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".
Input

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.

Output

The only line of the output should contains "YES" (without quotes) if the white king is in check and "NO" (without quotes) otherwise.

题意:就是给你一个很大的棋盘,给你一个白棋的位置还有n个黑棋的位置,问你黑棋能否一步就吃掉白棋

给你如下规则

1.‘B'只能对角线移动,而且不能越过其他黑棋。

2.’R'只能上下左右移动,而且不能越过其他黑棋。

3.‘Q’既能对角线移动又能左右移动,但是不能越过其他黑棋。

其实也就是个模拟题,但也有一些技巧,只要考虑白棋的正上方,正下方,正左方,正右方距离白棋最小的是否为‘R'or’Q'。

斜右上角,斜右下角,斜左下角,斜左上角距离白棋最小的是否为‘B'or’Q'。即可。还有就是要注意一下小细节。

#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
using namespace std;
typedef long long ll;
const int M = 5e5 + 10;
struct TnT {
char cp[2];
int x , y;
}s[M];
int main()
{
int t;
scanf("%d" , &t);
int x0 , y0;
scanf("%d%d" , &x0 , &y0);
int flag = 0;
for(int i = 0 ; i < t ; i++) {
scanf("%s %d %d" , s[i].cp , &s[i].x , &s[i].y);
if(s[i].x == x0 && s[i].y == y0)
flag = 1;
//cout << s[i].cp << ' ' << s[i].x << ' ' << s[i].y << endl;
}
for(int i = 0 ; i < 8 ; i++) {
int temp = 0;
int MIN = 2e9 + 10;
if(i == 0) {
for(int j = 0 ; j < t ; j++) {
if(s[j].x == x0 && s[j].y > y0) {
int gg = abs(s[j].y - y0);
if(MIN > gg) {
temp = j;
MIN = gg;
}
}
}
if((MIN != 2e9 + 10) && (s[temp].cp[0] == 'R' || s[temp].cp[0] == 'Q')) {
flag = 1;
}
if(flag == 1)
break;
else
continue;
}
if(i == 1) {
for(int j = 0 ; j < t ; j++) {
if(s[j].x == x0 && s[j].y < y0) {
int gg = abs(s[j].y - y0);
if(MIN > gg) {
temp = j;
MIN = gg;
}
}
}
if((MIN != 2e9 + 10) && (s[temp].cp[0] == 'R' || s[temp].cp[0] == 'Q')) {
flag = 1;
}
if(flag == 1)
break;
else
continue;
}
if(i == 2) {
for(int j = 0 ; j < t ; j++) {
if(s[j].y == y0 && s[j].x > x0) {
int gg = abs(s[j].x - x0);
if(MIN > gg) {
temp = j;
MIN = gg;
}
}
}
if((MIN != 2e9 + 10) && (s[temp].cp[0] == 'R' || s[temp].cp[0] == 'Q')) {
flag = 1;
}
if(flag == 1)
break;
else
continue;
}
if(i == 3) {
for(int j = 0 ; j < t ; j++) {
if(s[j].y == y0 && s[j].x < x0) {
int gg = abs(s[j].x - x0);
if(MIN > gg) {
temp = j;
MIN = gg;
}
}
}
if((MIN != 2e9 + 10) && (s[temp].cp[0] == 'R' || s[temp].cp[0] == 'Q')) {
flag = 1;
}
if(flag == 1)
break;
else
continue;
}
ll MIN2 = 9e18;
if(i == 4) {
for(int j = 0 ; j < t ; j++) {
if((s[j].y - y0) == -1 * (s[j].x - x0) && s[j].y > y0 && s[j].x < x0) {
int x1 = abs(s[j].x - x0);
int y1 = abs(s[j].y - y0);
ll gg = (ll)x1 * x1 + (ll)y1 * y1;
if(MIN2 > gg) {
temp = j;
MIN2 = gg;
} }
}
if((MIN2 != 9e18) && (s[temp].cp[0] == 'B' || s[temp].cp[0] == 'Q')) {
flag = 1;
}
if(flag == 1)
break;
else
continue;
}
if(i == 5) {
for(int j = 0 ; j < t ; j++) {
if((s[j].y - y0) == -1 * (s[j].x - x0) && s[j].y < y0 && s[j].x > x0) {
int x1 = abs(s[j].x - x0);
int y1 = abs(s[j].y - y0);
ll gg = (ll)x1 * x1 + (ll)y1 * y1;
if(MIN2 > gg) {
temp = j;
MIN2 = gg;
} }
}
if((MIN2 != 9e18) && (s[temp].cp[0] == 'B' || s[temp].cp[0] == 'Q')) {
flag = 1;
}
if(flag == 1)
break;
else
continue;
}
if(i == 6) {
for(int j = 0 ; j < t ; j++) {
if((s[j].y - y0) == (s[j].x - x0) && s[j].y < y0 && s[j].x < x0) {
int x1 = abs(s[j].x - x0);
int y1 = abs(s[j].y - y0);
ll gg = (ll)x1 * x1 + (ll)y1 * y1;
if(MIN2 > gg) {
temp = j;
MIN2 = gg;
} }
}
if((MIN2 != 9e18) && (s[temp].cp[0] == 'B' || s[temp].cp[0] == 'Q')) {
flag = 1;
}
if(flag == 1)
break;
else
continue;
}
if(i == 7) {
for(int j = 0 ; j < t ; j++) {
if((s[j].y - y0) == (s[j].x - x0) && s[j].y > y0 && s[j].x > x0) {
int x1 = abs(s[j].x - x0);
int y1 = abs(s[j].y - y0);
ll gg = (ll)x1 * x1 + (ll)y1 * y1;
if(MIN2 > gg) {
temp = j;
MIN2 = gg;
} }
}
if((MIN2 != 9e18) && (s[temp].cp[0] == 'B' || s[temp].cp[0] == 'Q')) {
flag = 1;
}
if(flag == 1)
break;
else
continue;
}
if(flag == 1)
break;
}
if(flag == 0)
printf("NO\n");
else
printf("YES\n");
return 0;
}

Codeforces 734D. Anton and Chess(模拟)的更多相关文章

  1. Codeforces Round #379 (Div. 2) D. Anton and Chess 模拟

    题目链接: http://codeforces.com/contest/734/problem/D D. Anton and Chess time limit per test4 secondsmem ...

  2. D. Anton and Chess 模拟题 + 读题

    http://codeforces.com/contest/734/problem/D 一开始的时候看不懂题目,以为象是中国象棋那样走,然后看不懂样例. 原来是走对角线的,长知识了. 所以我们就知道, ...

  3. Codeforces Round #379 (Div. 2) D. Anton and Chess 水题

    D. Anton and Chess 题目连接: http://codeforces.com/contest/734/problem/D Description Anton likes to play ...

  4. Codeforces Round #379 (Div. 2) D. Anton and Chess —— 基础题

    题目链接:http://codeforces.com/contest/734/problem/D D. Anton and Chess time limit per test 4 seconds me ...

  5. Anton and Chess

    Anton and Chess time limit per test 4 seconds memory limit per test 256 megabytes input standard inp ...

  6. CodeForces.158A Next Round (水模拟)

    CodeForces.158A Next Round (水模拟) 题意分析 校赛水题的英文版,坑点就是要求为正数. 代码总览 #include <iostream> #include &l ...

  7. 【29.89%】【codeforces 734D】Anton and Chess

    time limit per test4 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...

  8. Anton and Chess(模拟+思维)

    http://codeforces.com/group/1EzrFFyOc0/contest/734/problem/D 题意:就是给你一个很大的棋盘,给你一个白棋的位置还有n个黑棋的位置,问你黑棋能 ...

  9. Codeforces 747C:Servers(模拟)

    http://codeforces.com/problemset/problem/747/C 题意:有n台机器,q个操作.每次操作从ti时间开始,需要ki台机器,花费di的时间.每次选择机器从小到大开 ...

随机推荐

  1. Mac相关快捷键操作

    拷贝: shift + option + 拖动拖动至目的地 创建快捷方式: option + command + 拖动至目的地

  2. 浅析scrapy与scrapy_redis区别

    最近在工作中写了很多 scrapy_redis 分布式爬虫,但是回想 scrapy 与 scrapy_redis 两者区别的时候,竟然,思维只是局限在了应用方面,于是乎,搜索了很多相关文章介绍,这才搞 ...

  3. Android UI绘制流程及原理

    一.绘制流程源码路径 1.Activity加载ViewRootImpl ActivityThread.handleResumeActivity() --> WindowManagerImpl.a ...

  4. Apache之——多虚拟主机多站点配置的两种实现方案

    Apache中配置多主机多站点,可以通过两种方式实现: 将同一个域名的不同端口映射到不同的虚拟主机,不同端口映射到不同的站点: 将同一个端口映射成不同的域名,不同的域名映射到不同的站点. 我们只需要修 ...

  5. 【Laravel】 安装及常用的artisan命令

    composer Laravel 安装 cmd composer create-project laravel/laravel Laravel5 之后自动创建 常用的artisan命令 全局篇 查看a ...

  6. 保存localStorage并访问

    将用户的输入保存至localStorage对象的属性中,这些属性在再次访问时还会继续保持在原位置. 如果你在浏览器中按照fil://URL的方式直接打开本地文件,则文法在某些浏览器中使用存储功能(比如 ...

  7. 学习Qt的一点小感想

    作为一名电子信息工程的学生,嵌入式似乎是不二的选择,然后我便学习了一下在嵌入式广泛应用的QT软件,刚开始就是学学控件,觉得还是简单,也觉得比较新颖,可是到了做一些具体的小东西就会发现学的东西远远不够, ...

  8. Redis批量删除key的小技巧,你知道吗?

    在使用redis的过程中,经常会遇到要批量删除某种规则的key,但是redis提供了批量查询一类key的命令keys或scan,没有提供批量删除某种规则key的命令,怎么办?看完本文即可,哈哈. 本文 ...

  9. 缓存的有效期和淘汰策略【Redis和其他缓存】【刘新宇】

    缓存有效期与淘汰策略 有效期 TTL (Time to live) 设置有效期的作用: 节省空间 做到数据弱一致性,有效期失效后,可以保证数据的一致性 Redis的过期策略 过期策略通常有以下三种: ...

  10. 08_代码块丶继承和final

    Day07笔记 课程内容 1.封装 2.静态 3.工具类 4.Arrays工具类 封装 概述 1.封装:隐藏事物的属性和实现细节,对外提供公共的访问方式 2.封装的好处: 隐藏了事物的实现细节 提高了 ...