习题4-3 黑白棋 UVa220

1.题目描述：点击打开链接

2.解题思路：本题是一道模拟题，要求模拟黑白棋的三种操作。对于这类题，肯定是分情况编写相应的函数处理。不过本题的难点在于如何方便地寻找到合法位置。思考了一会儿，可以枚举每一个空位置，同时利用一个向量数组方便来后续讨论，由于实现的细节较多，详细过程见代码注释。本题还有一个细节：在执行M命令时记得要更换当前的游戏者，题目中说了是两个人轮流下棋。

3.代码：

#define _CRT_SECURE_NO_WARNINGS #include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<functional>using namespace std;typedef pair<int, int> P;#define N 8char g[N][N];vector<P>pos;char player;//当前玩家，设为全局变量，便于后续讨论int dx[] = { -1, 1, 0, 0, -1, 1, -1, 1 };//利用向量数组，方便处理各种情况int dy[] = { 0, 0, -1, 1, -1, 1, 1, -1 };bool inside(int x, int y){if (x<0 || x>7 || y<0 || y>7)return false;return true;}bool find(int d,int x,int y, char c)//沿着路径dx[d],dy[d]寻找目标棋子{int w = dx[d], v = dy[d], i = x, j = y;while (inside(i, j) && g[i][j] == 'W' + 'B' - c){ i += w; j += v; }if (inside(i, j) && g[i][j] == c)return true; return false;}bool is_valid(char c, int x, int y)//判断该位置是否为合法放置位置{if (c == 'W'){for (int d = 0; d < 8; d++)if (inside(x + dx[d], y + dy[d]) && g[x + dx[d]][y + dy[d]]=='B'){int i = x + dx[d], j = y + dy[d];if (find(d, i, j, 'W'))return true;}return false;}else{for (int d = 0; d < 8; d++)if (inside(x + dx[d], y + dy[d]) && g[x + dx[d]][y + dy[d]] == 'W'){int i = x + dx[d], j = y + dy[d];if (find(d, i, j, 'B'))return true;}return false;}}bool find_pos()//寻找合法放置位置{pos.clear();for (int i = 0; i < 8; i++)for (int j = 0; j < 8; j++)if (g[i][j] == '-'&&is_valid(player, i, j))//枚举每一个位置pos.push_back(P(i + 1, j + 1));int len = pos.size();if (len>0)return true;else return false;}void place(int x, int y)//放置棋子{if (player == 'W'){g[x][y] = 'W';for (int d = 0; d < 8;d++)if (inside(x + dx[d], y + dy[d]) && g[x + dx[d]][y + dy[d]] == 'B'){int i = x + dx[d], j = y + dy[d];if (find(d, i, j, 'W')){int w = dx[d], v = dy[d];while (g[i][j] != 'W'){ g[i][j] = 'W'; i += w; j += v; }}}}else{g[x][y] = 'B';for (int d = 0; d < 8; d++)if (inside(x + dx[d], y + dy[d]) && g[x + dx[d]][y + dy[d]] == 'W'){int i = x + dx[d], j = y + dy[d];if (find(d, i, j, 'B')){int w = dx[d], v = dy[d];while (g[i][j] != 'B'){ g[i][j] = 'B'; i += w; j += v; }}}}}void print_num()//打印黑白棋的个数{int n1, n2;n1 = n2 = 0;for (int i = 0; i < 8;i++)for (int j = 0; j < 8;j++)if (g[i][j] == 'W')n1++;else if (g[i][j]=='B')n2++;printf("Black - %2d White - %2d\n", n2, n1);}void print_board()//打印当前棋盘{for (int i = 0; i < 8; i++)for (int j = 0; j < 8; j++)printf("%c%s", g[i][j], j == 7 ? "\n":"");}int main(){//freopen("t.txt", "r", stdin);int T;cin >> T;int rnd = 0;while (T--){memset(g, '\0', sizeof(g));for (int i = 0; i < 8; i++)scanf("%s", g[i]);cin >> player;char cmd;if (rnd++)cout << endl;while (cin >> cmd){if (cmd == 'L'){if (find_pos()){int len = pos.size();for (int i = 0; i < len; i++)printf("(%d,%d)%c", pos[i].first, pos[i].second, i == len - 1 ? '\n' : ' ');}else puts("No legal move.");}if (cmd == 'M'){char str[10];scanf("%s", str);int r, c;r = str[0] - '0' - 1, c = str[1] - '0' - 1;if (g[r][c]=='-'&&!is_valid(player, r, c)){ player = 'W' + 'B' - player;  }place(r, c);print_num();player = 'W' + 'B' - player;}if (cmd == 'Q'){print_board();break;}}}return 0;}