LEETCODE: Minimum Path Sum-优快云博客

本文链接：https://blog.youkuaiyun.com/anyicheng2015/article/details/42076919

本文探讨了如何使用Dijkstra算法解决在给定的m x n网格中找到从左上角到右下角的最短路径问题。通过实现类Solution中的minPathSum方法，我们展示了如何在网格中移动并计算所有路径的最小和。

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Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

求最短路径的方法！！！最短路径dijkstra算法！！！

class Solution {
public:
    int minPathSum(vector<vector<int> > &grid) {
        if(grid.size() == 0 || grid[0].size() == 0) return 0;
        int dx = grid[0].size();
        int dy = grid.size();

        vector<vector<int>> distance(dy, vector<int>(dx, INT_MAX));
        distance[0][0] = grid[0][0];
        
        queue<pair<int,int>> thequeue;
        thequeue.push(make_pair(0,0));
        while(!thequeue.empty()) {
            pair<int,int> current = thequeue.front();
            thequeue.pop();
            
            if(current.first + 1 < dy) {
                if(distance[current.first + 1][current.second] 
                > distance[current.first][current.second] + grid[current.first + 1][current.second]) {
                    distance[current.first + 1][current.second] 
                    = distance[current.first][current.second] + grid[current.first + 1][current.second];
                    thequeue.push(make_pair(current.first + 1, current.second));
                }
            }
            
            if(current.second + 1 < dx) {
                if(distance[current.first][current.second + 1]
                > distance[current.first][current.second] + grid[current.first][current.second + 1]) {
                    distance[current.first][current.second + 1]
                = distance[current.first][current.second] + grid[current.first][current.second + 1];
                thequeue.push(make_pair(current.first, current.second + 1));
                }
            }
        }
        
        return distance[dy - 1][dx - 1];
    }
};