Uva1589 - Xiangqi

思路：把‘黑将’不能走的地方用'*'填掉，标记在map[][]上，map1[][]上存储棋子的位置。下棋中一般不会出现的情况如

3 1 5

R 4 4

R 4 6

G 10 5

这时黑棋未被将死.这种情况应该打印‘NO’，但原题的测试样例没有管这种不合下棋常理的情况，所以能AC.

开始输入用getchar(),老是有问题，改成cin>>...后就AC了。。。

//#define LOCAL
#include<stdio.h>
#include<string.h>
#include<time.h>
#include<iostream>
using namespace std;
int N, bgx, bgy, gx, gy;
char map[15][15];
char map1[15][15];
struct C{
	char ch;
	int x, y;
}C1,C2;
struct H
{
	char ch;
	int x, y;
}H1,H2;
struct R
{
	char ch;
	int x, y;
}R1,R2;
//输出中间结果调试用
void myprintf(char map[][15]){
	for (int i = 1; i <= 10; i++){
		for (int j =1; j <= 9; j++){
			printf("%c ", map[i][j]);
		}
		printf("\n");
	}
	printf("\n");
}
//马：先判断马是否为'hobbled horse'，再判断填'*'处是否越界
void Horse(H H3){
	if (H3.ch!='0')
	{
		if (map1[H3.x + 1][H3.y] != '1'){
			if (H3.y - 1>0) map[H3.x + 2][H3.y - 1] = '*';
			 map[H3.x + 2][H3.y + 1] = '*';
		}
		if (map1[H3.x - 1][H3.y] != '1'){
			if (H3.x - 2>0 && H3.y - 1>0) map[H3.x - 2][H3.y - 1] = '*';
			if (H3.x - 2>0) map[H3.x - 2][H3.y + 1] = '*';
		}
		if (map1[H3.x][H3.y - 1] != '1')	{
			if (H3.y - 2>0) map[H3.x + 1][H3.y - 2] = '*';
			if (H3.x - 1>0 && H3.y - 2>0)	map[H3.x - 1][H3.y - 2] = '*';
		}
		if (map1[H3.x][H3.y + 1] != '1') {
			map[H3.x + 1][H3.y + 2] = '*';
			if (H3.x - 1>0) map[H3.x - 1][H3.y + 2] = '*';
		}
	}
	
}
void CR(R R3){
	if (R3.ch!='0'){
		for (int i = R3.x - 1; i >0; i--){
			 map[i][R3.y] = '*';
			 if (map1[i][R3.y] == '1') break;
		}
		for (int i = R3.x + 1;  i <4; i++){
			map[i][R3.y] = '*';
			if (map1[i][R3.y] == '1') break;
		}
		for (int i = R3.y - 1;  i >3; i--){
			map[R3.x][i] = '*';
			if (map1[R3.x][i] == '1') break;
		}
		for (int i = R3.y + 1;  i<7; i++){
			map[R3.x][i] = '*';
			if (map1[R3.x][i] == '1') break;
		}
	}
}
void Cannon(C C3){
	if (C3.ch != '0'){
		if (C3.x>bgx){
			for (int i = C3.x-1;i>0; i--){
				int d4=0;
				if (map1[i][C3.y] == '1'){
					i--;
					for (i; i > 0; i--){						
						map[i][C3.y] = '*';
						if (map1[i][C3.y] == '1'){
							d4 = 1; break;
						}
					}
				}
				if (d4) break;
			}
		}
		else if (C3.x < bgx){
			if (map1[C3.x + 1][C3.y] == '1'&&map1[C3.x + 2][C3.y] != '1')
				map[C3.x + 2][C3.y] = '*';
		}
		if (C3.y < bgy){
			for (int i = C3.y+1; i<10; i++){
				int d4 = 0;
				if (map1[C3.x][i] == '1'){
					i++;
					for (i ; i <10; i++){
						map[C3.x][i] = '*';
						if (map1[C3.x][i] == '1'){
							d4 = 1; break;
						}
					}
				}
				if (d4) break;
			}
		}
		else if (C3.y > bgy){
			for (int i = C3.y-1; i>0; i--){
				int d4 = 0;
				if (map1[C3.x][i] == '1'){
					i--;
					for (i ; i > 0; i--){	
						map[C3.x][i] = '*';
						if (map1[C3.x][i] == '1'){
							d4 = 1; break;
						}
					}
				}
				if (d4) break;
			}
		}
	}
}
int main(){
#ifdef LOCAL
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
#endif
	while (scanf("%d%d%d", &N, &bgx, &bgy) == 3 && N ||bgx||bgy){
		memset(map, '0', sizeof(map));
		memset(map1, '0', sizeof(map1));
		bool d2=1;
		int first1,first2,first3;
		first1 = first2 = first3 = 1;
		C1.ch = C2.ch = H1.ch = H2.ch = R1.ch = R2.ch = '0';
		while (N--){
			char ch;
			cin >> ch;
			int x, y;
			if (ch == 'G'){
				scanf("%d%d", &gx, &gy);
				map1[gx][gy] = '1';
			}
			else if (ch == 'R'){
				scanf("%d%d", &x, &y);
				map1[x][y] = '1'; 
				if (first1) { first1 = 0; R1.ch = 'R', R1.x = x, R1.y = y; }
				else{ R2.ch = 'R', R2.x = x,R2.y = y; }
			}
			else if (ch == 'H'){
				scanf("%d%d", &x, &y);
				map1[x][y] = '1';
				if (first2) { first2 = 0; H1.ch = 'H', H1.x = x,H1.y = y; }
				else{ H2.ch = 'H',H2.x = x,H2.y = y; }
			}
			else if (ch == 'C'){
				scanf("%d%d", &x, &y);
				map1[x][y] = '1';
				if (first3) { first3 = 0; C1.ch = 'C', C1.x = x, C1.y = y; }
				else{ C2.ch = 'C', C2.x = x, C2.y = y; }
			}
		}
		//myprintf(map1);
		int d1 = 0;
		for (int i = gx-1; i >0; i--){									
			if (map1[i][gy] == '1'){
				d1 = i; break;
			}
		}
		if (d1 == bgx) d1 = 0;
		if (!d1){
			for (int i = 1; i < 4; i++){
				map[i][gy] = '*';			//将与帅相对的情况
			}
		}
		else {
			for (int i = d1; i < 4; i++){
				map[i][gy] = '*';
			}
		}
		Horse(H1);
		Horse(H2);
		CR(R1);
		CR(R2);
		Cannon(C1); 
		Cannon(C2);
		//myprintf(map);
		//把将的palace用'*'围起来，方便下面判断
		for (int i = 3; i <= 7; i++){
			map[0][i] = '*';
			map[4][i] = '*';
		}
		for (int i = 1; i <= 3; i++){
			map[i][3] = '*';
			map[i][7] = '*';
		}
		if (map[bgx][bgy-1] == '*'&&map[bgx][bgy+1] == '*'&&map[bgx-1][bgy] == '*'&&map[bgx+1][bgy] == '*')
			d2 = 0;
		//myprintf(map);
		if (!d2) printf("YES\n");
		else printf("NO\n");
	}
	//printf("Time used=%.3f\n", (double)clock() / CLOCKS_PER_SEC);
	return 0;
}