CodeForces - 734D Anton and Chess （BFS搜索+技巧）_codeforces 734d anton and chess(1)

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2
4 2
R 1 1
B 1 5





 Output

YES





 Input

2
4 2
R 3 3
B 1 5





 Output







 Note 
 
Picture for the first sample: 



![](http://codeforces.com/predownloaded/94/08/9408be50f47e693db5ec6e7290235a3bc878e147.png)
 White king is in check, because the black bishop can reach the cell with the white king in one move. The answer is " 
 YES". 
 Picture for the second sample: 



![](http://codeforces.com/predownloaded/2b/62/2b625b4a73e1b0f091760b1a5d2c6e549be6a8d7.png)
 Here bishop can't reach the cell with the white king, because his path is blocked by the rook, and the bishop cant "leap" over it. Rook can't reach the white king, because it can't move diagonally. Hence, the king is not in check and the answer is " 
 NO". 
   

  
 题目大意，给你一个白色的王，问在同一水平同一垂直上是否有黑车或者黑皇后（不能有其他棋子挡住） 
   
 在同一对角线上是否有黑象或者黑皇后（不能有其他棋子挡住） 
   

  
 思路 
   
 直接找白王周边的棋子，取各个方向最近的棋子，判断这些棋子是否满足上文说的下棋方向。 
   

  
 膜拜一位大佬写的简单代码，自己在他的基础上稍改动一点，看上去很清爽

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
const double eps=1e-8;
const double PI=acos(-1.0);
using namespace std;
int xx,yy;
struct point{
char kind;
int x,y;
}rounds[8];

int dir(point now){
if(now.xxx&&now.y<yy) return 0;
if(now.xxx&&now.y>yy) return 1;

if(now.x<xx&&now.y==yy) return 2;
if(now.x>xx&&now.y==yy) return 3;

if(xx+yy==now.x+now.y&&now.y<yy) return 4;
if(xx+yy==now.x+now.y&&now.y>yy) return 5;

if(xx-yy==now.x-now.y&&now.y<yy) return 6;
if(xx-yy==now.x-now.y&&now.y>yy) return 7;

return -1;

}

int dis(point a){
return max(abs(a.x-xx),abs(a.y-yy));
}

void update(point &old,point now){
if(old.kind==‘?’||dis(now)<dis(old))
{
old=now;
}
}

int main()
{
int n;
scanf(“%d%d%d”,&n,&xx,&yy);

for(int i=0;i<8;i++)
{
    rounds[i].kind='?';
}
for(int i=0;i<n;i++){
    point now;
     getchar();
    scanf("%c%d%d",&now.kind,&now.x,&now.y);
    int dire=dir(now);
    if(dire>=0){
        update(rounds[dire],now);
    }
}
bool flag=false;
for(int i=0;i<8;i++){
    if(i<4&&rounds[i].kind=='R'||rounds[i].kind=='Q')
    {
        flag=true;
        break;
    }
    if(i>=4&&rounds[i].kind=='B'||rounds[i].kind=='Q')
    {
        flag=true;
        break;
    }
}