2015 CCPC Ancient Go
来源:互联网 发布:淘宝o2o平台 编辑:程序博客网 时间:2024/05/01 01:42
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.
题意:给定一个9*9的围棋局面,要求你先除去对方的死棋,再除去己方的死棋。对于当前局面,问能否通过一手棋吃掉对面至少一个子。
直接暴力,代码很low;
#include <iostream>#include <cstdio>#include <cstring>using namespace std;char a[10][10];int b[10][10];int flag=0;void xm(int x,int y)//消灭x;{ b[x][y]=1; if(a[x+1][y]=='x'&&b[x+1][y]==0) xm(x+1,y); if(a[x][y+1]=='x'&&b[x][y+1]==0) xm(x,y+1); if(a[x][y-1]=='x'&&b[x][y-1]==0) xm(x,y-1); if(a[x-1][y]=='x'&&b[x-1][y]==0) xm(x-1,y); if(a[x+1][y]=='.') flag=1; if(a[x][y+1]=='.') flag=1; if(a[x-1][y]=='.') flag=1; if(a[x][y-1]=='.') flag=1;}void xm1(int x,int y)//消灭o;{ b[x][y]=1; if(a[x+1][y]=='o'&&b[x+1][y]==0) xm1(x+1,y); if(a[x][y+1]=='o'&&b[x][y+1]==0) xm1(x,y+1); if(a[x][y-1]=='o'&&b[x][y-1]==0) xm1(x,y-1); if(a[x-1][y]=='o'&&b[x-1][y]==0) xm1(x-1,y); if(a[x+1][y]=='.') flag=1; if(a[x][y+1]=='.') flag=1; if(a[x-1][y]=='.') flag=1; if(a[x][y-1]=='.') flag=1;}int main(){ int T; cin>>T; for(int k=1; k<=T; k++) { for(int i=0; i<9; i++) { for(int j=0; j<9; j++) { cin>>a[i][j]; } } for(int i=0; i<9; i++)//先把死掉的x给清除,变为o; { for(int j=0; j<9; j++) { if(a[i][j]=='x') { flag=0; memset(b,0,sizeof(b)); xm(i,j); if(flag==0) a[i][j]='o'; } } } int t=0; for(int i=0; i<9; i++) { for(int j=0; j<9; j++) { if(a[i][j]=='.')//把能下子的地方赋为x;看o是否死掉; { a[i][j]='x'; if(a[i+1][j]=='o') { flag=0; memset(b,0,sizeof(b)); xm1(i+1,j); if(flag==0) t=1; } if(a[i][j+1]=='o') { flag=0; memset(b,0,sizeof(b)); xm1(i,j+1); if(flag==0) t=1; } if(a[i-1][j]=='o') { flag=0; memset(b,0,sizeof(b)); xm1(i-1,j); if(flag==0) t=1; } if(a[i][j-1]=='o') { flag=0; memset(b,0,sizeof(b)); xm1(i,j-1); if(flag==0) t=1; } a[i][j]='.'; } } } if(t==1) printf("Case #%d: Can kill in one move!!!\n",k); if(t==0) printf("Case #%d: Can not kill in one move!!!\n",k); } return 0;}
- 2015 CCPC Ancient Go
- 2015 南阳理工CCPC Ancient Go
- 2015 南阳 CCPC G.Ancient Go
- Ancient Go【ccpc dfs】
- 2015 南阳 CCPC hdu 5546 Ancient Go(DFS,暴力)
- HDU5546 Ancient Go(深搜DFS)(2015CCPC)
- 南阳CCPC G题 Ancient Go
- hdu 5546 Ancient Go(2016ccpc) 暴力DFS
- 【HDU5546 2015 CCPC 南阳国赛G】【DFS】Ancient Go 棋盘围杀 优化写法O(n^2)
- Ancient Go
- Ancient Go
- Ancient Go
- ccpc Ancient Go(dfs求连通块周围'.'的数目)
- uestc 1221 Ancient Go
- UESTC 1221 Ancient Go
- hdu 5546 Ancient Go
- dfs ancient go
- HDU 5546 Ancient Go
- 【bzoj2982】combination
- eclipse没有(添加)”Dynamic Web Project”选项的方法
- Spring @Transactional 到底是怎么工作的?
- 30.自定义UITableViewCell第一章
- 《Linux命令、编辑器与Shell编程》读书笔记4.3-其他数据文件处理命令(tr,sort,cut,paste,join,uniq,split)
- 2015 CCPC Ancient Go
- 2016 JAVA与Android面试题整理
- C#(WPF)按钮图片不显示,壁纸切换功能
- java集合
- magento 中常见的js冲突
- Node.js(4) -- 七天学会node.js(2)
- hdu 1385 Minimum Transport Cost(Floyd打印路径)
- vijos--P1211--生日日数(纯模拟)
- MIT算法导论-第11讲-动态规划