• 广搜(棋)


    题目网址: http://acm.uestc.edu.cn/#/problem/show/1221

    Ancient Go

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

    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(1T100). 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 InputSample 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.

    Source

    The 2015 China Collegiate Programming Contest
     
    #include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <cstdio>
    #include <cmath>
    ///http://acm.hust.edu.cn/vjudge/contest/view.action?cid=100505#problem/D
    using namespace std;
    char map[15][15];
    bool vis[15][15];
    int qx[100],qy[100];
    int dir[4][2]= {{1,0},{-1,0},{0,1},{0,-1}};
    int s1,s2;
    int sum=0;
    
    int bfs(int i,int j)
    {
        int flag;
        flag=1;
        sum=0;
        int l=0,r=0;
        qx[r]=i;
        qy[r++]=j;
        vis[i][j]=1;
        while(l<r)
        {
            int curx=qx[l],cury=qy[l++];
            for(int t=0; t<4; t++)
            {
                int nx=curx+dir[t][0];
                int ny=cury+dir[t][1];
                if(sum>=1&&nx>=0&&nx<=8&&ny>=0&&ny<=8&&!vis[nx][ny]&&map[nx][ny]=='.')
                {
                    flag=0;
                    vis[s1][s2]=0;
                }
                if(nx>=0&&nx<=8&&ny>=0&&ny<=8&&!vis[nx][ny]&&map[nx][ny]=='o')
                {
                    vis[nx][ny]=1;
                    qx[r]=nx;
                    qy[r++]=ny;
                }
                if(nx>=0&&nx<=8&&ny>=0&&ny<=8&&!vis[nx][ny]&&map[nx][ny]=='.')
                {
                    if(flag)
                    {
                        sum++;
                        s1=nx;
                        s2=ny;
                        vis[nx][ny]=1;
                    }
                }
            }
        }
        if(flag)
        return 1;
        return 0;
    }
    
    int main()
    {
        int T,ca=1;
        int i,j,f;
        cin>>T;
        while(T--)
        {
            f=0;
            for(i=0;i<9;i++)
            cin>>map[i];
            memset(vis,0,sizeof(vis));
            for(i=0;i<9;i++)
            for(j=0;j<9;j++)
            {
                if(map[i][j]=='o'&&!vis[i][j])
                {
                    f=bfs(i,j);
                    if(f) goto endw;
                }
            }
            endw:;
            if(f)
                printf("Case #%d: Can kill in one move!!!
    ",ca++);
            else
                printf("Case #%d: Can not kill in one move!!!
    ",ca++);
        }
        return 0;
    }
  • 相关阅读:
    蓝盾杯writeup
    记一次被吊打的排位赛(writeup)
    记一次简单的PHP代码审计(SSRF案例)
    记一次简单的GetShell案例
    斯坦福cs231n计算机视觉经典课程笔记(更新中)
    centos8 下配置 uwsgi + Django
    C++坑点随笔
    matlab调教日记 ---- 语法问题汇总
    matlab调教日记 --- debug篇
    MySQL解决中文编码问题
  • 原文地址:https://www.cnblogs.com/chen9510/p/5027104.html
走看看 - 开发者的网上家园