LintCode 573: Build Post Office II (BFS经典题）

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原文链接：blog.csdn.net/roufoo/arti…

Build Post Office II Given a 2D grid, each cell is either a wall 2, an house 1 or empty 0 (the number zero, one, two), find a place to build a post office so that the sum of the distance from the post office to all the houses is smallest.

Return the smallest sum of distance. Return -1 if it is not possible.

Example Example 1:

Input：[[0,1,0,0,0],[1,0,0,2,1],[0,1,0,0,0]] Output：8 Explanation： Placing a post office at (1,1), the distance that post office to all the house sum is smallest. Example 2:

Input：[[0,1,0],[1,0,1],[0,1,0]] Output：4 Explanation： Placing a post office at (1,1), the distance that post office to all the house sum is smallest. Challenge Solve this problem within O(n^3) time.

Notice You cannot pass through wall and house, but can pass through empty. You only build post office on an empty.

解法1：BFS 注意这题跟Build Post Office I 不一样，那道题目没有wall，并且可以穿过房屋和空地，所以肯定只要有房屋和空地就肯定有解，而这道题不能穿墙和房屋，有可能有的房屋永远到不了某个空地，所以就无解。 另外Build Post Office I可以简单的把4个方向的累加房屋数加起来，然后再根据Manhattan距离加上相应的累加距离即可。这题不能用这个方法，因为不能穿墙和房屋，所以用BFS比较好，因为BFS可以算最短距离。 房屋和墙的区别是：房屋要累加其到起始点的距离。

class Solution {
public:
    /**
     * @param grid: a 2D grid
     * @return: An integer
     */
    int shortestDistance(vector<vector<int>> &grid) {
        int rowSize = grid.size();
        int colSize = grid[0].size();
        if (rowSize == 0 || colSize == 0) return 0;
        gMinTotalLen = INT_MAX;
        gTotalHouseCount = 0;
        
        for (int i = 0; i < rowSize; ++i) {
            for (int j = 0; j < colSize; ++j) {
                if (grid[i][j] == 1) gTotalHouseCount++;
            }
        }
        
        for (int i = 0; i < rowSize; ++i) {
            for (int j = 0; j < colSize; ++j) {
                if (grid[i][j] == 0) bfs(grid, i, j);
        }
        
        return gMinTotalLen == INT_MAX ? -1 : gMinTotalLen;
    }
    
private:
    int gMinTotalLen;
    int gTotalHouseCount;
    void bfs(vector<vector<int>> &grid, int x, int y) {
        vector<int> dx = {1, -1, 0, 0};
        vector<int> dy = {0, 0, 1, -1};
        queue<pair<int, int>> q;
        int rowSize = grid.size();
        int colSize = grid[0].size();
        vector<vector<int>> visited(rowSize, vector<int>(colSize, 0));
        q.push({x, y});
        visited[x][y] = 1;
        int totalLen = 0;
        int step = 0;
        int totalHouseCount = 0;
        while(!q.empty()) {
            step++;  
            int qSize = q.size();
            for (int i = 0; i < qSize; ++i) {
                pair<int, int> topNode = q.front();
                q.pop();
                for (int j = 0; j < 4; ++j) {
                    int newX = topNode.first + dx[j];
                    int newY = topNode.second + dy[j];
                    if (newX >= 0 && newX < rowSize && newY >= 0 && newY < colSize && !visited[newX][newY]) {
                        visited[newX][newY] = 1;
                        if (grid[newX][newY] == 2) {
                            continue;
                        }
                        else if (grid[newX][newY] == 1) {
                            totalLen += step;
                            totalHouseCount++;
                        }
                        else {   
                            q.push({newX, newY});
                        }
                    }
                }
            }

        }
        
        if (gTotalHouseCount == totalHouseCount) gMinTotalLen = min(gMinTotalLen, totalLen);

        return;
    }
};