使用我访问过的节点查找到原点的最短路径的图表

Question

Given the directions and a route, I want to know the shortest path to, in the end, get back to where I started(origin) only using places I have visited before

(example of the image below, N is north, S south etc)

Sample Input: NNEEEENWNENNSEEEWWWSWSESWWWNNNNWWWSSNNNNEE

Sample Output: 14

image with example, red is shortest path and gives the 14 output

I am working in java and want to use a Breadth-First Search algorithm and want to use a graph, which is the best way I can construct this graph and look a this problem?

Can someone help?

Answer 1

You can generate a graph of visited cells by reading the input and then run a BFS on this graph from the last position.

To generate the graph you can do:

for each direction :
   create node for the cell if not already existing
   for each adjacent cell, if visited then connect them together

Answer 2

Just run a BFS, starting from the end and going to positions you have already passed, stopping once you reach the origin position:

#include <bits/stdc++.h>
using namespace std;

int m[9][10];
int v[9][10];

struct point{
    int x, y, dist;
    point(){}
    point(int x, int y){this->x = x; this->y = y; dist = 0;}
    point(int x, int y, int dist){this->x = x; this->y = y; this->dist = dist;}
    bool operator == (const point &rhs) const { return rhs.x == x && rhs.y == y; }
};

point origin(2, 8);
point target(2, 0);

int BFS(){
    queue<point> q;
    q.push(target);
    while(!q.empty()){
        point p = q.front();
        q.pop();
        v[p.y][p.x] = 1;
        if(p == origin) return p.dist;

        if(p.x > 0 && !v[p.y][p.x - 1] && m[p.y][p.x - 1]) q.push(point(p.x - 1, p.y, p.dist + 1)); //left
        if(p.x < 9 && !v[p.y][p.x + 1] && m[p.y][p.x + 1]) q.push(point(p.x + 1, p.y, p.dist + 1)); //right
        if(p.y > 0 && !v[p.y - 1][p.x] && m[p.y - 1][p.x]) q.push(point(p.x, p.y - 1, p.dist + 1)); //up
        if(p.y < 8 && !v[p.y + 1][p.x] && m[p.y + 1][p.x]) q.push(point(p.x, p.y + 1, p.dist + 1)); //down
    }
    return -1;
}

int main(){
    memset(m, 0, sizeof(m));
    memset(v, 0, sizeof(v));
    m[origin.y][origin.x] = 1;

    string path = "NNEEEENWNENNSEEEWWWSWSESWWWNNNNWWWSSNNNNEE";

    point o(origin.x, origin.y);

    for(int i = 0; i < path.size(); i++){
        if(path[i] == 'N') o.y--;
        else if(path[i] == 'S') o.y++;
        else if(path[i] == 'W') o.x--;
        else if(path[i] == 'E') o.x++;
        m[o.y][o.x] = 1;
    }

    cout<<"dist = "<<BFS()<<endl;
}

OUTPUT: dist = 14 .