习题7-2 黄金图形（Golygons, ACM/ICPC World Finals 1993, UVa225）

唯一需要注意的就是每个点只经过一次。回溯法求解即可。

我写的好像有点慢，把不能过的点存数组应该能快些。1.03s

#include <iostream>#include <string>#include <vector>#include <stack>#include <queue>#include <deque>#include <set>#include <map>#include <algorithm>#include <sstream>#include <utility>#include <cstring>#include <cstdio>#include <cstdlib>#include <ctime>#include <cmath>#include <cctype>#define CLEAR(a, b) memset(a, b, sizeof(a))#define IN() freopen("in.txt", "r", stdin)#define OUT() freopen("out.txt", "w", stdout)#define LL long long#define mod 1000000007#define INF 1000000007#define eps 1e-5#define PI 3.1415926535898using namespace std;//-------------------------CHC------------------------------//const int maxn = 25;const char dircs[] = "ensw";const int dx[] = { 0, 1, -1, 0 };const int dy[] = { 1, 0, 0, -1 };const int walk[4][2] = { 1, 2, 0, 3, 0, 3, 1, 2 };int n, k, cnt;char ans[maxn];pair<int, int> points[10005], cities[10005];int fp, fc;bool inside(int x, int y, int x1, int y1, int x2, int y2) {if ((x - x1)*(y - y2) + (x - x2) * (y - y1) != 0) return false;if ((x - x2)*(x - x1) > 0 || (y - y1) * (y - y2) > 0) return false;return true;}bool check(int x1, int y1, int x2, int y2) {for (int i = 0; i < fc; ++i)if (x1 == cities[i].first && y1 == cities[i].second)return false;for (int i = 0; i < fp; ++i) {int x = points[i].first, y = points[i].second;if (inside(x, y, x1, y1, x2, y2)) return false;}return true;}void dfs(int x, int y, int dir, int d) {int nx = x + dx[dir] * d, ny = y + dy[dir] * d;//printf("x = %d, y = %d\n", nx, ny);if (!check(nx, ny, x, y)) return;if (d == n) {if (nx == 0 && ny == 0) {++cnt;for (int i = 0; i < n; ++i)putchar(ans[i]);puts("");}return;}for (int i = 0; i < 2; ++i) {int ndir = walk[dir][i];ans[d] = dircs[ndir];cities[fc++] = make_pair(nx, ny);dfs(nx, ny, ndir, d + 1);fc--;}}int main() {//IN(); OUT();int t;scanf("%d", &t);while (t--) {fp = fc = 0;scanf("%d%d", &n, &k);while (k--) {int x, y;scanf("%d%d", &y, &x);points[fp++] = make_pair(x, y);}cnt = 0;for (int i = 0; i < 4; ++i) {ans[0] = dircs[i];dfs(0, 0, i, 1);}printf("Found %d golygon(s).\n\n", cnt);}return 0;}