码迷，mamicode.com
首页 > 其他好文 > 详细

poj--2706--Connect(极限大模拟)

时间：2015-01-30 22:41:51      阅读：293      评论：0      收藏：0      [点我收藏+]

标签：

Connect
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 1148   Accepted: 375

Description

技术分享 技术分享 技术分享 技术分享 技术分享
Figure 1 Figure 2 Figure 3a Figure 3b Figure 4

Your task is to decide if a specified sequence of moves in the board game Twixt ends with a winning move.

In this version of the game, different board sizes may be specified. Pegs are placed on a board at integer coordinates in the range [0, N]. Players Black and White use pegs of their own color. Black always starts and then alternates with White, placing a peg at one unoccupied position (x,y). Black‘s endzones are where x equals 0 or N, and White‘s endzones are where y equals 0 or N. Neither player may place a peg in the other player‘s endzones. After each play the latest position is connected by a segment to every position with a peg of the same color that is a chess knight‘s move away (2 away in one coordinate and 1 away in the other), provided that a new segment will touch no segment already added, except at an endpoint. Play stops after a winning move, which is when a player‘s segments complete a connected path between the player‘s endzones.

For example Figure 1 shows a board with N=4 after the moves (0,2), (2,4), and (4,2). Figure 2 adds the next move (3,2). Figure 3a shows a poor next move of Black to (2,3). Figure 3b shows an alternate move for Black to (2,1) which would win the game.

Figure 4 shows the board with N=7 after Black wins in 11 moves:
(0, 3), (6, 5), (3, 2), (5, 7), (7, 2), (4, 4), (5, 3), (5, 2), (4, 5), (4, 0), (2, 4).

Input

The input contains from 1 to 20 datasets followed by a line containing only two zeroes, "0 0". The first line of each dataset contains the maximum coordinate N and the number of total moves M where 3 < N < 21, 4 < M < 250, and M is odd. The rest of the dataset contains a total of M coordinate pairs, with one or more pairs per line. All numbers on a line will be separated by a space. M being odd means that Black will always be the last player. All data will be legal. There will never be a winning move before the last move.

Output

The output contains one line for each data set: "yes" if the last move is a winning move and "no" otherwise.

Sample Input

4 5
0 2 2 4 4 2 3 2 2 3
4 5
0 2 2 4 4 2 3 2 2 1
7 11
0 3 6 5 3 2 5 7 7 2 4 4
5 3 5 2 4 5 4 0 2 4
0 0

Sample Output

no
yes
yes

Source

Mid-Central USA 2005

 

黑白棋的游戏，在n*n的棋盘上，黑棋先手，且最后一步是黑棋。当黑棋从x=0连接的到x=n，时黑棋获胜，问最后一步是不是致胜的一步

每个点都可以和它周围的八个点连接，类似象棋中的马，但是在连接两个点的时候，在连线之间不能有其他的连线（包括黑棋自身的连线）。所以在连接一条线的时候，要判断其他的可能会切割这条线的九条线是否存在。

先判断不包含最后一步的黑棋能不能胜利，再判断加上最后一步后能不能获胜，如果开始不能获胜，后来获胜，那么输出yes，否则，输出no

 

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std ;
int vis[500] ;
int Map[25][25] , c[500][500] ;//ºÚΪ1£¬°×Ϊ-1
queue <int> que ;
void solve1(int i,int j,int n,int temp)
{
    if( (i-1) >= 0 && (i+1) < n && (j-1) >= 0 && c[ (i-1)*n+j ][ (i+1)*n+(j-1) ] != 0 )
        return ;
    if( (i-1) >= 0 && (j-2) >= 0 && (i+1) < n && (j-1) >= 0 && c[ (i-1)*n+(j-2) ][ (i+1)*n+(j-1) ] != 0 )
        return ;
    if( (j-3) >= 0 && (i+1) < n && (j-1) >= 0 && c[ i*n+(j-3) ][ (i+1)*n+(j-1) ] != 0 )
        return ;
    if( (j-1) >= 0 && (i+1) < n && (j+1) < n && c[ i*n+(j-1) ][ (i+1)*n+(j+1) ] != 0 )
        return ;
    if( (j-1) >= 0 && (i+2) < n && c[ i*n+(j-1) ][ (i+2)*n+j ] != 0 )
        return ;
    if( (j-1) >= 0 && (i+2) < n && (j-2) >= 0 && c[ i*n+(j-1) ][ (i+2)*n+(j-2) ] != 0 )
        return ;
    if( (i-1) >= 0 && (j-1) >= 0 && (i+1) < n && c[ (i-1)*n+(j-1) ][ (i+1)*n+j ] != 0 )
        return ;
    if( (i+1) < n && (j-2) >= 0 && c[ (i+1)*n+j ][ i*n+(j-2) ] != 0 )
        return ;
    if( (j-2) >= 0 && (i+2) < n && (j-1) >= 0 && c[ i*n+(j-2) ][ (i+2)*n+(j-1) ] != 0 )
        return ;
    c[ i*n+j ][ (i+1)*n+(j-2) ] = c[ (i+1)*n+(j-2) ][ i*n+j ] = temp ;
    return ;
}
void solve2(int u,int v,int n,int temp)
{
    if( (u-1) >= 0 && (v-1) >= 0 )
    {
        if( (u+1) < n && c[ (u-1)*n+(v-1) ][ (u+1)*n+v ] != 0 ) return ;
        if( (u+1) < n && (v-2) >= 0 && c[ (u-1)*n+(v-1) ][ (u+1)*n+(v-2) ] != 0 ) return ;
        if( (v-3) >= 0 && c[ (u-1)*n+(v-1) ][ u*n+(v-3) ] != 0 ) return ;
    }
    if( (v-1) >= 0 )
    {
        if( (u-1) >= 0 && (v+1) < n && c[ u*n+(v-1) ][ (u-1)*n+(v+1) ] != 0 ) return ;
        if( (u-2) >= 0 && c[ u*n+(v-1) ][ (u-2)*n+v ] != 0 ) return ;
        if( (u-2) >= 0 && (v-2) >= 0 && c[ u*n+(v-1) ][ (u-2)*n+(v-2) ] != 0 ) return ;
    }
    if( (u+1) < n && (v-1) >= 0 && (u-1) >= 0 && c[ (u+1)*n+(v-1) ][ (u-1)*n+v ] != 0 ) return ;
    if( (u-1) >= 0 && (v-2) >= 0 && c[ (u-1)*n+v ][ u*n+(v-2) ] != 0 ) return ;
    if( (v-2) >= 0 && (u-2) >= 0 && (v-1) >= 0 && c[ u*n+(v-2) ][ (u-2)*n+(v-1) ] != 0 ) return ;
    c[ u*n+v ][ (u-1)*n+(v-2) ] = c[ (u-1)*n+(v-2) ][ u*n+v ] = temp ;
    return ;
}
void solve3(int u,int v,int n,int temp)
{
    if( (u+1) < n )
    {
        if( (u-1) >= 0 && (v-1) >= 0 && c[ (u+1)*n+v ][ (u-1)*n+(v-1) ] != 0 ) return ;
        if( (v-2) >= 0 && c[ (u+1)*n+v ][ u*n+(v-2) ] != 0 ) return ;
        if( (u+2) < n && (v-2) >= 0 && c[ (u+1)*n+v ][ (u+2)*n+(v-2) ] != 0 ) return ;
    }
    if( (u+1) < n && (v-1) >= 0 )
    {
        if( (v+1) < n && c[ (u+1)*n+(v-1) ][ u*n+(v+1) ] != 0 ) return ;
        if( (u+2) < n && (v+1) < n && c[ (u+1)*n+(v-1) ][ (u+2)*n+(v+1) ] != 0 ) return ;
        if( (u+3) < n && c[ (u+1)*n+(v-1) ][ (u+3)*n+v ] != 0 ) return ;
    }
    if( (u+1) < n && (v+1) < n && (v-1) >= 0 && c[ (u+1)*n+(v+1) ][ u*n+(v-1) ] != 0 ) return ;
    if( (v-1) >= 0 && (u+2) < n && c[ u*n+(v-1) ][ (u+2)*n+v ] != 0 ) return ;
    if( (u+2) < n && (u+1) < n && (v-2) >= 0 && c[ (u+2)*n+v ][ (u+1)*n+(v-2) ] != 0 ) return ;
    c[ u*n+v ][ (u+2)*n+(v-1) ] = c[ (u+2)*n+(v-1) ][ u*n+v ] = temp ;
    return ;
}
void solve4(int u,int v,int n,int temp)
{
    if( (u-1) >= 0 )
    {
        if( (u-2) >= 0 && (v-2) >= 0 && c[ (u-1)*n+v ][ (u-2)*n+(v-2) ] != 0 ) return ;
        if( (v-2) >= 0 && c[ (u-1)*n+v ][ u*n+(v-2) ] != 0 ) return ;
        if( (u+1) < n && (v-1) >= 0 && c[ (u-1)*n+v ][ (u+1)*n+(v-1) ] != 0 ) return ;
    }
    if( (u-1) >= 0 && (v-1) >= 0 )
    {
        if( (u-3) >= 0 && c[ (u-1)*n+(v-1) ][ (v-3)*n+v ] != 0 ) return ;
        if( (u-2) >= 0 && (v+1) < n && c[ (u-1)*n+(v-1) ][ (u-2)*n+(v+1) ] != 0 ) return ;
        if( (v+1) < n && c[ (u-1)*n+(v-1) ][ u*n+(v+1) ] != 0 ) return ;
    }
    if( (u-1) >= 0 && (v+1) < n && (v-1) >= 0 && c[ (u-1)*n+(v+1) ][ u*n+(v-1) ] != 0 ) return ;
    if( (v-1) >= 0 && (u-2) >= 0 && c[ u*n+(v-1) ][ (u-2)*n+v ] != 0 ) return ;
    if( (u-2) >= 0 && (v-2) >= 0 && c[ (u-2)*n+v ][ (u-1)*n+(v-2) ] != 0 ) return ;
    c[ u*n+v ][ (u-2)*n+(v-1) ] = c[ (u-2)*n+(v-1) ][ u*n+v ] = temp ;
    return ;
}
int bfs(int u,int n)
{
    int v , i , j ;
    while( !que.empty() ) que.pop() ;
    vis[u] = 1 ;
    que.push(u) ;
    while( !que.empty() )
    {
        u = que.front() ;
        que.pop() ;
        if( u >= (n-1)*n )
            return 1 ;
        for(i = 0 ; i < n*n ; i++)
        {
            if( c[u][i] == 1 && vis[i] == 0 )
            {
                vis[i] = 1 ;
                que.push(i) ;
            }
        }
    }
    return 0 ;
}
void solve(int temp,int n)
{
    int u , v ;
    scanf("%d %d", &u, &v) ;
    Map[u][v] = temp ;
    if( (u+1) < n && (v-2) >= 0 && Map[u][v] == Map[u+1][v-2] )
        solve1(u,v,n,temp);
    if( (u-1) >= 0 && (v+2) < n && Map[u][v] == Map[u-1][v+2] )
        solve1(u-1,v+2,n,temp) ;
    if( (u-1) >= 0 && (v-2) >= 0 && Map[u][v] == Map[u-1][v-2] )
        solve2(u,v,n,temp) ;
    if( (u+1) < n && (v+2) < n && Map[u][v] == Map[u+1][v+2] )
        solve2(u+1,v+2,n,temp) ;
    if( (u+2) < n && (v-1) >= 0 && Map[u][v] == Map[u+2][v-1] )
        solve3(u,v,n,temp) ;
    if( (u-2) >= 0 && (v+1) < n && Map[u][v] == Map[u-2][v+1] )
        solve3(u-2,v+1,n,temp) ;
    if( (u-2) >= 0 && (v-1) >= 0 && Map[u][v] == Map[u-2][v-1] )
        solve4(u,v,n,temp) ;
    if( (u+2) < n && (v+1) < n && Map[u][v] == Map[u+2][v+1] )
        solve4(u+2,v+1,n,temp) ;
}
int main()
{
    int n , m , x , y , u , v , temp , i , j ;
    while( scanf("%d %d", &n, &m) != EOF )
    {
        if( m == 0 && n == 0 ) break ;
        n++ ;
        memset(Map,0,sizeof(vis)) ;
        memset(c,0,sizeof(c)) ;
        temp = 1 ;
        m-- ;
        while( m-- )
        {
            solve(temp,n) ;
            temp = -temp ;
        }
        int k1 = 0 , k2 = 0 ;
        memset(vis,0,sizeof(vis)) ;
        for(j = 0 ; j < n ; j++)
        {
            if(  Map[0][j] == 1 && vis[j] == 0 )
            {
                k1 = bfs(j,n) ;
                if( k1 == 1 ) break ;
            }
        }
        solve(temp,n) ;
        memset(vis,0,sizeof(vis)) ;
        for(j = 0 ; j < n ; j++)
        {
            if(  Map[0][j] == 1 && vis[j] == 0 )
            {
                k2 = bfs(j,n) ;
                if( k2 == 1 ) break ;
            }
        }
        if( k1 == 0 && k2 == 1)
            printf("yes\n") ;
        else
            printf("no\n") ;
    }
    return 0;
}

poj--2706--Connect(极限大模拟)

标签：

原文地址：http://blog.csdn.net/winddreams/article/details/43315469
踩
(0)
赞
(0)
   
举报
评论 一句话评论（0）
登录后才能评论！
分享档案
更多>
2021年07月29日 (22)
2021年07月28日 (40)
2021年07月27日 (32)
2021年07月26日 (79)
2021年07月23日 (29)
2021年07月22日 (30)
2021年07月21日 (42)
2021年07月20日 (16)
2021年07月19日 (90)
2021年07月16日 (35)
周排行
mamicode.com排行更多图片
更多
友情链接
兰亭集智   国之画   百度统计   站长统计   阿里云   chrome插件   新版天听网
关于我们 - 联系我们 - 留言反馈
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码！