习题3

#include <iostream> #include <vector> #include <stack> #include <list> #include <map> #include <set> #include <deque> #include <queue> #include <cstring> #include <unordered_map> #include <unordered_set> #include <algorithm> #include <numeric> #include <chrono> #include <ctime> #include <cmath> #include <cctype> #include <string> #include <cstdio> #include <iomanip> #include <thread> #include <mutex> #include <condition_variable> #include <functional> #include <iterator> using namespace std; char matrix[5][7], order[1024]; int x, y, len = 0; bool IsOk(char in) {int nx = x, ny = y;switch (in){case 'A': nx = x - 1; break;case 'B': nx = x + 1; break;case 'L': ny = y - 1; break;case 'R': ny = y + 1; break;}if (nx < 0 || ny < 0 || nx >= 5 || ny >= 5) return false;swap(matrix[x][y], matrix[nx][ny]);x = nx;y = ny;return true; } bool Init() {memset(matrix, ' ', sizeof(matrix));//gets 也是可以的fgets(matrix[0],7,stdin);return true; } void GetStart() {for (int i = 0; i < 5; ++i) {for (int j = 0; j < 5; ++j) {//vs2013 上面 换行前一个是空格 不输入到字符串中 但是提交代码 可以不用判断 似乎空格也输入到字符串中了 gets fges getline 都是这种情况if (matrix[i][j] == ' ' || matrix[i][j] == 'n') {x = i;y = j;matrix[i][j] = ' ';return;}}} } int main() {int nCount = 0;while (Init() && matrix[0][0] != 'Z') {int canDo = 1;x = y = len = 0;for (int i = 1; i < 5; ++i) {fgets(matrix[i], 7, stdin);}GetStart();while (cin >> order[len] && order[len] != '0') ++len;cin.get();if (nCount) cout << endl;cout << "Puzzle #" << ++nCount << ":" << endl;for (int i = 0; i < len; ++i) {if (canDo && !IsOk(order[i])) {cout << "This puzzle has no final configuration." << endl; canDo = 0; break;}}if (!canDo) { continue; }for (int i = 0; i < 5; ++i) {cout << matrix[i][0];for (int j = 1; j < 5; ++j) {cout << " "<<matrix[i][j] ;}cout << endl;}}return 0; }