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 and output
| Sample Input | Sample Output |
|---|---|
2 .......xo ......... ......... ..x...... .xox....x .o.o...xo ..o...... .....xxxo ....xooo. ......ox. .......o. ...o..... ..o.o.... ...o..... ......... .......o. ...x..... ........o |
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.
#include<cstdio>
#include<cstring>
#include<stack>
#include<vector>
#include<queue>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
const int oo = 1e9+7;
const int maxn = 1e6+7;
typedef long long LL;
int dir[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};
int times, vis[200][200];
char maps[30][30];
void judge(int x, int y, int & cnt)
{
int i, sx, sy;
vis[x][y] = 1;
for(i = 0; i < 4; i++)
{
sx = x+dir[i][0];
sy = y+dir[i][1];
if(maps[sx][sy] == '.' && vis[sx][sy] != times)///一个点不能被多次访问
{
cnt++;
vis[sx][sy] = times;///标记这个点被访问的时间
}
if(maps[sx][sy] == 'o' && !vis[sx][sy])
judge(sx, sy, cnt);
}
}
int solve()
{
memset(vis, 0, sizeof(vis));
int i, j, cnt; times= 0;
for(i = 1; i <= 9; i++)
{
for(j = 1; j <= 9; j++)
{
if(maps[i][j] == 'o' && !vis[i][j])
{
++times;
cnt = 0;///统计‘o'周围可以放多少个‘x'使’o'被围住。
judge(i, j, cnt);
if(cnt == 1) return 1;
}
}
}
return 0;
}
int main()
{
int T, i, cas=1;
scanf("%d", &T);
while(T--)
{
for(i = 1; i <= 9; i++)
scanf("%s", maps[i]+1);
if(solve()) printf("Case #%d: Can kill in one move!!!
", cas++);
else printf("Case #%d: Can not kill in one move!!!
", cas++);
}
return 0;
}
/*
【题意】
给出一个9*9的棋盘，棋盘上有——
x（我方棋子）
o（敌方棋子）
.（空白位置）
现在问你，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'.'
【分析】
这题棋盘这么小。
于是我们可以直接枚举'.'填放。
更高效的是，我们枚举'o'周围的'.'填放。
然后放完'x'之后，再dfs看看这个'o'能否搜到'.'。
搜的到的话，这个放置就是失败的。
搜不到的话，这个放置就是成功的。
这题的一个优化策略是，对于每个'o'的联通块，我看其周围能探索到的'.'的个数，如果个数恰为1，那么就找到.
然而要小心重复计数。时间复杂度就可以降为O(n^2)
【时间复杂度&&优化】
O(n^2)
*/
#include<cstdio>
#include<cstring>
#include<stack>
#include<vector>
#include<queue>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
const int oo = 1e9+7;
const int maxn = 3*1e6+7;
int dir[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};
int vis[20][20];
char maps[20][20];
int judge(int x, int y)
{
vis[x][y] = 1;
int i, sx, sy;
for(i = 0; i < 4; i++)
{
sx = x + dir[i][0];
sy = y + dir[i][1];
if(sx < 0 || sx >= 9 || sy < 0 || sy >= 9 || vis[sx][sy]) continue;
if(maps[sx][sy] == '.') return 0;
if(maps[sx][sy] == 'o' && !judge(sx, sy)) return 0;
}
return 1;
}
int dfs(int x, int y)
{
int i;
for(i = 0; i < 4; i++)
{
int sx, sy;
sx = x+dir[i][0];
sy = y+dir[i][1];
if(maps[sx][sy] == 'o')
if(sx >= 0 && sx < 9 && sy >= 0 && sy < 9)
{
memset(vis, 0, sizeof(vis));
if(judge(sx, sy))
return 1;
}
}
return 0;
}
int main()
{
int T, i, j, cas = 1, ok;
scanf("%d", &T);
while(T--)
{
ok = 0;
for(i = 0; i < 9; i++)
scanf("%s", maps[i]);
for(i = 0; i < 9; i++)
{
for(j = 0; j < 9; j++)
{
if(maps[i][j] == '.')
{
maps[i][j] = 'x';
ok = dfs(i, j);
maps[i][j] = '.';
}
if(ok == 1) break;
}
if(ok == 1)break;
}
if (ok == 1) printf ("Case #%d: Can kill in one move!!!
", cas++);
else printf ("Case #%d: Can not kill in one move!!!
", cas++);
}
return 0;
}