Anton likes to play chess. Also, he likes to do programming. That is why he decided to write the program that plays chess. However, he finds the game on 8 to 8 board to too simple, he uses an infinite one instead.
The first task he faced is to check whether the king is in check. Anton doesn't know how to implement this so he asks you to help.
Consider that an infinite chess board contains one white king and the number of black pieces. There are only rooks, bishops and queens, as the other pieces are not supported yet. The white king is said to be in check if at least one black piece can reach the cell with the king in one move.
Help Anton and write the program that for the given position determines whether the white king is in check.
Remainder, on how do chess pieces move:
- Bishop moves any number of cells diagonally, but it can't "leap" over the occupied cells.
- Rook moves any number of cells horizontally or vertically, but it also can't "leap" over the occupied cells.
- Queen is able to move any number of cells horizontally, vertically or diagonally, but it also can't "leap".
The first line of the input contains a single integer n (1 ≤ n ≤ 500 000) — the number of black pieces.
The second line contains two integers x0 and y0 ( - 109 ≤ x0, y0 ≤ 109) — coordinates of the white king.
Then follow n lines, each of them contains a character and two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — type of the i-th piece and its position. Character 'B' stands for the bishop, 'R' for the rook and 'Q' for the queen. It's guaranteed that no two pieces occupy the same position.
The only line of the output should contains "YES" (without quotes) if the white king is in check and "NO" (without quotes) otherwise.
题目大意 ， R B Q 三种黑色棋子，R只能攻击同行同列，B只能攻击对角线上的棋子，C能攻击对角线和同行同列
问黑色能否将白色将军,（如果满足要求，但在攻击的路线中有其他棋子，就不行
思路 ，先将黑色棋子到白色棋子的距离排序**（核心），然后我是分成 8个区域一点点判断，就好了
1 #include <algorithm> 2 #include <stack> 3 #include <istream> 4 #include <stdio.h> 5 #include <map> 6 #include <math.h> 7 #include <vector> 8 #include <iostream> 9 #include <queue> 10 #include <string.h> 11 #include <set> 12 #include <cstdio> 13 #define FR(i,n) for(int i=0;i<n;i++) 14 #define MAX 2005 15 #define mkp pair <int,int> 16 using namespace std; 17 #include <bits/stdc++.h> 18 const int maxn = 5e6 + 40; 19 typedef long long ll; 20 const int inf = 0x3fffff; 21 void read(ll &x) { 22 char ch; bool flag = 0; 23 for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar()); 24 for (x = 0; isdigit(ch); x = (x << 1) + (x << 3) + ch - 48, ch = getchar()); 25 x *= 1 - 2 * flag; 26 } 27 int vis[105]; 28 struct inof{ 29 int flag; 30 ll x,y; 31 ll res; 32 }p[maxn]; 33 bool cmp(inof a,inof b) 34 { 35 return a.res<b.res; 36 } 37 int getdir(ll x,ll y,ll z,ll m) 38 { 39 if(x==z&&y>m)return 0; 40 if(x==z&&y<m)return 1; 41 if(y==m&&x>z)return 2; 42 if(y==m&&x<z)return 3; /// shang xia zuo you 43 44 if(abs((m-y)/(z-x))!=1||(m-y)%(z-x))return -1;///这里wa了几发... 45 if(x<z&&y<m)return 4; 46 if(x<z&&y>m)return 5; 47 if(x>z&&y<m)return 6; 48 if(x>z&&y>m)return 7; 49 } 50 int main() { 51 ll n; 52 read(n); 53 ll xx,yy; 54 memset(vis,0,sizeof(vis)); 55 read(xx),read(yy); 56 for(int i=0;i<n;i++){ 57 char c; 58 scanf("%c",&c); 59 if(c=='R')p[i].flag=0;/// 同行同列 60 else if(c=='B')p[i].flag=1;/// 对角线 61 else if(c=='Q')p[i].flag=2;/// 对角线同行同列 62 read(p[i].x),read(p[i].y); 63 p[i].res=abs(p[i].x-xx)+abs(p[i].y-yy); 64 } 65 sort(p,p+n,cmp); 66 for(int i=0;i<n;i++){ 67 int flag = getdir(p[i].x,p[i].y,xx,yy); 68 if(flag==-1)continue; 69 if(vis[flag])continue; 70 if(flag<=3&&(p[i].flag==0||p[i].flag==2)){return 0*printf("YES");} 71 if(flag>3&&(p[i].flag==1||p[i].flag==2)){return 0*printf("YES");} 72 vis[flag]=1; 73 } 74 puts("NO"); 75 return 0; 76 }