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 10×9 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”.
InputThe 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 (2<=N<=7) 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.OutputFor each test case, if the situation is checkmate, output a single word ‘YES’, otherwise output the word ‘NO’.
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
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 above.
题意：
就是给出一个象棋残局，问是否绝杀，并且黑方只有一个帅（其他规则与象棋走法一致）
分析：
开始存入将能走到的四个点，判断这四个点是否有未被其他棋子将军到的。
一道细节题，不同的人有不同的错误，一些方面没有考虑，或者细节上就是有问题，当然，也有可能是逻辑上的问题。
这份代码是我WA多次后的，做的心烦，导致自己无法静心。下面错误数据就是导致我错N次的原因。
4 1 4
C 3 4
R 5 4
C 7 4
G 10 5
4 1 4
C 3 4
G 10 4
R 10 5
R 7 5
细节决定成败，一个逻辑错误，导致一晚上的心烦。
HDU上数据与本题并不一致，一开始就飞将，本题要输出NO，而HDU上的4121却不用
（这个情况题目没说明，但根据象棋的走法，应该是输出NO的）
1 #include <cstdio>
2 #include <cstdlib>
3 #include <cmath>
4 #include <cstring>
5 #include <time.h>
6 #include <iostream>
7 #include <algorithm>
8 #include <string>
9 #include <vector>
10 #include <map>
11 #include <stack>
12 #include <queue>
13 #include <deque>
14 #include <set>
15 #define ll long long
16 #define eps 1e-6
17 #define pi (acos(-1))
18 #define e exp(1.0)
19 #define cei(n) ((int)ceil((n)))
20 #define rou(n) ((int)round((n)))
21 #define qclear(q) while(!q.empty())q.pop();
22
23 template<class T> T gcd(T a, T b) { return b ? GCD(b, a%b) : a; }
24 template<class T> T lcm(T a, T b) { return a / GCD(a,b) * b; }
25 using namespace std;
26
27 char a[22][22];
28 bool panduan(int x,int y)//判断是否超过中间3*3的格子
29 {
30 if(x<1||x>3||y<4||y>6)return false;
31 return true;
32 }
33 int fx[8][2]={1,2,1,-2,-1,2,-1,-2,2,1,2,-1,-2,1,-2,-1};//马的路线
34 int matui[8][2]={1,1,1,-1,-1,1,-1,-1,1,1,1,-1,-1,1,-1,-1};//马脚的位置
35 bool ma(int x,int y)//判断是否能打到这个点
36 {
37 for(int i=0;i<8;i++)
38 {
39 if(x+fx[i][0]>=1&&y+fx[i][1]>=1&&x+fx[i][0]<=10&&y+fx[i][1]<=10)
40 {
41 if(a[x+fx[i][0]][y+fx[i][1]]==‘H‘&&a[x+matui[i][0]][y+matui[i][1]]==‘.‘)return true;
42 }
43 }
44 return false;
45 }
46 bool kk(int x,int y)//判断车、将、跑是否能打到这个点
47 {
48 int i=x-1,k=0;
49 for(i=x-1;i>=1;i--)
50 {
51 if(((a[i][y]==‘R‘||a[i][y]==‘G‘)&&k==0)||(a[i][y]==‘C‘&&k==1))
52 {
53 return true;
54 }
55 else if(a[i][y]!=‘.‘)
56 {
57 k++;
58 if(k>=2)
59 break;
60 }
61 }
62 i=x+1,k=0;
63 for(i=x+1;i<=10;i++)
64 {
65 if(((a[i][y]==‘R‘||a[i][y]==‘G‘)&&k==0)||(a[i][y]==‘C‘&&k==1))
66 {
67 return true;
68 }
69 else if(a[i][y]!=‘.‘)
70 {
71 k++;
72 if(k>=2)
73 break;
74 }
75 }
76 i=y+1,k=0;
77 for(i=y+1;i<=10;i++)
78 {
79 if(((a[x][i]==‘R‘||a[x][i]==‘G‘)&&k==0)||(a[x][i]==‘C‘&&k==1))
80 {
81 return true;
82 }
83 else if(a[x][i]!=‘.‘)
84 {
85 k++;
86 if(k>=2)
87 break;
88 }
89 }
90 i=y-1,k=0;
91 for(i=y-1;i>=1;i--)
92 {
93 if(((a[x][i]==‘R‘||a[x][i]==‘G‘)&&k==0)||(a[x][i]==‘C‘&&k==1))
94 {
95 return true;
96 }
97 else if(a[x][i]!=‘.‘)
98 {
99 k++;
100 if(k>=2)break;
101 }
102 }
103 return false;
104 }
105 bool fj(int x,int y)//一开始将是否在面对面
106 {
107 for(int i=x+1;i<=10;i++)
108 {
109 if(a[i][y]==‘G‘)return 1;
110 if(a[i][y]!=‘.‘)return 0;
111 }
112 }
113 int main()
114 {
115 #ifndef ONLINE_JUDGE
116 freopen("in.txt","r",stdin);
117 freopen("out.txt","w",stdout);
118 ll _btime=clock();
119 #endif
120 char ch[2];
121 int T,x,y,x1,y1;
122 queue< pair <int,int> >q; pair <int,int> t;
123 while(~scanf("%d%d%d",&T,&x,&y),T||x||y)
124 {
125 //初始化
126 while(!q.empty())q.pop();
127 x1=x;
128 y1=y;
129 memset(a,‘.‘,sizeof(a));
130 a[x][y]=‘#‘;
131 //判断将旁边4个位置
132 if(panduan(x-1,y))
133 {
134 q.push(make_pair(x-1,y));
135 }
136 if(panduan(x+1,y))
137 {
138 q.push(make_pair(x+1,y));
139 }
140 if(panduan(x,y+1))
141 {
142 q.push(make_pair(x,y+1));
143 }
144 if(panduan(x,y-1))
145 {
146 q.push(make_pair(x,y-1));
147 }
148 #ifndef ONLINE_JUDGE
149 for(int i=0;i<kyz.size();i++)
150 {
151 cout<<kyz[i].first<<"*"<<kyz[i].second<<endl;
152 }
153 xintengzhiji,c-free moumingdeguangbi;
154 #endif
155 while(T--)
156 {
157 scanf("%s%d%d",ch,&x,&y);
158 a[x][y]=ch[0];
159 }
160 #ifndef ONLINE_JUDGE
161 for(int i=1;i<=10;i++)
162 {
163 for(int j=1;j<=10;j++)
164 {
165 cout<<a[i][j];
166 }
167 puts("");
168 }
169 #endif
170 if(fj(x1,y1))//一开始飞将的情况
171 {
172 puts("NO");
173 continue;
174 }
175 int llll=0;
176 while(!q.empty())
177 {
178 t=q.front();
179 int xx=t.first;
180 int yy=t.second;
181 if(kk(xx,yy)||ma(xx,yy))//判断这个点是否能走
182 {
183 q.pop();
184 }
185 else{llll=1;
186 break;
187 }
188
189 }
190 //本来不想有llll的，一直错，就改成这样了
191 if(!llll)puts("YES");
192 else puts("NO");
193 }
194 #ifndef ONLINE_JUDGE
195 ll _etime=clock();
196 printf("time = %lld ms",_etime-_btime);
197 #endif
198 return 0;
199 }
200 /*
201 2 1 4
202 R 2 2
203 G 10 5
204
205 2 1 5
206 H 3 5
207 C 10 5
208
209 2 1 4
210 G 10 5
211 R 6 4
212
213 3 1 5
214 H 4 5
215 G 10 5
216 C 7 5
217
218 2 1 5
219 R 4 4
220 G 10 5
221
222 3 1 5
223 G 10 4
224 R 5 5
225 H 3 7
226
227 4 1 5
228 G 10 4
229 C 6 5
230 H 5 5
231 R 1 1
232
233 5 1 5
234 G 10 4
235 C 6 5
236 H 5 5
237 H 4 5
238 R 1 1
239
240 3 1 5
241 G 10 4
242 C 2 7
243 H 3 7
244
245 3 1 5
246 G 10 4
247 R 5 5
248 R 1 6
249
250 4 1 5
251 G 10 4
252 R 5 5
253 R 1 6
254 H 3 7
255
256
257 1
258 1 4
259 G 10 4
260
261 3 1 4
262 G 10 4
263 R 10 3
264 R 10 5
265
266
267 0 0 0
268
269 */