HDOJ 5546 Ancient Go (DFS)

Ancient Go

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 49    Accepted Submission(s): 25

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 are9×9 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.题意：一个9*9的棋盘，判断能否把接下来的一个x，放到地图中的一个空白位置上。 使得出现一个联通块内的o，找不到任何一个可以扩展的联通块位置'.' 思路：枚举o周围的x放置，如果周围可放置数为1，表示成功，注意重复记录，所以用两个数组个标记ac代码： 
#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>#include<algorithm>#define MAXN 100100#define MOD 1000000007#define LL long long#define INF 0xfffffffusing namespace std;char map[10][10];int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};int v[10][10];int v1[10][10];int cnt;int check(int x,int y){if(x<0||x>=9||y<0||y>=9)return 1;return 0;}void dfs(int x,int y){for(int i=0;i<4;i++){int nx=x+dir[i][0];int ny=y+dir[i][1];if(check(nx,ny)==0&&map[nx][ny]=='.'&&!v1[nx][ny]){v1[nx][ny]=1;cnt++;}}for(int i=0;i<4;i++){int nx=x+dir[i][0];int ny=y+dir[i][1];if(check(nx,ny)==0&&map[nx][ny]=='o'&&!v[nx][ny]){v[nx][ny]=1;dfs(nx,ny);}}}int main(){    int t,n,i,j;    int cas=0;    scanf("%d",&t);    while(t--)    {    for(i=0;i<9;i++)    scanf("%s",map[i]);    int bz=0;memset(v,0,sizeof(v));    for(i=0;i<9;i++)    {    for(j=0;j<9;j++)    {    if(map[i][j]=='o'&&!v[i][j])    {    memset(v1,0,sizeof(v1));    cnt=0;    v[i][j]=1;    dfs(i,j);    if(cnt==1)    {    bz=1;    break;}    }}if(bz)break;}if(bz)printf("Case #%d: Can kill in one move!!!\n",++cas);elseprintf("Case #%d: Can not kill in one move!!!\n",++cas);}    return 0;}