CSU-ACM2017暑假集训比赛2 D - ^(●゚∀゚○)ノ
2017-08-02 20:07
381 查看
D - ^(●゚∀゚○)ノ
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.
Example
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".
一道模拟题,情况很多,需要仔细。策略是按黑棋与国王的距离为关键值,值小的优先级高,压入优先队列。于是优先检查离国王近的黑棋是否可将军。若国王的八个方向都被堵死或被将军,可以中断判断,输出结果。
#include <iostream> #include <cstdio> #include <algorithm> #include <vector> #include <queue> #include <cmath> #define TEST using namespace std; long long n, king[4], x, y;// king: [0] horizontal; [1] vertical; [2] main diagonal; [3] counter diagonal; bool ckmt[4][2];//国王的八向能否被将军 // [0][0] left; [0][1] right; [1][0] up; [1][1] down; [2][0] upper left; [2][1] upper right; [3][0] lower right; [3][1] lower left; struct chess_piece{ long long x, y, md, cd; char type; chess_piece(int a, int b, char c):x(a), y(b), type(c), md(a+b), cd(a-b){} bool operator < (const chess_piece &p)const{ return abs(x-king[0])+abs(y-king[1]) > abs(p.x-king[0])+abs(p.y-king[1]); } }; priority_queue<chess_piece> q; bool checkmate(){ while(!q.empty()){ int sz = q.size(); chess_piece temp = q.top(); q.pop(); long long vx = temp.x, vy = temp.y, vm = temp.md, cd = temp.cd; char typ = temp.type; switch (typ){ case 'R': if(vm == king[2] || cd == king[3]){//on diagonal if(vx < king[0] && vy > king[1])//upper left ckmt[2][0] = false; else if(vx > king[0] && vy > king[1])//upper right ckmt[3][0] = false; else if(vx < king[0] && vy < king[1])//lower left ckmt[3][1] = false; else if(vx > king[0] && vy < king[1])//lower right ckmt[2][1] = false; } else{//on vertical or horizontal line if(vx == king[0] && vy > king[1] && ckmt[1][0])//up return true; else if(vx == king[0] && vy < king[1] && ckmt[1][1])//down return true; else if(vx < king[0] && vy == king[1] && ckmt[0][0])//left return true; else if(vx > king[0] && vy == king[1] && ckmt[0][1])//right return true; } break; case 'B': if(vm == king[2] || cd == king[3]){//on diagonal if(vx < king[0] && vy > king[1] && ckmt[2][0])//upper left return true; else if(vx > king[0] && vy > king[1] && ckmt[3][0])//upper right return true; else if(vx < king[0] && vy < king[1] && ckmt[3][1])//lower left return true; else if(vx > king[0] && vy < king[1] && ckmt[2][1])//lower right return true; } else{//on vertical or horizontal line if(vx == king[0] && vy > king[1])//up ckmt[1][0] = false; else if(vx == king[0] && vy < king[1])//down ckmt[1][1] = false; else if(vx < king[0] && vy == king[1])//left ckmt[0 d1f7 ][0] = false; else if(vx > king[0] && vy == king[1])//right ckmt[0][1] = false; } break; case 'Q': if(vm == king[2] || cd == king[3]){//on diagonal if(vx < king[0] && vy > king[1] && ckmt[2][0])//upper left return true; else if(vx > king[0] && vy > king[1] && ckmt[3][0])//upper right return true; else if(vx < king[0] && vy < king[1] && ckmt[3][1])//lower left return true; else if(vx > king[0] && vy < king[1] && ckmt[2][1])//lower right return true; } else{//on vertical or horizontal line if(vx == king[0] && vy > king[1] && ckmt[1][0])//up return true; else if(vx == king[0] && vy < king[1] && ckmt[1][1])//down return true; else if(vx < king[0] && vy == king[1] && ckmt[0][0])//left return true; else if(vx > king[0] && vy == king[1] && ckmt[0][1])//right return true; } break; } int cnt = 0; for(int z = 0; z < 4; z++) for(int v = 0; v < 2; v++) if(ckmt[z][v] == false) cnt++; if(cnt == 8) return false; } return false; } int main(){ #ifdef TEST freopen("test.txt", "r", stdin); #endif // TEST while(cin >> n){ cin >> king[0] >> king[1]; king[2] = king[0] + king[1], king[3] = king[0] - king[1]; for(int i = 0; i < 4; i++) for(int j = 0; j < 2; j++) ckmt[i][j] = true; while(!q.empty()) q.pop(); char typ; getchar(); for(int i = 0; i < n; i++){ scanf("%c%d%d", &typ, &x, &y); q.push(chess_piece(x,y,typ)); getchar(); } if(checkmate()) cout << "YES" << endl; else cout << "NO" << endl; } return 0; }
相关文章推荐
- CSU-ACM2017暑假集训比赛3D - D CodeForces - 557C
- CSU-ACM2017暑假集训比赛1 TD POJ3111
- CSU-ACM2017暑假集训比赛7 C - Pasha and Tea - CodeForces - 557B
- CSU-ACM2017暑假集训比赛2 B - : )
- CSU-ACM2017暑假集训比赛1 B - R2D2 and Droid Army
- CSU-ACM2017暑假集训比赛2 A - _(:з」∠)_
- CSU-ACM2017暑假集训比赛7 - D - Bicoloring - UVA - 10004
- CSU-ACM2017暑假集训比赛8 - E - Escape - HDU - 3605
- CSU-ACM2017暑假集训比赛1 A - I Can Guess the Data Structure! -uva111995
- CSU-ACM2017暑假集训比赛7 - E - Courses - HDU - 1083
- CSU-ACM2017暑假集训比赛2 CodeForces - 724D
- CSU-ACM2017暑假集训比赛1 C - Gourmet and Banquet CodeForces - 589F
- CSU-ACM2017暑假集训比赛8 - A - Xor Sum - HDU - 4825
- CSU-ACM2017暑假集训比赛1 C - Gourmet and Banquet
- CSU-ACM2017暑假集训比赛2 E - ( ̄▽ ̄)/
- CSU-ACM2017暑假集训比赛7 - F - 凑数字 - 51Nod - 1385
- CSU-ACM2017暑假集训比赛1 C - Gourmet and Banquet
- CSU-ACM2017暑假集训比赛8 - B - The Two Routes - CodeForces - 601A
- CSU-ACM2017暑假集训比赛2 HDU - 1597 find the nth digit
- CSU-ACM2017暑假集训比赛2 C - (╯°口°)╯(┴—┴