Uvs-225 golygons

Question: take steps 1, 2,... n from the origin point to the number of solutions to return to the origin point. No obstacles are allowed. You must turn around each time.

Train of Thought: a relatively simple DFS, the results have been done for a long time

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>using namespace std;const int MAXN = 250;const int Add = 100;int n, ans;int G[MAXN][MAXN], step[MAXN], sum[MAXN];int dx[4]={1, 0, 0, -1};int dy[4]={0, 1, -1, 0};char sign[5]="ensw";bool check(int x, int y, int d, int k) {for (int i = 1; i <= k; i++) {x += dx[d];y += dy[d];if (abs(x) > Add || abs(y) > Add)continue;if (G[x+Add][y+Add] == -1)return true;}if (abs(x)+abs(y) > sum[20] - sum[k])return true;return false;}void dfs(int x, int y, int cnt, int last) {if (cnt > n) {if (x == 0 && y == 0) {for (int i = 1; i <= n; i++)printf("%c", sign[step[i]]);printf("\n");ans++;}return;}int &i = step[cnt];for (i = 0; i < 4; i++) {if (i == last || i+last == 3)continue;if (check(x, y, i, cnt))continue;int nx = x + dx[i]*cnt;int ny = y + dy[i]*cnt;if (G[nx+Add][ny+Add])continue;G[nx+Add][ny+Add] = 1;dfs(nx, ny, cnt+1, i);G[nx+Add][ny+Add] = 0;}}int main() {sum[0] = 0;for (int i = 1; i <= 20; i++)sum[i] = sum[i-1] + i;int t, k;scanf("%d", &t);while (t--) {scanf("%d%d", &n, &k);memset(G, 0, sizeof(G));ans = 0;for (int i = 0; i < k; i++) {int a, b;scanf("%d%d", &a, &b);if (abs(a) > Add || abs(b) > Add)continue;G[a+Add][b+Add] = -1;}dfs(0, 0, 1, -1);printf("Found %d golygon(s).\n\n", ans);}return 0;}