c_cpp 找到从矩阵的一个单元格移动到另一个单元格所需的最小移动次数

Posted 2021-05-14

//https://www.geeksforgeeks.org/find-minimum-numbers-moves-needed-move-one-cell-matrix-another/
#include <bits/stdc++.h>
using namespace std;
#define N 4
class graph {
    int v;
    list<int> *adj;
public:
    graph (int v) {
        this->v= v;
        adj= new list<int> [v];
    };
    void addEdge (int u, int w) {
        adj[u].push_back(w);
        adj[w].push_back(u);
    }
    void bfs(int , int d);
};
void graph::bfs (int s, int d) {
    queue<int> q;
    int level[v];
    for (int i=0;i<v;i++)
        level[i]= -1;
    q.push(s);
    level[s]=0;
    while (!q.empty()) {
        s= q.front();
        q.pop();
        list<int> ::iterator i;
        for (i= adj[s].begin();i!= adj[s].end(); ++i ) {
            if (level[*i]<0 || level[*i]>level[s]+1) {
                level[*i]= level[s]+1;
                q.push(*i);
            }
        }
    }

    cout<<level[d];
}
bool isSafe (int i, int j, int m[][N]) {
    if (i<0 || i>N-1 || j<0 || j>=N || m[i][j]==0)
        return false;
    return true;
}
int minPath (int m[][N]) {
    int s,d,k=0;
    int v= N*N;
    graph g(v);

    for (int i=0;i<N;i++) {
        for (int j=0;j<N;j++) {
            if (m[i][j]!= 0) {
                if (isSafe (i, j+1, m))
                    g.addEdge(k,k+1);
                if (isSafe (i, j-1, m))
                    g.addEdge(k-1,k);
                if (j<N-1 && isSafe (i+1,j,m))
                    g.addEdge(k,k+N);
                if (i>0 && isSafe (i-1,j,m))
                    g.addEdge(k,k-N);
            }
            if (m[i][j]== 1)
                s=k;
            if (m[i][j]== 2)
                d=k;
            k++;
        }
    }
    g.bfs(s,d);
}

int main() {
    int M[N][N] = {{ 3 , 3 , 1 , 0 },
        { 3 , 0 , 3 , 3 },
        { 2 , 3 , 0 , 3 },
        { 0 , 3 , 3 , 3 }
    };
    minPath(M);
}