HDU-5546（简单DFS）

Ancient Go

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2301    Accepted Submission(s): 718

Problem Description

Yu Zhou likes to play Go with Su Lu. From the historical research, we found that there are much difference on the rules between ancient go and modern go.

Here is the rules for ancient go they were playing:

⋅The game is played on a 8×8 cell board, the chess can be put on the intersection of the board lines, so there are 9×9 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 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′ represents a cell with black chess which owned by Yu Zhou. ′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 x is the test case number (starting from 1) and y 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.

题解：

题目大意：

x和o分别代表两个人的棋，游戏规则就是围棋，问x再下一个棋能不能杀死o的至少一个棋子（被杀死的条件就是o完全被x包围，在x的包围圈里已经一个棋都下不了了）。

我的思路：一道简单的dfs，就是找到一个o棋子，如果他旁边只有一个位置可以下，那么就可以被杀死。如果是有好多个连起来的o棋子，就判断这些连起来所有的边界有多少个可以下的位置如果只有一个，就可以被杀死。

下面附上代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;char mp[12][12];int temp, vis[12][12], vis2[12][12];int fx[4][2] = {1,0,-1,0,0,1,0,-1};void dfs(int x, int y){//    if(temp >= 2) return;    for(int i = 0; i < 4; i++)    {        int nx = x+fx[i][0], ny = y+fx[i][1];        if(mp[nx][ny] == '.' && !vis2[nx][ny])        {            temp++;            vis2[nx][ny] = 1;        }        if(mp[nx][ny] == 'o' && !vis[nx][ny])        {            vis[nx][ny] = 1;            dfs(nx, ny);        }    }}int main(){    int T, num = 1;    scanf("%d", &T);    while(T--)    {        int flag = 0;        for(int i = 0; i < 12; i++)            for(int j = 0; j < 12; j++)                mp[i][j] = 'x';        memset(vis, 0, sizeof(vis));        for(int i = 1; i <= 9; i++)            scanf("%s", mp[i]+1);        for(int i = 1; i <= 9; i++)        {            for(int j = 1; j <= 9; j++)            {                if(mp[i][j] == 'o' && !vis[i][j])                {                    memset(vis2, 0, sizeof(vis2));                    vis[i][j] = 1;                    temp = 0;                    dfs(i, j);                    if(temp<=1) // 这个一定是小于等于1，小于1的时候是x不需要下棋o已经被包围了。                    {                        flag = 1;                        break;                    }                }            }            if(flag) break;        }        if(flag) printf("Case #%d: Can kill in one move!!!\n", num++);        else printf("Case #%d: Can not kill in one move!!!\n", num++);    }    return 0;}