看到这个题目基本的想法是，将黑将能走的四个方向都遍历一遍，然后再判断判断红方能否吃掉它。如果所有可能的走向都会被将死，则黑将被将死。需要注意越界，吃子，和马被别脚的情况。附上一张自己画的一张简图可能会清楚一些：
思路导图
下面贴一下代码，更详细的解释在代码的注释里，代码有点冗长，有能力的可以把相同的地方写成函数（比如说帅和车的判断是可以放一起的）
#include<iostream>
#include<fstream>
using namespace std;
struct qizhi
{
char kind;
int row;
int column;
};
int panduan(int b_row, int b_column, int size, const qizhi* p) //用来判断黑将走某一步会不会被将死,如果会死返回0，不会返回1
{
for (int i = 0; i < size; ++i)
{
switch (p[i].kind)
{
case 'G':
if (p[i].column == b_column)
{
int key = 0; //标记，
for (int j = 0; j < size; ++j) //如果黑将和红帅在同列，判断中间有没有其他的旗子
{
if (j == i)
continue;
if (p[j].column == p[i].column&&p[j].row > b_row&&p[j].row < p[i].row)
{
key = 1;
break;
}
}
if (0 == key)
return 0; //如果key为0，表面被将死直接返回0
}
break;
case 'R':
if (p[i].column == b_column)
{
int key = 0, key1; //标记，
p[i].row > b_row ? key1 = 1 : key1 = 0;
for (int j = 0; j < size; ++j) //如果黑将和车在同列，判断中间有没有其他的旗子
{
if (j == i)
continue;
if (key1 == 1)
{
if (p[j].column == p[i].column&&p[j].row > b_row&&p[j].row < p[i].row)
{
key = 1;
break;
}
}
else
{
if (p[j].column == p[i].column&&p[j].row <b_row&&p[j].row > p[i].row)
{
key = 1;
break;
}
}
}
if (0 == key)
return 0; //如果key为0，表明被将死直接返回0
}
if (p[i].row == b_row)
{
int key = 0, key1; //标记，
p[i].column > b_column ? key1 = 1 : key1 = 0;
for (int j = 0; j < size; ++j) //如果黑将和车在同行，判断中间有没有其他的旗子
{
if (j == i)
continue;
if (key1 == 1)
{
if (p[j].row == p[i].row&&p[j].column > b_column&&p[j].column < p[i].column)
{
key = 1;
break;
}
}
else
{
if (p[j].row == p[i].row&&p[j].column < b_column&&p[j].column > p[i].column)
{
key = 1;
break;
}
}
}
if (0 == key)
return 0; //如果key为0，表明被将死直接返回0
}
break;
case 'C':
if (p[i].column == b_column)
{
int key = 0, key1; //标记，
p[i].row > b_row ? key1 = 1 : key1 = 0;
for (int j = 0; j < size; ++j) //如果黑将和炮在同列，判断中间有多少个旗子
{
if (j == i)
continue;
if (1 == key1)
{
if (p[j].column == p[i].column&&p[j].row > b_row&&p[j].row < p[i].row)
{
key += 1;
}
}
else
{
if (p[j].column == p[i].column&&p[j].row<b_row&&p[j].row>p[i].row)
{
key += 1;
}
}
}
if (1 == key)
return 0; //如果key为1，表明被将死直接返回0
}
if (p[i].row == b_row)
{
int key = 0, key1; //标记，
p[i].column > b_column ? key1 = 1 : key1 = 0;
for (int j = 0; j < size; ++j) //如果黑将和炮在同行，判断中间有多少个旗子
{
if (j == i)
continue;
if (1 == key1)
{
if (p[j].row == p[i].row&&p[j].column > b_column&&p[j].column < p[i].column)
{
key += 1;
}
}
else
{
if (p[j].row == p[i].row&&p[j].column<b_column&&p[j].column>p[i].column)
{
key += 1;
}
}
}
if (1 == key)
return 0; //如果key为1，表明被将死直接返回0
}
break;
case 'H':
int up = 0, down = 0, left = 0, right = 0;
for (int j = 0; j < size; ++j) //判断向上是否会别马脚
{
if (j == i)
continue;
if (p[j].row == p[i].row - 1 && p[j].column == p[i].column)
{
up = 1;
break; //如果别马脚，就结束
}
}
if (0 == up)
{
if (p[i].column - 1 == b_column&&p[i].row - 2 == b_row)
return 0;
else if (p[i].column + 1 == b_column&&p[i].row - 2 == b_row)
return 0;
}
for (int j = 0; j < size; ++j) //判断向左是否会别马脚
{
if (j == i)
continue;
if (p[j].row == p[i].row && p[j].column == p[i].column - 1)
{
left = 1;
break; //如果别马脚，就结束
}
}
if (0 == left)
{
if (p[i].column - 2 == b_column&&p[i].row - 1 == b_row)
return 0;
else if (p[i].column - 2 == b_column&&p[i].row + 1 == b_row)
return 0;
}
for (int j = 0; j < size; ++j) //判断向右是否会别马脚
{
if (j == i)
continue;
if (p[j].row == p[i].row && p[j].column == p[i].column + 1)
{
right = 1;
break; //如果别马脚，就结束
}
}
if (0 == right)
{
if (p[i].column + 2 == b_column&&p[i].row - 1 == b_row)
return 0;
else if (p[i].column + 2 == b_column&&p[i].row + 1 == b_row)
return 0;
}
for (int j = 0; j < size; ++j) //判断向下是否会别马脚
{
if (j == i)
continue;
if (p[j].row == p[i].row + 1 && p[j].column == p[i].column)
{
down = 1;
break; //如果别马脚，就结束
}
}
if (0 == down)
{
if (p[i].column - 1 == b_column&&p[i].row + 2 == b_row)
return 0;
else if (p[i].column + 1 == b_column&&p[i].row + 2 == b_row)
return 0;
}
break;
}
}
return 1;
}
int yes_or_no(int b_row, int b_column, int size, qizhi* p) //判断有没有被将死，将死返回false，没死返回true
{
int up = 0, down = 0, left = 0, right = 0;
if (b_row - 1 > 0) //判断向上走会不会死
{
for (int i = 0; i < size; ++i)
{
if (p[i].column == b_column&&p[i].row == b_row - 1) //判断黑将去的地方有没有红棋的子如果有吃掉
{
p[i].kind = ' ';
break;
}
}
up = panduan(b_row - 1, b_column, size, p);
}
if (b_row + 1 < 4) //判断向下走会不会死
{
for (int i = 0; i < size; ++i)
{
if (p[i].column == b_column&&p[i].row == b_row + 1) //判断黑将去的地方有没有红棋的子如果有吃掉
{
p[i].kind = ' ';
break;
}
}
down = panduan(b_row + 1, b_column, size, p);
}
if (b_column - 1 > 3) //判断向左走会不会死
{
for (int i = 0; i < size; ++i)
{
if (p[i].column == b_column - 1 && p[i].row == b_row) //判断黑将去的地方有没有红棋的子如果有吃掉
{
p[i].kind = ' ';
break;
}
}
left = panduan(b_row, b_column - 1, size, p);
}
if (b_column + 1 < 7) //判断向右走会不会死
{
for (int i = 0; i < size; ++i)
{
if (p[i].column == b_column + 1 && p[i].row == b_row) //判断黑将去的地方有没有红棋的子如果有吃掉
{
p[i].kind = ' ';
break;
}
}
right = panduan(b_row, b_column + 1, size, p);
}
return up + down + left + right;
}
int main()
{
///*文件重定向，输出测试*/
//ifstream fin;
//fin.open("data.in");
//cin.rdbuf(fin.rdbuf());
//ofstream out;
//out.open("data.out");
//cout.rdbuf(out.rdbuf());
int r_number = 0, b_row = 0, b_column = 0; //红方棋字个数，黑方将的位置
while (cin >> r_number >> b_row >> b_column&&r_number != 0 && b_column != 0 && b_row != 0)
{
auto p = new qizhi[r_number];
for (int i = 0; i < r_number; ++i)
{
cin >> p[i].kind >> p[i].row >> p[i].column;
}
if (yes_or_no(b_row, b_column, r_number, p))
cout << "NO" << endl;
else
cout << "YES" << endl;
delete[] p;
}
}
