Xiangqi is one of the most popular two-player board games in China. The game represents a battle between two armies with the goal of capturing the enemy's ``general" piece. In this problem, you are given a situation of later stage in the game. Besides, the red side has already ``delivered a check". Your work is to check whether the situation is ``checkmate".

Now we introduce some basic rules of Xiangqi. Xiangqi is played on a 10 x 9 board and the pieces are placed on the intersections (points). The top left point is (1,1) and the bottom right point is (10,9). There are two groups of pieces marked by black or red Chinese characters, belonging to the two players separately. During the game, each player in turn moves one piece from the point it occupies to another point. No two pieces can occupy the same point at the same time. A piece can be moved onto a point occupied by an enemy piece, in which case the enemy piece is``captured" and removed from the board. When the general is in danger of being captured by the enemy player on the enemy player's next move, the enemy player is said to have ``delivered a check". If the general's player can make no move to prevent the general's capture by next enemy move, the situation is called ``checkmate".

We only use 4 kinds of pieces introducing as follows:

General: the generals can move and capture one point either vertically or horizontally and cannot leave the `` palace" unless the situation called `` flying general" (see the figure above). ``Flying general" means that one general can ``fly" across the board to capture the enemy general if they stand on the same line without intervening pieces.

Chariot: the chariots can move and capture vertically and horizontally by any distance, but may not jump over intervening pieces

Cannon: the cannons move like the chariots, horizontally and vertically, but capture by jumping exactly one piece (whether it is friendly or enemy) over to its target.

Horse: the horses have 8 kinds of jumps to move and capture shown in the left figure. However, if there is any pieces lying on a point away from the horse horizontally or vertically it cannot move or capture in that direction (see the figure below), which is called `` hobbling the horse's leg".

Now you are given a situation only containing a black general, a red general and several red chariots, cannons and horses, and the red side has delivered a check. Now it turns to black side's move. Your job is to determine that whether this situation is ``checkmate".

Input

The input contains no more than 40 test cases. For each test case, the first line contains three integers representing the number of red pieces N ( 2N7) and the position of the black general. The following N lines contain details of N red pieces. For each line, there are a char and two integers representing the type and position of the piece (type char `G' for general, `R' for chariot, `H' for horse and `C' for cannon). We guarantee that the situation is legal and the red side has delivered the check.

There is a blank line between two test cases. The input ends by `0 0 0'.

Output

For each test case, if the situation is checkmate, output a single word ` YES', otherwise output the word ` NO'.

Hint: In the first situation, the black general is checked by chariot and ``flying general''. In the second situation, the black general can move to (1, 4) or (1, 6) to stop check. See the figure below.

Sample Input

2 1 4
G 10 5
R 6 4 3 1 5
H 4 5
G 10 5
C 7 5 0 0 0

Sample Output

YES
NO
看到这么个问题确实有点头疼,但是仔细想过之后不难用模拟来解决问题,附上代码(略长);
 #include<stdio.h>
#include<string.h>
struct L
{
int a,b;
};
struct L General;
struct L Chariot[];
struct L Horse[];
struct L Cannon[];
int Qipan[][];
int n,x,y;
int tempx,tempy;
char type[];
int i1,i2,i3;
int mainflag;
int checkGenerals(int,int);
int checkChariot(int,int,int,int);
int checkHorse(int,int,int,int);
int checkCannon(int,int,int,int);
int checkPoint(int,int);
int main()
{
while(scanf("%d%d%d",&n,&x,&y)==&&n)
//Pay attention to the point (x,y),which is the black general located.
{
memset(Qipan,,sizeof(Qipan));
//Create a Qipan and define all of it are 0.
Qipan[x][y]=;
// When put some point on the broad,transfer it into 1.
//input the date,and store it into struct General,Chariot,Horse,Cannon.
i1=i2=i3=;
for(int i=;i<n;i++)
{
scanf("%s%d%d",type,&tempx,&tempy);
Qipan[tempx][tempy]=;
if(type[]=='G')
{
General.a=tempx;
General.b=tempy;
}
else if(type[]=='R')
{
Chariot[i1].a=tempx;
Chariot[i1++].b=tempy;
}
else if(type[]=='H')
{
Horse[i2].a=tempx;
Horse[i2++].b=tempy;
}
else if(type[]=='C')
{
Cannon[i3].a=tempx;
Cannon[i3++].b=tempy;
}
}
// Now it's high time to deal with the problem,with all of the dates are stored in order.
// The first we must check whether the two generals face to face directly.
if(checkGenerals(x,y))
{
printf("NO\n");
continue;
}
mainflag=;
if(x+<)
{
Qipan[x][y]=;
Qipan[x+][y]++;
if(checkPoint(x+,y))
mainflag=;
Qipan[x][y]=;
Qipan[x+][y]--;
}
if(mainflag)
{
printf("NO\n");
continue;
}
if(x->)
{
Qipan[x][y]=;
Qipan[x-][y]++;
if(checkPoint(x-,y))
mainflag=;
Qipan[x][y]=;
Qipan[x-][y]--;
}
if(mainflag)
{
printf("NO\n");
continue;
}
if(y+<)
{
Qipan[x][y]=;
Qipan[x][y+]++;
if(checkPoint(x,y+))
mainflag=;
Qipan[x][y]=;
Qipan[x][y+]--;
}
if(mainflag)
{
printf("NO\n");
continue;
}
if(y->)
{
Qipan[x][y]=;
Qipan[x][y-]++;
if(checkPoint(x,y-))
mainflag=;
Qipan[x][y]=;
Qipan[x][y-]++;
}
if(mainflag)
{
printf("NO\n");
continue;
}
printf("YES\n");
}
}
int checkGenerals(int x,int y)
{
if(General.b==y)
{
int Flag=;
for(int i=x+;i<General.a;i++)
if(Qipan[i][y])
Flag++;
if(Flag==)
return ;
}
return ;
}
int checkChariot(int x1,int y1,int x2,int y2)
{
int Flag;
int xj1,yj1,xj2,yj2;
xj1=x1<x2?x1:x2;
xj2=x1>x2?x1:x2;
yj1=y1<y2?y1:y2;
yj2=y1>y2?y1:y2;
if(x1==x2&&y1==y2)
return ;
if(x1==x2)
{
Flag=;
for(int j=yj1+;j<yj2;j++)
if(Qipan[x1][j])
Flag=;
if(Flag==) return ;
}
if(y1==y2)
{
Flag=;
for(int i=xj1+;i<xj2;i++)
if(Qipan[i][y1])
Flag=;
if(Flag==) return ;
}
return ;
}
int checkHorse(int x1,int y1,int x2,int y2)
{
if(x1+==x2&&y1+==y2)
if(Qipan[x1][y1+]==)
return ;
if(x1+==x2&&y1-==y2)
if(Qipan[x1][y1-]==)
return ;
if(x1-==x2&&y1+==y2)
if(Qipan[x1][y1+]==)
return ;
if(x1-==x2&&y1-==y2)
if(Qipan[x1][y1-]==)
return ;
if(x1+==x2&&y1+==y2)
if(Qipan[x1+][y1]==)
return ;
if(x1+==x2&&y1-==y2)
if(Qipan[x1+][y1]==)
return ;
if(x1-==x2&&y1+==y2)
if(Qipan[x1-][y1]==)
return ;
if(x1-==x2&&y1-==y2)
if(Qipan[x1-][y1]==)
return ;
return ;
}
int checkCannon(int x1,int y1,int x2,int y2)
{
int Flag=;
int xj1,yj1,xj2,yj2;
xj1=x1<x2?x1:x2;
xj2=x1>x2?x1:x2;
yj1=y1<y2?y1:y2;
yj2=y1>y2?y1:y2;
if(x1==x2&&y1==y2) return ;
if(x1==x2)
{
for(int j=yj1+;j<yj2;j++)
if(Qipan[x1][j]) Flag++;
if(Flag==) return ;
}
if(y1==y2)
{
for(int i=xj1+;i<xj2;i++)
if(Qipan[i][y1]) Flag++;
if(Flag==) return ;
}
return ;
}
int checkPoint(int x,int y)
{
if(checkGenerals(x,y))
return ;
for(int i=;i<i1;i++)
if(checkChariot(Chariot[i].a,Chariot[i].b,x,y))
return ;
for(int i=;i<i2;i++)
if(checkHorse(Horse[i].a,Horse[i].b,x,y))
return ;
for(int i=;i<i3;i++)
{
if(checkCannon(Cannon[i].a,Cannon[i].b,x,y))
return ;
}
return ;
}

UVA1589 Xiangqi的更多相关文章

  1. [刷题]算法竞赛入门经典(第2版) 4-1/UVa1589 - Xiangqi

    书上具体所有题目:http://pan.baidu.com/s/1hssH0KO 代码:(Accepted,0 ms) //UVa1589 #include<iostream> #incl ...

  2. UVA1589——xiangqi

    开始碰到这个题时觉得太麻烦了直接跳过没做,现在放假了再次看这个题发现没有想象中那么麻烦,主要是题目理解要透彻,基本思路就是用结构体数组存下红方棋子,让黑将军每次移动一下,看移动后是否有一个红方棋子可以 ...

  3. 【习题4-1 Uva1589】Xiangqi

    [链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 车是可以被吃掉的... 注意这个情况. 其他的模拟即可. [代码] #include <bits/stdc++.h> u ...

  4. HDU 4121 Xiangqi 我老了?

    Xiangqi Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Sub ...

  5. HDU 4121 Xiangqi 模拟题

    Xiangqi Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4121 ...

  6. Uva - 1589 - Xiangqi

    Xiangqi is one of the most popular two-player board games in China. The game represents a battle bet ...

  7. [算法竞赛入门经典] 象棋 ACM/ICPC Fuzhou 2011, UVa1589 较详细注释

    Description: Xiangqi is one of the most popular two-player board games in China. The game represents ...

  8. Xiangqi(简单模拟)

    4746: Xiangqi 时间限制(普通/Java):1000MS/3000MS     内存限制:65536KByte 总提交: 15            测试通过:2 描述 Xiangqi i ...

  9. TZOJ 4746 Xiangqi(模拟棋盘数组)

    描述 Xiangqi is one of the most popular two-player board games in China. The game represents a battle ...

随机推荐

  1. 数据挖掘-Python基本模块

    1.numpy:基础模块,高效处理数据.提供数组支持 2.pandas:数据探索和数据分析 3.matplotlib:数据成图模块,解决数据可视化 4.scipy:支持数值计算,支持矩阵运算,提供高等 ...

  2. ubuntu配置android开发环境和编译源码遇到的一些问题

    ---------------------------------------------环境变量设置--------------------------------------------- 1.设 ...

  3. Linux 删除空行

    在Linux上处理一些数据文件时,有时候需要将其中的空行过滤掉,系统中提供的各种工具都可以完成这个功能.将常用的介绍如下吧:1. grep grep . data.txt grep -v '^$' d ...

  4. BaseAdapter自定义适配器——思路详解

    BaseAdapter自定义适配器——思路详解 引言: Adapter用来把数据绑定到扩展了AdapterView类的视图组.系统自带了几个原生的Adapter. 由于原生的Adapter视图功能太少 ...

  5. 能取悦生理期的女性吗?Le Parcel提供女性卫生用品按月订购服务,不是按包出售而是可以按片自由搭配 | 36氪

    能取悦生理期的女性吗?Le Parcel提供女性卫生用品按月订购服务,不是按包出售而是可以按片自由搭配 | 36氪 能取悦生理期的女性吗?Le Parcel提供女性卫生用品按月订购服务,不是按包出售而 ...

  6. qt tablewidget中单个和批量删除代码如下(部分)截图如下

    def coltable(self):#行删除    row=self.downwidget.currentRow()    select=self.downwidget.isItemSelected ...

  7. MVVM架构~knockoutjs系列之表单添加(验证)与列表操作源码开放

    返回目录 本文章应该是knockoutjs系列的最后一篇了,前几篇中主要讲一些基础知识,这一讲主要是一个实际的例子,对于一个对象的添加与编辑功能的实现,并将项目源代码公开了,共大家一起学习! knoc ...

  8. 编写jeb插件打印目标方法的交叉引用

    jeb插件的编写,文档在文件夹\jeb_1.5\doc\apidoc. 我这里的目的是回溯sendTextMessage等敏感api的调用路径,实现代码如下: from jeb.api import ...

  9. Android开发有用的站点

    在github上面找到一个个人认为比較好的站点,好在能够方便下载开发工具.我的AndroidStudio就是在上面下载的.安装了一直在使用.该 网址主要收集整理Android开发所需的Android ...

  10. H264编码技术

    H.264的目标应用涵盖了眼下大部分的视频服务,如有线电视远程监控.交互媒体.数字电视.视频会议.视频点播.流媒体服务等.H.264为解决不同应用中的网络传输的差异.定义了两层:视频编码层(VCL:V ...