标签：ted 细节 hit dash pre 左右 又能 single white
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:
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 Round #379 (Div. 2) D. Anton and Chess（模拟）
标签：ted 细节 hit dash pre 左右 又能 single white
原文地址：http://www.cnblogs.com/TnT2333333/p/6068036.html
