HDU

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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.

题目大意：两个人在下棋，如果一个o周围全为x则x赢，判断x一步会不会赢。

DFS深度优先搜索：从一种状态出发，不断地转移状态直到状态无法转移，然后回退到之前的状态，直到找到所求的解，是对所有状态的一种遍历列举。

这个题只需要逐一搜索为'o'的点，然后判断其周围有几个‘.’,如果不存在一个‘o’周围只有一个‘.’,则x不会赢，在这里还要注意‘o’连通的问题。起初的时候WA了，仔细分析发现是vis数组初始的地方不对，不是在每组样例初始，而是在发现每一个‘o’之后dfs前初始。

附上AC代码：

#include<bits/stdc++.h>using namespace std;const int maxn=9;int dx[]={0,1,0,-1},dy[]={1,0,-1,0};int vis[maxn][maxn];char c[maxn][maxn]; int po=0; void  dfs(int x,int y){    vis[x][y]=1;    for(int i=0;i<4;i++)    {        int nx=dx[i]+x;        int ny=dy[i]+y;        if(nx<0||nx>8||ny<0||ny>8||vis[nx][ny]==1||c[nx][ny]=='x')            continue;        if(c[nx][ny]=='.')        {            vis[nx][ny]=1;            po++;            continue;        }        dfs(nx,ny);    }}int main(){    int t,cases=0;    cin>>t;    while(t--)    {        int flag=0;        for(int i=0;i<9;i++)            cin>>c[i];        for(int i=0;i<9;i++)            for(int j=0;j<9;j++)            {                if(c[i][j]=='o')                {                    memset(vis,0,sizeof(vis));                    po=0;                    dfs(i,j);                    if(po<2){flag=1;break;}                }            }            if(flag)                printf("Case #%d: Can kill in one move!!!\n",++cases);            else                printf("Case #%d: Can not kill in one move!!!\n",++cases);    }    return 0;}