sicily 1172. Queens, Knights and Pawns

Description

You all are familiar with the famous 8-queens problem which asks you to place 8 queens on a chess board so no two attack each other. In this problem, you will be given locations of queens and knights and pawns and asked to find how many of the unoccupied squares on the board are not under attack from either a queen or a knight (or both). We'll call such squares "safe" squares. Here, pawns will only serve as blockers and have no capturing ability. The board below has 6 safe squares. (The shaded squares are safe.) 

Recall that a knight moves to any unoccupied square that is on the opposite corner of a 2x3 rectangle from its current position; a queen moves to any square that is visible in any of the eight horizontal, vertical, and diagonal directions from the current position. Note that the movement of a queen can be blocked by another piece, while a knight's movement can not.

Input

There will be multiple test cases. Each test case will consist of 4 lines. The first line will contain two integers n and m, indicating the dimensions of the board, giving rows and columns, respectively. Neither integer will exceed 1000. The next three lines will each be of the form 
k r1 c1 r2 c2 ... rk ck 
indicating the location of the queens, knights and pawns, respectively. The numbering of the rows and columns will start at one. There will be no more than 100 of any one piece. Values of n = m = 0 indicate end of input.

Output

Each test case should generate one line of the form 
Board b has s safe squares. 
where b is the number of the board (starting at one) and you supply the correct value for s.

Sample Input

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4 4
2 1 4 2 4
1 1 2
1 2 3
2 3
1 1 2
1 1 1
0
1000 1000
1 3 3
0
0
0 0

Sample Output

Board 1 has 6 safe squares.
Board 2 has 0 safe squares.
Board 3 has 996998 safe squares.

这题可能有理解错误的地方就是对queen，它在八个方向的移动是不限的，除非遇到一个其他对象。

#include <iostream>
#include <vector>

using namespace std;

char location[1001][1001];

int king[16] = {-2, -1, -2, 1, -1, 2, 1, 2, 2, 1, 2, -1, 1, -2, -1, -2};
int queueL[16] = {-1, -1, -1, 0, -1, 1, 0, -1, 0, 1, 1, -1, 1, 0, 1, 1};

int main(int argc, char const *argv[])
{
    int m, n, num, x, y;
    int testCase = 0;
    while (cin >> m >> n && m != 0 && n != 0) {
        ++testCase;
        memset(location, '0', sizeof(location));
        vector<int> queue;

        // queen
        cin >> num;
        queue.resize(num * 2 + 1);
        int i = 0, j = 0;
        while (i++ < num) {
            cin >> x >> y;
            location[x - 1][y - 1] = '2';
            queue[j] = x - 1;
            queue[j + 1] = y - 1;
            j += 2;
        }

        // knight
        i = 0;
        cin >> num;
        while (i++ < num) {
            cin >> x >> y;
            location[x - 1][y - 1] = '2';
            for (int k = 0; k < 16; k += 2) {  // 处理8个方向，也即八个“日”字形路线
                int xT = x - 1 + king[k];
                int yT = y - 1 + king[k + 1];
                if (xT >= 0 && xT < m && yT >= 0 && yT < n && location[xT][yT] == '0')
                    location[xT][yT] = '1';
            }
        }

        // pawn
        i = 0;
        cin >> num;
        while (i++ < num) {
            cin >> x >> y;
            location[x - 1][y - 1] = '2';
        }

        // Queen 的处理
        for (i = 0; i < j; i += 2) {
            for (int p = 0; p < 16; p += 2) {
                int xT = queue[i];
                int yT = queue[i + 1];
                for (int q = 0; ; ++q) {
                    xT += queueL[p];
                    yT += queueL[p + 1];
                    if (xT >= 0 && xT < m && yT >= 0 && yT < n && location[xT][yT] == '0') {
                        location[xT][yT] = '1';
                    } else if (xT >= 0 && xT < m && yT >= 0 && yT < n && location[xT][yT] == '2') { // 在该直线上遇到了其他对象，所以不需要再走了
                        break;
                    } else if(xT < 0 || xT >= m || yT < 0 || yT >= n) {  // 跑出了棋盘，那么说明在这条直线上已经不需要再走了
                        break;
                    }
                }
            }
        }

        int result = 0;
        for (i = 0; i != m; ++i) {
            for (j = 0; j != n; ++j)
                if (location[i][j] == '0')
                    ++result;
        }

        cout << "Board " << testCase << " has " << result << " safe squares." << endl;
    }
    return 0;
}