Shortest Distance from All Buildings

最新推荐文章于 2024-02-20 02:56:58 发布

小菜鸟的leetcode

最新推荐文章于 2024-02-20 02:56:58 发布

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本文链接：https://blog.youkuaiyun.com/Scarlett_leetcode/article/details/50979880

本文介绍了一个寻找在给定网格中建造房子以使到达所有建筑物总距离最短的问题解决方案。通过广度优先搜索算法，计算从每个建筑出发到达空地的距离，并最终确定最佳建造位置。

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Problem

You want to build a house on an empty land which reaches all buildings in the shortest amount of distance. You can only move up, down, left and right. You are given a 2D grid of values 0, 1 or 2, where:

Each 0 marks an empty land which you can pass by freely.
Each 1 marks a building which you cannot pass through.
Each 2 marks an obstacle which you cannot pass through.

For example, given three buildings at (0,0), (0,4), (2,2), and an obstacle at (0,2):

1 - 0 - 2 - 0 - 1
|   |   |   |   |
0 - 0 - 0 - 0 - 0
|   |   |   |   |
0 - 0 - 1 - 0 - 0

The point (1,2) is an ideal empty land to build a house, as the total travel distance of 3+3+1=7 is minimal. So return 7.

Note:
There will be at least one building. If it is not possible to build such house according to the above rules, return -1.

Solution

对于每一个building，计算它能到达的每一个空地的距离。具体来讲用 dist 数组纪录，最后找最小的就好。

到下一个building时，上一个如果没有到达这个空地，那这个空地就不考虑了。这个用 canReach 数组实现。

class Solution {
    vector<pair<int,int>> dirs = { make_pair(1,0),make_pair(-1,0),make_pair(0,1),make_pair(0,-1)} ;
    
    void bfs( int curRow,int curCol,const vector<vector<int>>& grid,int numBuild,vector<vector<int>>& dist,vector<vector<int>>& canReach) {
        
        const int M = grid.size(), N = grid[0].size();
        queue<int> curLvl, nextLvl;
        unordered_set<int> visited;
        
        curLvl.push(curRow*N + curCol);
        visited.insert(curRow*N + curCol);
        int curDist = 0;
        while(!curLvl.empty()) {
            int curBuild = curLvl.front(); curLvl.pop();
            int curRow = curBuild/N, curCol = curBuild%N;
            
            for( int i = 0; i < dirs.size(); i++) {        
                int nextRow = curRow + dirs[i].first, nextCol = curCol + dirs[i].second;
                int nextBuild = nextRow * N + nextCol;
                if( nextRow >= 0 && nextRow < M && nextCol >= 0 && nextCol < N && visited.find(nextBuild) == visited.end() && grid[nextRow][nextCol] == 0 && canReach[nextRow][nextCol] == numBuild ) {
                    nextLvl.push(nextBuild);
                    visited.insert(nextBuild);
                    dist[nextRow][nextCol] += curDist + 1;
                    canReach[nextRow][nextCol] = numBuild + 1;
                }
            }
            
            if(curLvl.empty()) {
                swap(curLvl, nextLvl);
                curDist++;
            }
        }
        return;
    }
public:
    int shortestDistance(vector<vector<int>>& grid) {
        if(grid.empty() || grid[0].empty()) return 0;
        
        const int M = grid.size(), N = grid[0].size();
        vector<vector<int>> dist( M, vector<int>(N, 0) );
        vector<vector<int>> canReach( M, vector<int>(N, 0) );
        
        int numBuild = 0;
        for( int row = 0; row < M; row++ ){
            for( int col = 0; col < N; col++){
                if( grid[row][col] == 1 ) {
                    bfs(row, col, grid, numBuild++, dist, canReach );
                }
            }
        }
        
        int rst = INT_MAX;
        for( int row = 0; row < M; row++ ){
            for( int col = 0; col < N; col++){
                if(canReach[row][col] == numBuild ) {
                    rst = min( rst, dist[row][col]);
                }
            }
        }
        return rst==INT_MAX ? -1 : rst;
    }
};