题解 | #迷宫问题#

加上回溯，就可以100%通过测试用例了， let n = readline().split(" ").map(Number), maze = []; //读数据 while ((line = readline())) maze.push(line.split(" ").map(Number)); let v = [...Array(n[0])].map(() => Array(n[1]).fill(0)); function find(i, j, road) { road.push([i, j]); //先推入0,0 maze[i][j] = 1; if (i == n[0] - 1 && j == n[1] - 1) { for (let i of road) print(`(${i[0]},${i[1]})`); } if (j + 1 < n[1] && maze[i][j + 1] == 0 && v[i][j + 1] == 0) { v[i][j + 1] = 1; find(i, j + 1, road); v[i][j + 1] = 0; road.pop(); } //右 if (i + 1 < n[0] && maze[i + 1][j] == 0 && v[i + 1][j] == 0) { v[i + 1][j] = 1; find(i + 1, j, road); v[i + 1][j] = 0; road.pop(); } //下 if (j - 1 >= 0 && maze[i][j - 1] == 0 && v[i][j - 1] == 0) { v[i][j - 1] = 1; find(i, j - 1, road); v[i][j - 1] = 0; road.pop(); } //左 if (i - 1 >= 0 && maze[i - 1][j] == 0 && v[i - 1][j] == 0) { v[i - 1][j] = 1; find(i - 1, j, road); v[i - 1][j] = 0; road.pop(); } //上 } find(0, 0, []);

数组需要深拷贝，第4行改成，let tmp = [...road]; tmp.push([i,j]);

let n=readline().split(' ').map(Number),maze=[]//读数据 while(line=readline()) maze.push(line.split(' ').map(Number)) function find(i,j,road){ road.push([i,j]);//先推入0,0 maze[i][j]=1; if(i==n[0]-1 && j==n[1]-1){ for (let i of road) print(`(${i[0]},${i[1]})`); } if(j+1<n>= 0 && maze[i][j-1]==0) find(i,j-1,road.slice())//左 if(i-1 >= 0 && maze[i-1][j]==0) find(i-1,j,road.slice())//上 } find(0,0,[])</n>

递归调用需要把road数组拷贝一个