题目大意
智能手机九点屏幕滑动解锁，如果给出某些连接线段，求出经过所有给出线段的合法的滑动解锁手势的总数。题目链接：
题目分析
首先，尝试求解没有给定线段情况下，所有合法的路径的总数。可以使用dfs进行搜索。代码如下：
void dfs(int row, int col, int cur_len) { visited[row][col] = true; if (cur_len >= 4) { //到达该点时，走过的路径长度大于等于4，则为合法的一个解锁手势 total_count++; } if (cur_len == 10) return; for (int i = 0; i < 3; i++) { for (int j = 0; j < 3; j++) { int next_row = (row + i) % 3; int next_col = (col + j) % 3; if (!visited[next_row][next_col]) { //可能出现当前点和下一个点的连线上经过了另一个点的情况，进行判断。如果 //经过的另一个点之前被访问过，则这次仍然是合法的。 if (abs(row - next_row) == 2 && abs(col - next_col) != 1 || abs(col - next_col) == 2 && abs(row - next_row) != 1) { int mid_row = (row + next_row) / 2; int mid_col = (col + next_col) / 2; if (!visited[mid_row][mid_col]) { continue; } } int cur = 3 * row + col; int next = 3 * next_row + next_col; dfs(next_row, next_col, cur_len + 1, need_use, cur_used + 1); } } } visited[row][col] = false;}
在上面的dfs搜索基础上，添加对已有线段的限制。9个点，维护 connected[9][9]， connected[i][j] 表示已经有线段将i和j连接。搜索的时候，还需要维护状态，cur_used表示当前已经经过了已有线段的数目，如果已经经过了所有的线段。且当前路径经过点数大于等于4，则是一个合法的解锁路径。
实现
#include#include #include using namespace std;bool visited[3][3];bool connected[9][9];int total_count;//cur_len 为到达row，col点时候，路径的长度； cur_used表示已经走过的路径所覆盖的已有线段的个数void dfs(int row, int col, int cur_len, int need_use, int cur_used) { visited[row][col] = true; if (cur_len >= 4 && need_use == cur_used) { total_count++; } if (cur_len == 10) return; for (int i = 0; i < 3; i++) { for (int j = 0; j < 3; j++) { int next_row = (row + i) % 3; int next_col = (col + j) % 3; if (!visited[next_row][next_col]) { if (abs(row - next_row) == 2 && abs(col - next_col) != 1 || abs(col - next_col) == 2 && abs(row - next_row) != 1) { int mid_row = (row + next_row) / 2; int mid_col = (col + next_col) / 2; if (!visited[mid_row][mid_col]) { continue; } } int cur = 3 * row + col; int next = 3 * next_row + next_col; if(connected[cur][next]) dfs(next_row, next_col, cur_len + 1, need_use, cur_used + 1); else dfs(next_row, next_col, cur_len + 1, need_use, cur_used); } } } visited[row][col] = false;}void GetCount(int need_use) { for (int row = 0; row < 3; row++) { for (int col = 0; col < 3; col++) { dfs(row, col, 1, need_use, 0); } }}int main() { memset(visited, false, sizeof(visited)); int T, need_use, u, v; scanf("%d", &T); while (T--) { total_count = 0; memset(connected, false, sizeof(connected)); scanf("%d", &need_use); for (int i = 0; i < need_use; i++) { scanf("%d %d", &u, &v); connected[u-1][v-1] = connected[v-1][u-1] = true; } GetCount(need_use); printf("%d\n", total_count); } return 0;}