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

time limit per test4 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

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.

Examples

input

2

4 2

R 1 1

B 1 5

output

YES

input

2

4 2

R 3 3

B 1 5

output

NO

Note

Picture for the first sample:

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:

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”.

【题目链接】:http://codeforces.com/contest/734/problem/D

【题解】



一开始想错了；还以为是一个棋子一个棋子地摆下去；后来才发现；原来是全部摆完之后才判断；

做法是把那个白皇后的横、竖、两个对角线上的坐标记录一下;(i+j)就在’/’型对角线上,(i-j)在’\’型对角线上.用4个vector记录;

然后按照y坐标排序(除了竖的线按照x坐标之外);

(当然，你要把那个白皇后一开始也加到这4个vector中);

然后用个lower_bound查找那个白皇后在4个vector中的位置;

然后看4个vector中该白皇后相邻的两个棋子会不会攻击到这个棋子;

横的和竖的有皇后和车能攻击到

对角线则有皇后和象能攻击到;

具体的看代码吧；



【完整代码】

#include <bits/stdc++.h>
#define LL long long

using namespace std;

struct abc
{
    LL y;
    char key;
};

int n;
LL a0,b0;
vector <abc> heng,shu;
vector <abc> xpy,xsy;
abc temp;

bool cmp(abc a,abc b)
{
    return a.y < b.y;
}

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    cin >> n;
    cin >> a0 >>b0;
    temp.y = b0;
    heng.push_back(temp);
    temp.y = a0;
    shu.push_back(temp);
    temp.y = b0;temp.key = '*';
    xpy.push_back(temp);
    xsy.push_back(temp);
    for (int i = 1;i <= n;i++)
    {
        LL x,y;
        char key;
        cin >> key >> x >> y;
        temp.y = y;temp.key = key;
        if (x == a0)
            heng.push_back(temp);
        temp.y = x;
        if (y == b0)
            shu.push_back(temp);
        temp.y = y;
        if (x+y==a0+b0)
            xpy.push_back(temp);
        if (x-y==a0-b0)
            xsy.push_back(temp);
    }
    sort(heng.begin(),heng.end(),cmp);
    sort(shu.begin(),shu.end(),cmp);
    sort(xpy.begin(),xpy.end(),cmp);
    sort(xsy.begin(),xsy.end(),cmp);

    int pos;
    //横线
    temp.y = b0;
    pos = lower_bound(heng.begin(),heng.end(),temp,cmp)-heng.begin();
    int len = heng.size();
    if (pos-1>=0)
    {
        char key = heng[pos-1].key;
        if (key == 'Q' || key == 'R')
        {
            puts("YES");
            return 0;
        }
    }

    if (pos+1 <= len-1)
    {
        char key = heng[pos+1].key;
        if (key == 'Q' || key == 'R')
        {
            puts("YES");
            return 0;
        }
    }

    len =  shu.size();
    //竖线
    temp.y = a0;
    pos = lower_bound(shu.begin(),shu.end(),temp,cmp)-shu.begin();
    if (pos-1>=0)
    {
        char key = shu[pos-1].key;
        if (key == 'Q' || key == 'R')
        {
            puts("YES");
            return 0;
        }
    }

    if (pos+1 <= len-1)
    {
        char key = shu[pos+1].key;
        if (key == 'Q' || key == 'R')
        {
            puts("YES");
            return 0;
        }
    }

    len = xpy.size();
    // '/'型对角线
    temp.y = b0;
    pos = lower_bound(xpy.begin(),xpy.end(),temp,cmp)-xpy.begin();
    if (pos-1>=0)
    {
        char key = xpy[pos-1].key;
        if (key == 'Q' || key == 'B')
        {
            puts("YES");
            return 0;
        }
    }

    if (pos+1 <= len-1)
    {
        char key = xpy[pos+1].key;
        if (key == 'Q' || key == 'B')
        {
            puts("YES");
            return 0;
        }
    }

    len = xsy.size();
    // '\'型对角线
    pos = lower_bound(xsy.begin(),xsy.end(),temp,cmp)-xsy.begin();
    if (pos-1>=0)
    {
        char key = xsy[pos-1].key;
        if (key == 'Q' || key == 'B')
        {
            puts("YES");
            return 0;
        }
    }

    if (pos+1 <= len-1)
    {
        char key = xsy[pos+1].key;
        if (key == 'Q' || key == 'B')
        {
            puts("YES");
            return 0;
        }
    }

    puts("NO");

    return 0;
}