矩阵的最短路径问题

今天字节二面，面试官给了一个笔试题，因为时间仓促，也没有提前准备，就挂掉了，回来仔细思考了一下做了个简单的解答，

题目是这样的，一个mxn的二维矩阵，均为正元素，且对应的元素代表该处的代价，求从（0，0）到（m,n)的最短路径。

code如下：

#include <iostream>
#include <cstring>
#include <vector>
#include <queue>

using namespace std;

class Solution {
public:
    typedef struct Index{
        Index(){x = 0; y = 0;};
        Index(int xx, int yy):x(xx),y(yy){}
        int x, y;
    };
    std::vector<std::pair<int ,int>> shortestRoad(const vector<vector<int>> &matrix) {
        std::vector<std::pair<int, int>> answer;
        int m = matrix.size();
        if(m == 0) return answer;
        const int maxvale = 10000000;
        int n = matrix[0].size();
        vector<vector<int>> table = matrix;
        for(size_t i = 0; i < table.size(); i++){
            for(size_t j = 0; j < table[i].size(); j++){
                table[i][j] = maxvale;
            }
        }

        std::queue<Index> q;

        Index target(m-1, n-1);
        table[m-1][n-1] = std::min(maxvale, matrix[m-1][n-1]);

        q.push(target);


        while (!q.empty()){
            Index top = q.front();
            q.pop();

            if(top.x == 0 && top.y == 0){
                break;
            }

            if(top.x - 1 >= 0){
                if(table[top.x-1][top.y] > table[top.x][top.y] + matrix[top.x-1][top.y]){
                    table[top.x - 1][top.y] = table[top.x][top.y] + matrix[top.x-1][top.y];
                    q.push(Index(top.x-1, top.y));
                }
            }
            if(top.x + 1 < m){
                if(table[top.x+1][top.y] > table[top.x][top.y] + matrix[top.x+1][top.y]){
                    table[top.x + 1][top.y] = table[top.x][top.y] + matrix[top.x+1][top.y];
                    q.push(Index(top.x+1, top.y));
                }
            }
            if(top.y - 1 >= 0){
                if(table[top.x][top.y-1] > table[top.x][top.y] + matrix[top.x][top.y-1]){
                    table[top.x][top.y-1] = table[top.x][top.y] + matrix[top.x][top.y-1];
                    q.push(Index(top.x, top.y-1));
                }
            }
            if(top.y + 1 < n){
                if(table[top.x][top.y+1] > table[top.x][top.y] + matrix[top.x][top.y+1]){
                    table[top.x][top.y+1] = table[top.x][top.y] + matrix[top.x][top.y+1];
                    q.push(Index(top.x, top.y+1));
                }
            }
        }

        Index index(0, 0);
        answer.push_back(std::pair<int, int>(0, 0));
        while (index.x != m-1 || index.y != n-1){
            int maxcost = maxvale;
            Index cur = index;
            if(index.x - 1 >= 0){
                if(maxcost > table[index.x-1][index.y]){
                    maxcost = table[index.x-1][index.y];
                    cur = index;
                    cur.x = cur.x - 1;
                }
            }
            if(index.x + 1 < m){
                if(maxcost > table[index.x+1][index.y]){
                    maxcost = table[index.x+1][index.y];
                    cur = index;
                    cur.x = cur.x + 1;
                }
            }
            if(index.y - 1 >= 0){
                if(maxcost > table[index.x][index.y-1]){
                    maxcost = table[index.x][index.y-1];
                    cur = index;
                    cur.y = cur.y - 1;
                }
            }
            if(index.y + 1 < n){
                if(maxcost > table[index.x][index.y+1]){
                    maxcost = table[index.x][index.y+1];
                    cur = index;
                    cur.y = cur.y + 1;
                }
            }
            index = cur;
            answer.push_back(std::pair<int, int>(index.x, index.y));
        }
        return answer;
    }
};

vector<vector<int>> matrix = {
        {1, 4, 7, 5, 3, 6, 2, 2},
        {5, 3, 6, 8, 9, 7, 6, 1},
        {3, 2, 5, 1, 6, 4, 7, 8},
        {9, 5, 3, 6, 8, 7, 4, 1},
        {6, 3, 8, 6, 3, 2, 5, 8}}
;

int main() {    
Solution s = Solution();
vector<std::pair<int, int>> answer = s.shortestRoad(matrix);
    for(size_t i = 0; i < answer.size(); i++){        
std::cout<<"("<<answer[i].first<<", "<<answer[i].second<<")"<<std::endl;
    }    
return 0;}