HDU - 5546 Ancient Go
Here is the rules for ancient go they were playing:
The game is played on a 8×88×8 cell board, the chess can be put on the intersection of the board lines, so there are 9×99×9 different positions to put the chess.
Yu Zhou always takes the black and Su Lu the white. They put the chess onto the game board alternately.
The chess of the same color makes connected components(connected by the board lines), for each of the components, if it's not connected with any of the empty cells, this component dies and will be removed from the game board.
When one of the player makes his move, check the opponent's components first. After removing the dead opponent's components, check with the player's components and remove the dead components.
One day, Yu Zhou was playing ancient go with Su Lu at home. It's Yu Zhou's move now. But they had to go for an emergency military action. Little Qiao looked at the game board and would like to know whether Yu Zhou has a move to kill at least one of Su Lu's chess.
Input
The first line of the input gives the number of test cases, T(1≤T≤100)T(1≤T≤100). TT test cases follow. Test cases are separated by an empty line. Each test case consist of 9 lines represent the game board. Each line consists of 9 characters. Each character represents a cell on the game board. ′.′′.′ represents an empty cell. ′x′′x′ represents a cell with black chess which owned by Yu Zhou. ′o′′o′ represents a cell with white chess which owned by Su Lu.
Output
For each test case, output one line containing Case #x: y, where xx is the test case number (starting from 1) and yy is Can kill in one move!!! if Yu Zhou has a move to kill at least one of Su Lu's components. Can not kill in one move!!! otherwise.
Sample Input
2 .......xo ......... ......... ..x...... .xox....x .o.o...xo ..o...... .....xxxo ....xooo. ......ox. .......o. ...o..... ..o.o.... ...o..... ......... .......o. ...x..... ........o
Sample Output
Case #1: Can kill in one move!!! Case #2: Can not kill in one move!!!
Hint
In the first test case, Yu Zhou has 4 different ways to kill Su Lu's component. In the second test case, there is no way to kill Su Lu's component.
题意：
所给9*9大小的棋盘，'.'为空，问'x'再下一棋，围住一堆'o'（围成的一块是没有'.'的）
思路：
枚举每个空点，判断能否成功，
那么怎么判断是否成功呢？
四个方向走，看做四个块，再写另一个函数，判断块是否封闭
怎么判断是否封闭呢？
块里不能有空点就是封闭，每层函数，上||下||左||右 走一格
1 #include <stdio.h> 2 #include <string.h> 3 4 int n = 9; 5 int t[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}}; 6 char a[15][15]; 7 int vis[15][15]; 8 9 int ff(int x, int y) //递归判断该部分'o'是否完全被围起来了 10 { 11 int i, xx, yy; 12 for(i=0;i<4;i++) 13 { 14 xx = x + t[i][0]; 15 yy = y + t[i][1]; 16 if(xx >= 0 && xx < 9 && yy >= 0 && yy < 9 && vis[xx][yy]==0) 17 { 18 vis[xx][yy] = 1; 19 if(a[xx][yy]=='.') return 1; //有空的位置，则确定没有成功围起来 20 else if(a[xx][yy]=='o'&&ff(xx, yy)) return 1; //点为'o'，需判断周围，再下结论 21 } 22 } 23 return 0; //完全围起 24 } 25 26 int f(int x, int y) 27 { 28 int i, xx, yy; 29 for(i=0;i<4;i++) //四个方向都有围成功的可能 30 { 31 xx = x + t[i][0]; 32 yy = y + t[i][1]; 33 if(xx >= 0 && xx < 9 && yy >= 0 && yy < 9 && a[xx][yy]=='o') 34 { 35 memset(vis, 0, sizeof(vis)); //四个方向的块，分开判断，vis每次都需初始化 36 if(ff(xx, yy)==0) return 1; 37 } 38 } 39 return 0; 40 } 41 42 int main() 43 { 44 int t, i, j, num, ans; 45 scanf("%d", &t); 46 getchar(); 47 num = 0; 48 while(t--) 49 { 50 for(i=0; i<n; i++) 51 { 52 scanf("%s", a[i]); 53 } 54 55 ans = 0; 56 memset(vis, 0, sizeof(vis)); 57 for(i=0; i<n; i++) 58 { 59 for(j=0; j<n; j++) 60 { 61 if(a[i][j]=='.') 62 { 63 a[i][j] = 'x'; 64 if(f(i, j)) 65 { 66 ans = 1; 67 break; 68 } 69 a[i][j] = '.'; 70 } 71 } 72 } 73 if(ans) printf("Case #%d: Can kill in one move!!!\n", ++num); 74 else printf("Case #%d: Can not kill in one move!!!\n", ++num); 75 } 76 return 0; 77 }
