) { return (uf[u] == u ? u : uf[u] = find(uf[u])); } void Union(int x, int y) { uf[find(x)] = find(y); return; } void solve(int x) { int w = 0, a, b, c; par p; for(int i=x; i<=2*n; i++) uf[i] = i; for(int i=x; i<=2*n; i++) { c = 1; for(int j=0; j<3; j++) { a = (invtab[i].fi + moves[j].fi) % 2; b = (invtab[i].se + moves[j].se) % n; if(b < 0) b += n; if(tab[a][b] >= x && tab[a][b] < i && find(tab[a][b]) != find(i)) { Union(tab[a][b], i); c--; } } w += c; wyn[w]++; } return; } int main() { ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0); cin>>n>>k; for(int i=0; i<=1; i++) { for(int j=0; j<n; j++) { cin>>tab[i][j]; invtab[tab[i][j]] = par(i, j); } } for(int i=1; i<=2*n; i++) solve(i); for(int i=1; i<=k; i++) cout<<wyn[i]<<" "; return 0; }
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 | #include<bits/stdc++.h> #define fi first #define se second #define pb push_back using namespace std; typedef pair<int,int> par; long long n, k, uf[100009], tab[2][100009], wyn[200009]; par invtab[200009], moves[3] = {{1, 0}, {0, 1}, {0, -1}}; int find(int u) { return (uf[u] == u ? u : uf[u] = find(uf[u])); } void Union(int x, int y) { uf[find(x)] = find(y); return; } void solve(int x) { int w = 0, a, b, c; par p; for(int i=x; i<=2*n; i++) uf[i] = i; for(int i=x; i<=2*n; i++) { c = 1; for(int j=0; j<3; j++) { a = (invtab[i].fi + moves[j].fi) % 2; b = (invtab[i].se + moves[j].se) % n; if(b < 0) b += n; if(tab[a][b] >= x && tab[a][b] < i && find(tab[a][b]) != find(i)) { Union(tab[a][b], i); c--; } } w += c; wyn[w]++; } return; } int main() { ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0); cin>>n>>k; for(int i=0; i<=1; i++) { for(int j=0; j<n; j++) { cin>>tab[i][j]; invtab[tab[i][j]] = par(i, j); } } for(int i=1; i<=2*n; i++) solve(i); for(int i=1; i<=k; i++) cout<<wyn[i]<<" "; return 0; } |