Ancient Go（模拟+搜索）

Ancient Go

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)

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：给你一个9x9的方阵，让你判断其中是不是至少存在一个O，他的周围被X环绕或为边界，只有一个方向是'.'

     思路：
            首先用个二维矩阵，存图，遍历每一个O，遍历到一个O就判断它是否符合条件，符合直接跳出，否
            则遍历到底……



代码如下

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<stdlib.h>
#include<time.h>
#include<string>
#include<math.h>
#include<map>
#include<queue>
#include<stack>
#define INF 0x3f3f3f3f
#define ll long long
#define For(i,a,b) for(int i=a;i<b;i++)
#define sf(a)  scanf("%d",&a)
#define sfs(a)  scanf("%s",a)
#define sff(a,b)  scanf("%d%d",&a,&b)
#define sfff(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define pf(a) printf("%d\n",a)
#define P()  printf("\n")
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
char w[120][120];
int v[120][120];
int d[4][2]= {0,1,1,0,0,-1,-1,0};
int check(int x,int y)                 // 判断此时的坐标是否符合范围和其他条件
{
    if(x>=0&&x<9&&y>=0&&y<9)
    {
        if(w[x][y]!='x')
            return 1;
        else return 0;
    }
    else return 0;
}
int ans=0;
int find(int x,int y)                  // 查找此时O点有几个出口
{
    v[x][y]=1;
    int s=ans;
    for(int i=0; i<4; i++)
    {

        int xx=x+d[i][0],yy=y+d[i][1];
        if(check(xx,yy)&&!v[xx][yy])
        {
            v[xx][yy]=1;
            if(w[xx][yy]=='.')
                s++;
            ans=max(ans,s);
            if(w[xx][yy]=='o')
            {
                s=max(s,find(xx,yy));
            }
        }
    }
    return s;
}
int main()
{
    int t,s=0;
    sf(t);
    while(t--)
    {
        s++;
        For(i,0,9)sfs(w[i]);
        int flag=0;
        for(int i=0; i<9; i++)
        {
            for(int j=0; j<9; j++)
            {
                if(w[i][j]=='o')        // 遍历
                {
                    mem(v,0);
                    ans=0;
                    int d=find(i,j);
                    if(d==1)            // 找到跳出
                    {
                        flag=1;
                        break;
                    }
                }
            }
            if(flag)                    // 找到跳出
                break;
        }
        printf("Case #%d: ",s);
        if(flag)
            printf("Can kill in one move!!!\n");
        else
            printf("Can not kill in one move!!!\n");
    }
}