LA 7747 Appearance Analysis——模拟

LA 7747 Appearance Analysis——模拟

题意：给定一个大窗户（严格按照规则给出），窗户上有很多种玻璃，相同玻璃可以通过旋转而得，问一共有多少种玻璃

思路：存下每块玻璃的左上角坐标，配合玻璃的尺寸来表示每块玻璃，然后遍历即可求解，坐标的旋转规则：（i，j）（j，m-i+1）（m-i+1，n-j+1）（n-j+1，i）

#include #include #include #include using namespace std;const int maxn = 120;int m, n, x, y, cnt, ans, vis[maxn * maxn];char str[maxn][maxn];struct Window { int x, y;}window[maxn * maxn];void init() { cnt = ans = 0; memset(vis, 0, sizeof(vis));}int main() { while (scanf("%d %d", &m, &n) == 2) { getchar(); init(); for (int i = 1; i <= m; i++) gets(str[i] + 1); for (int i = 2; i <= m; i++) if (str[i][2] == '#') { x = i - 2; break; } for (int i = 2; i <= n; i++) if (str[2][i] == '#') { y = i - 2; break; } for (int i = 2; i <= m; i += (x + 1)) { for (int j = 2; j <= n; j += (y + 1)) { window[++cnt].x = i, window[cnt].y = j; } } for (int w1 = 1; w1 <= cnt; w1++) { if (vis[w1]) continue; vis[w1] = 1, ans++; for (int w2 = w1 + 1; w2 <= cnt; w2++) { if (vis[w2]) continue; bool ok1 = true, ok2 = true, ok3 = true, ok4 = true; if (x == y) { for (int i = 1; i <= x; i++) { for (int j = 1; j <= y; j++) { int x1 = window[w1].x + i - 1, y1 = window[w1].y + j - 1, x2 = window[w2].x + i - 1, y2 = window[w2].y + j - 1; if (str[x1][y1] != str[x2][y2]) ok1 = false; if (str[x1][y1] != str[x-i+window[w2].x][y-j+window[w2].y]) ok2 = false; if (str[x1][y1] != str[j+window[w2].x-1][x-i+window[w2].y]) ok3 = false; if (str[x1][y1] != str[y-j+window[w2].x][i+window[w2].y-1]) ok4 = false; } } if (ok1 || ok2 || ok3 || ok4) vis[w2] = 1; } else { for (int i = 1; i <= x; i++) { for (int j = 1; j <= y; j++) { int x1 = window[w1].x + i - 1, y1 = window[w1].y + j - 1, x2 = window[w2].x + i - 1, y2 = window[w2].y + j - 1; if (str[x1][y1] != str[x2][y2]) ok1 = false; if (str[x1][y1] != str[x-i+window[w2].x][y-j+window[w2].y]) ok2 = false; } } if (ok1 || ok2) vis[w2] = 1; } } } printf("%d\n", ans); }}

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