Careercup - Google面试题 - 4716965625069568

2014-05-06 00:17

题目链接

原题：

Given a 2-D matrix represents the room, obstacle and guard like the following (0 is room, B->obstacle, G-> Guard): 
0 0 0 
B G G 
B 0 0 

calculate the steps from a room to nearest Guard and set the matrix, like this 
2 1 1 
B G G 
B 1 1 
Write the algorithm, with optimal solution.

题目：有一个二维矩阵表示的迷宫，其中G表示保安人员，B表示不可穿越的墙，其他位置均为空地。如果每次可以向东南西北四个方向移动一格，请计算出每个空格离最近的保安有多远。题目给出了一个示例。

解法：既然矩阵里可能有多个保安，我的思路是以各个保安为中心，进行广度优先搜索，搜索的过程中随时更新每个空格位置的最短距离，保证全部搜索完之后所有的结果都是最小的。单次BFS的时间代价是O(n^2)级别的。

代码：

// http://www.careercup.com/question?id=4716965625069568
#include <queue>
#include <vector>
using namespace std;

class Solution {
public:
    void solve(vector<vector<int> > &matrix) {
        // -1 for guard
        // -2 for blockade
        // 0 for room
        // > 0 for distance
        
        n = (int)matrix.size();
        if (n == 0) {
            return;
        }
        m = (int)matrix[0].size();
        if (m == 0) {
            return;
        }
        
        int i, j;
        
        for (i = 0; i < n; ++i) {
            for (j = 0; j < m; ++j) {
                if (matrix[i][j] == -1) {
                    doBFS(matrix, i, j);
                }
            }
        }
    };
private:
    int n, m;
    
    bool inBound(int x, int y) {
        return x >= 0 && x <= n - 1 && y >= 0 && y <= m - 1;
    }
    
    void doBFS(vector<vector<int> > &matrix, int x, int y) {
        queue<int> q;
        static int dd[4][2] = {{-1, 0}, {+1, 0}, {0, -1}, {0, +1}};
        int tmp;
        int xx, yy;
        int i;
        int dist;
        
        q.push(x * m + y);
        while (!q.empty()) {
            tmp = q.front();
            q.pop();
            x = tmp / m;
            y = tmp % m;
            dist = matrix[x][y] > 0 ? matrix[x][y] : 0;
            for (i = 0; i < 4; ++i) {
                xx = x + dd[i][0];
                yy = y + dd[i][1];
                if (!inBound(xx, yy) || matrix[xx][yy] < 0 || 
                    (matrix[xx][yy] > 0 && matrix[xx][yy] <= dist + 1)) {
                    // out of boundary
                    // a guard or a blockade
                    // the distance is no shorter
                    continue;
                }
                matrix[xx][yy] = dist + 1;
                q.push(xx * m + yy);
            }
        }
    }
};