vivo游戏中心的运营小伙伴最近接到一款新游戏的上架申请,为了保障用户体验,运营同学将按运营流程和规范对其做出分析评估。经过初步了解后分析得知,该游戏的地图可以用一个大小为 n*n 的矩阵表示,每个元素可以视为一个格子,根据游戏剧情设定其中某些格子是不可达的(比如建筑、高山、河流或者其它障碍物等),现在请你设计一种算法寻找从起点出发到达终点的最优抵达路径,以协助运营小伙伴评估该游戏的可玩性和上手难度。
数据范围:
进阶:时间复杂度,空间复杂度
第一行表示矩阵大小 n,5 第二行表示起点和终点的坐标第三行起是一个用矩阵表示的游戏地图,其中#或者@表示障碍物,其他字母、非0数字、以及符号+、-、* 等等均表示普通可达格子,共有 n 行 n 列
输出描述:
输出最优路径的长度;若无法到达,则输出-1示例1输入
15 0 7 7 7 *5#++B+B+++++$3 55#+++++++###$$ ###$++++++#+*#+ ++$@$+++$$$3+#+ +++$$+++$+4###+ A++++###$@+$++A +++++#++$#$$+++ A++++#+5+#+++++ +++$$#$++#++++A +++$+@$###+++++ +###4+$+++$$+++ +#+3$$$+++$##++ +#*+#++++++#$$+ $####+++++++$## 3$+++B++B++++#5输出
13
备注:
最优即最短,寻找最优路径时只能按上下左右四个方向前进。
深度优先搜索DFS(与剑指offer机器人思想类似)但这题需要考虑上下左右
def dfs(x, y, grid, endx, endy, visited, count): if x < 0 or y < 0 or x >= len(grid) or y >= len(grid[0]) or grid[x][y] in "#@" \ or (visited[x][y] != 0 and visited[x][y] <= count): return visited[x][y] = count if x == endx and y == endy: return dfs(x, y + 1, grid, endx, endy, visited, count + 1) dfs(x, y - 1, grid, endx, endy, visited, count + 1) dfs(x - 1, y, grid, endx, endy, visited, count + 1) dfs(x + 1, y, grid, endx, endy, visited, count + 1) if __name__ == "__main__": n = int(input()) [starty,startx,endy,endx] = list(map(int,input().split())) road = [] for i in range(n): road.append(list(input())) visited = [[0]*n for _ in range(n)] dfs(startx,starty,road,endx,endy,visited,1) print(visited[endx][endy]-1)
package com.cnblogs; import java.util.*; public class Main{ private static Queue<Posi> q = new LinkedList<Posi>(); private static int[][] dir = {{0,1},{0,-1},{-1,0},{1,0}};//方向 public static void main(String[] args) { Scanner in = new Scanner(System.in); int N = in.nextInt(); String[] map = new String[N]; int [][] vis = new int[N][N];//访问位... // for(int i = 0;i<vis.length;i++){ // System.out.println(Arrays.toString(vis[i])); // } // System.out.println("--------------------"); int y1 = in.nextInt(),x1 = in.nextInt(),y2 = in.nextInt(),x2 = in.nextInt(); // System.out.println(x1 + "," + y1 + "," + x2 + "," + y2); // System.out.println("--------------------"); in.nextLine(); for(int i = 0;i<N;i++){ map[i] = in.nextLine(); } System.out.println(bfs(map,vis,x1,y1,x2,y2)); // for(int i = 0;i<vis.length;i++){ // System.out.println(Arrays.toString(vis[i])); // } // for(int i = 0;i<dir.length;i++){ // System.out.println(Arrays.toString(dir[i])); // } } private static int bfs(String[] map,int[][] vis,int x1,int y1,int x2,int y2){ q.add(new Posi(x1,y1,0)); vis[x1][y1] = 1; Posi local = null; int x,y; while(!q.isEmpty()){ local = q.poll(); for(int i = 0;i<4;i++){ x = local.x + dir[i][0];//下一个要迭代的位置 y = local.y + dir[i][1]; // System.out.println(x + "," + y); if(inBoundary(x,y,map.length) && vis[x][y] == 0 && canThrough(map,x,y)){//can访问 vis[x][y] = 1; if(x == x2 && y == y2){ return local.z + 1; } q.offer(new Posi(x,y,local.z + 1)); // System.out.println(x + "," + y + "," + (local.z + 1)); } } } return -1; } private static boolean canThrough(String[] map,int x,int y){ if(map[x].charAt(y) !='#' && map[x].charAt(y) != '@'){ return true; } return false; } private static boolean inBoundary(int x,int y,int N){ if(x >=0 && x<N && y >=0 && y<N){ return true; }else{ return false; } } private static class Posi{ public int x; public int y; public int z; public Posi(){} public Posi(int x, int y,int z) { this.x = x; this.y = y; this.z = z; } } }
import java.io.BufferedReader; import java.io.InputStreamReader; import java.io.IOException; import java.util.Queue; import java.util.LinkedList; class Node { int x; int y; int steps; public Node(int x, int y) { this.x = x; this.y = y; } } public class Main { private static int path = Integer.MAX_VALUE; private static int[] dx = new int[]{0, 0, 1, -1}; private static int[] dy = new int[]{1, -1, 0, 0}; public static void main(String[] args) throws IOException { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); int n = Integer.parseInt(br.readLine()); String[] params = br.readLine().split(" "); int startCol = Integer.parseInt(params[0]); int startRow = Integer.parseInt(params[1]); int endCol = Integer.parseInt(params[2]); int endRow = Integer.parseInt(params[3]); char[][] graph = new char[n][n]; for(int i = 0; i < n; i++) graph[i] = br.readLine().toCharArray(); int m = graph[0].length; boolean[][] visited = new boolean[n][m]; Queue<Node> queue = new LinkedList<>(); queue.offer(new Node(startRow, startCol)); // 队列用于存储下一步所有可能的位置 visited[startRow][startCol] = true; int steps = 0; while(!queue.isEmpty()){ Node cur = queue.poll(); if(cur.x == endRow && cur.y == endCol){ System.out.println(cur.steps); return; } // 走一步(尝试4个方向) for(int i = 0; i < 4; i++){ int x = cur.x + dx[i]; int y = cur.y + dy[i]; if(x < 0 || x >= n || y < 0 || y >= m || visited[x][y] || graph[x][y] == '@' || graph[x][y] == '#'){ // 越界,走过或格子不可达就跳过 continue; }else{ // 否则当前的格子就是下一步的候选 Node next = new Node(x, y); next.steps = cur.steps + 1; queue.offer(next); visited[x][y] = true; } } } System.out.println(-1); } }
#include<iostream> #include<bits/stdc++.h> using namespace std; void dfs(vector<vector<char>>& in,int i,int j,int endx,int endy,int count,vector<vector<int>>& visited){ if(i>=in.size()||i<0||j>=in[0].size()||j<0) return ; if(in[i][j]=='#'||in[i][j]=='@'||(visited[i][j]!=0&&count>=visited[i][j])) return; visited[i][j]=count; if(i==endx&&j==endy){ return; } char c=in[i][j]; in[i][j]='#'; dfs(in,i+1,j,endx,endy,count+1,visited); dfs(in,i-1,j,endx,endy,count+1,visited); dfs(in,i,j+1,endx,endy,count+1,visited); dfs(in,i,j-1,endx,endy,count+1,visited); in[i][j]=c; } int main(){ int n; cin>>n; //输入输出 int startx=0,starty=0,endx=0,endy=0; cin>>starty>>startx>>endy>>endx; vector<vector<char>> in(n,vector<char>(n)); vector<vector<int>> visited(n,vector<int>(n)); for(int i=0;i<n;++i){ for(int j=0;j<n;++j){ cin>>in[i][j]; } } int res=INT_MAX; dfs(in,startx,starty,endx,endy,1,visited); if(visited[endx][endy]==0) cout<<-1<<endl; else cout<<visited[endx][endy]-1<<endl; return 0; }
#include<iostream>
#include<string>
#include<vector>
#include<queue>
using namespace std;
struct Step {
int x, y; //位置
int steps; //从起点开始走到x,y处需要的步数
Step(int xx, int yy, int st) :x(xx), y(yy), steps(st) {}
};
queue<Step> q;
int main()
{
int n;
cin >> n;
int a, b, c, d;
cin >> b >> a >> d >> c; //输入的数据是先列后行。。。
vector<vector<char> > v;
for (int i = 0; i < n; ++i)
{
vector<char> vv;
for (int j = 0; j < n; ++j)
{
char tmp;
cin >> tmp;
vv.push_back(tmp);
}
v.push_back(vv);
}
//判重标记
int visited[1000][1000];
for (int i = 0; i < n; ++i)
for (int j = 0; j < n; ++j)
visited[i][j] = 0;
int minlen = (1 << 30);
q.push(Step(a, b, 0)); //起点入队列
visited[a][b] = 1;
while (!q.empty())
{
Step s = q.front();
if (s.x == c && s.y == d) //到出口了
{
minlen = s.steps;
break;
}
else
{
if (s.y - 1>=0 && v[s.x][s.y - 1] != '#' && v[s.x][s.y - 1] != '@' && !visited[s.x][s.y - 1]) //如果可以往左走
{
q.push(Step(s.x, s.y - 1, s.steps + 1));
visited[s.x][s.y - 1] = 1;
}
if (s.y + 1 <= n - 1 && v[s.x][s.y + 1] != '#' && v[s.x][s.y + 1] != '@' && !visited[s.x][s.y + 1]) //如果可以往右走
{
q.push(Step(s.x, s.y + 1, s.steps + 1));
visited[s.x][s.y + 1] = 1;
}
if (s.x - 1>=0 && v[s.x-1][s.y] != '#' && v[s.x-1][s.y] != '@' && !visited[s.x-1][s.y]) //如果可以往上走
{
q.push(Step(s.x-1, s.y, s.steps + 1));
visited[s.x - 1][s.y] = 1;
}
if (s.x + 1 <=n-1 && v[s.x + 1][s.y] != '#' && v[s.x + 1][s.y] != '@' && !visited[s.x + 1][s.y]) //如果可以往下走
{
q.push(Step(s.x + 1, s.y, s.steps + 1));
visited[s.x + 1][s.y] = 1;
}
q.pop();
}
}
if (minlen == (1 << 30))
minlen = -1;
cout << minlen << endl;
return 0;
}
from collections import deque def solution(): n = int(input().strip()) y0, x0, y1, x1 = [int(x) for x in input().split(" ")] M = [] step = 0 flag = 0 for _ in range(n): M.append([x for x in input().strip()]) que = deque([[x0, y0]]) M[x0][y0] = "0" while que: c = len(que) step += 1 for _ in range(c): cur_x, cur_y = que.popleft() for a, b in [(1, 0), (-1, 0), (0, 1), (0, -1)]: next_x, next_y = cur_x + a, cur_y + b if next_x == x1 and next_y == y1: if M[next_x][next_y] not in "0#@": print(step) flag = 1 return else: print(-1) return if 0 <= next_x < n and 0 <= next_y < n and M[next_x][next_y] not in "0#@": que.append([next_x, next_y]) M[next_x][next_y] = "0" if flag == 0: print(-1) solution()
public class Main { private static int[][] posi = new int[][] { { 0, -1 }, { 0, 1 }, { 1, 0 }, { -1, 0 } }; static int len; static int start_y; static int start_x; static int end_y; static int end_x; static int[][] map; public static void main(String[] args) { Scanner in = new Scanner(System.in); len = in.nextInt(); start_y = in.nextInt(); start_x = in.nextInt(); end_y = in.nextInt(); end_x = in.nextInt(); int[][] grid = new int[len][len]; for (int i = 0; i < len; i++) { String row = in.next(); for (int j = 0; j < len; j++) { if (row.charAt(j) == '#' || row.charAt(j) == '@') grid[i][j] = -1; } } map = new int[len][len]; for (int i = 0; i < map.length; i++) Arrays.fill(map[i], Integer.MAX_VALUE); calcWayDistance(grid, start_x, start_y, 0); if(map[end_x][end_y] == Integer.MAX_VALUE){ System.out.println(-1); }else{ System.out.println(map[end_x][end_y]); } } private static void calcWayDistance(int[][] grid, int x, int y, int dis) { if(x<0||x>=len||y<0||y>=len||grid[x][y] == -1||dis>=map[x][y]) return; map[x][y] = dis; for(int[] dir:posi){ int xx = x+dir[0]; int yy = y+dir[1]; calcWayDistance(grid, xx, yy, dis+1); } } }
import java.util.*; import java.lang.*; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); int[][] map = new int[n][n]; int start_y = sc.nextInt(); int start_x = sc.nextInt(); int end_y = sc.nextInt(); int end_x = sc.nextInt(); for (int i = 0; i < n; i++) { String s = sc.next(); for (int j = 0; j < n; j++) { char c = s.charAt(j); if (c == '#' || c == '@') { map[i][j] = 1; } } } Solution sol = new Solution(map, n, end_x, end_y); sol.dfs(start_x, start_y, 0); System.out.println(sol.getAns()); } } class Solution { private int[][] record; private int[][] map; private int end_x; private int end_y; private int n; private int[][] direction = {{1,0},{0,1},{-1,0},{0,-1}}; public Solution(int[][] map, int n, int x, int y) { this.map = map; this.n = n; record = new int[n][n]; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) record[i][j] = Integer.MAX_VALUE; end_x = x; end_y = y; } public void dfs(int x, int y, int d) { record[x][y] = d; for (int[] dir : direction) { int xx = dir[0] + x; int yy = dir[1] + y; if (!isValid(xx, yy, d+1)) continue; dfs(xx, yy, d+1); } } public boolean isValid(int x, int y, int d) { return x >= 0 && x < n && y >= 0 && y < n && record[x][y] > d && map[x][y] == 0; } public int getAns() { return record[end_x][end_y] == Integer.MAX_VALUE? -1 : record[end_x][end_y]; } }
#include <bits/stdc++.h> using namespace std; using pii = pair<int, int>; int main() { int n; cin >> n; pii s; pii e; cin >> s.second >> s.first >> e.second >> e.first; string tmp; getline(cin, tmp); vector<string> m; for (int i = 0; i != n; ++i) { getline(cin, tmp); m.emplace_back(move(tmp)); } int step = 0; deque<deque<bool>> visited(n, deque<bool>(n, false)); pii direct[4] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}}; queue<pii> q({s}); visited[s.first][s.second] = true; while (!q.empty()) { const int sz = q.size(); for (int i = 0; i != sz; ++i) { const auto node = q.front(); q.pop(); if (node.first == e.first && node.second == e.second) { cout << step; return 0; } for (int j = 0; j != 4; ++j) { auto nn = node; nn.first += direct[j].first; nn.second += direct[j].second; if (0 <= nn.first && nn.first < n && 0 <= nn.second && nn.second < n && m[nn.first][nn.second] != '#' && m[nn.first][nn.second] != '@' && !visited[nn.first][nn.second]) { q.push(nn); visited[nn.first][nn.second] = true; } } } ++step; } cout << -1; }
输入给出的起点与终点坐标反了
#include <bits/stdc++.h> #define x first #define y second using namespace std; typedef pair<int, int> PII; const int N = 1010, INF = 0x3f3f3f3f; int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1}; int n, sx, sy, tx, ty; char g[N][N]; int dist[N][N]; int main() { cin >> n >> sy >> sx >> ty >> tx; for(int i = 0; i < n; i ++) cin >> g[i]; queue<PII> que; que.push({sx, sy}); memset(dist, 0x3f, sizeof dist); dist[sx][sy] = 0; while(que.size()) { auto tt = que.front(); que.pop(); for(int i = 0; i < 4; i ++) { int a = tt.x + dx[i], b = tt.y + dy[i]; if(a >= 0 && a < n && b >= 0 && b < n && g[a][b] != '#' && g[a][b] != '@') { if(dist[a][b] > dist[tt.x][tt.y] + 1) { que.push({a, b}); dist[a][b] = dist[tt.x][tt.y] + 1; } } } } if(dist[tx][ty] == INF) puts("-1"); else cout << dist[tx][ty] << "\n"; return 0; }
import java.util.*; public class Main { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int y1 = in.nextInt(), x1 = in.nextInt(), y2 = in.nextInt(), x2 = in.nextInt(); String[] board = new String[n]; in.nextLine(); for (int i = 0; i < n; i++) { board[i] = in.nextLine(); } int res = fun(x1, y1, x2, y2, board); System.out.println(res); } // bfs处理 static int[][] d = new int[][]{{1,0}, {0,1}, {-1,0}, {0,-1}}; public static int fun(int x1, int y1, int x2, int y2, String[] board) { ArrayDeque<int[]> q = new ArrayDeque<>(); boolean[][] v = new boolean[board.length][board.length]; v[x1][y1] = true; q.offer(new int[]{x1, y1}); int dis = 0; while (!q.isEmpty()) { int len = q.size(); for (int i = 0; i < len; i++) { int[] node = q.poll(); int x = node[0], y = node[1]; for (int j = 0; j < 4; j++) { int nx = x + d[j][0], ny = y + d[j][1]; if (nx == x2 && ny == y2) { // 到达终点 return dis+1; } else if (nx >= 0 && nx < board.length && ny >= 0 && ny < board.length && board[nx].charAt(ny) != '#' && board[nx].charAt(ny) != '@' && !v[nx][ny]) { v[nx][ny] = true; q.offer(new int[]{nx, ny}); } } } dis++; } return -1; } }就是这个输入坐标是反的很气人!!!
DFS 记忆化memory #include<iostream> using namespace std; #include<vector> void dfs(const vector<vector<char>>& vec,int beginX,int beginY,int endX,int endY,const int n,vector<vector<int>>& memory,int count){ if(beginX<0||beginX>=n||beginY<0||beginY>=n) return; // 越界 if(vec[beginX][beginY]=='#'||vec[beginX][beginY]=='@') return; // 障碍 if(memory[beginX][beginY]!=0 && memory[beginX][beginY]<=count) return ; // 已经访问过或者记忆数组记录的距离更小 memory[beginX][beginY]=count; // 记录当前距离 if(beginX==endX && beginY==endY) return; // 到达终点 dfs(vec, beginX-1, beginY, endX, endY, n, memory,count+1); dfs(vec, beginX+1, beginY, endX, endY, n, memory,count+1); dfs(vec, beginX, beginY+1, endX, endY, n, memory,count+1); dfs(vec, beginX, beginY-1, endX, endY, n, memory,count+1); } int main(){ int n=0;cin>>n; vector<vector<char>>vec(n,vector<char>()); int beginX=0;int beginY=0;int endX=0;int endY=0; cin>>beginY>>beginX>>endY>>endX; for(int i=0;i<n;i++){ string a;cin>>a; for(auto it:a){ vec[i].push_back(it); } } vector<vector<int>> memory(n,vector<int>(n,0)); dfs(vec,beginX,beginY,endX,endY,n,memory,1); if(memory[endX][endY]==0) cout<<-1<<endl; else cout<<memory[endX][endY]-1<<endl; return 0; }
#include<bits/stdc++.h> using namespace std; typedef pair<int,int> PII; int dx[4] = {0,1,0,-1}; int dy[4] = {1,0,-1,0}; int main() { int n; cin>>n; vector<string>g(n); //记录任一点到起点的距离,初始化为-1可以代替visited数组,表示未访问 vector<vector<int>>dist(n,vector<int>(n,-1)); int x1,y1,x2,y2; cin>>y1>>x1>>y2>>x2; dist[x1][y1] = 0; for(int i =0;i<n;i++) cin>>g[i]; queue<PII> q; q.push({x1,y1}); bool flag =false; while(q.size()) { auto t = q.front();q.pop(); int x = t.first, y = t.second; if(x == x2 && y == y2){ cout<< dist[x][y]; flag =true; break; } for(int i =0; i<4;i++) { int nx = x+dx[i]; int ny = y+dy[i]; if(nx>=0 && nx<n && ny>=0 && ny<n&& g[nx][ny]!='#'&&g[nx][ny]!='@' &&dist[nx][ny]==-1) { //st[nx][ny] = true; dist[nx][ny] =dist[x][y]+1; q.push({nx,ny}); } } } if(!flag) cout<< -1 <<endl; return 0; }
n = input() sc, sr, ec, er = map(int, raw_input().strip().split(' ')) N = n arr = [] while N > 0: arr.append(raw_input().strip()) N -= 1 mask = [[0]*len(arr[0]) for _ in range(len(arr))] if arr[er][ec] == '#'&nbs***bsp;arr[er][ec] == '@': print(-1) else: dirs = [[-1,0],[1,0],[0, -1], [0, 1]] step = 0 queue = [(er, ec)] flag = 0 mask[er][ec] = 1 while queue: try: l = len(queue) for i in range(l): node = queue.pop(0) for d in dirs: nr, nc = node[0] + d[0], node[1] + d[1] if nr <n and nr > -1 and nc < n and nc > -1 and mask[nr][nc] == 0: mask[nr][nc] = 1 if arr[nr][nc] not in ['#', '@']: if nr == sr and nc == sc: print(step + 1) flag = 1 raise else: queue.append((nr, nc)) step += 1 except: break if flag == 0: print(-1)
import java.util.Scanner; public class Main { private static int[][] map; private static char[][] chars; private static int n; public static void main(String[] args) { Scanner in = new Scanner(System.in); n = in.nextInt(); int startY = in.nextInt(),startX = in.nextInt(),endY = in.nextInt(),endX = in.nextInt(); chars = new char[n][n]; map = new int[n][n]; for (int i = 0; i < n; i++) chars[i] = in.next().toCharArray(); dfs(startX,startY,1,endX,endY); System.out.println(map[endX][endY] - 1); } public static void dfs(int x,int y,int count,int endX,int endY){ if ((x < 0 || x >= n || y < 0 || y >= n)||(chars[x][y]=='#'||chars[x][y]=='@')||(map[x][y] != 0 && map[x][y] <= count)) return; map[x][y] = count; if (x == endX && y == endY) return; count ++; dfs(x - 1,y,count,endX,endY); dfs(x + 1,y,count,endX,endY); dfs(x,y - 1,count,endX,endY); dfs(x,y + 1,count,endX,endY); } } //暴力递归加一个动态规划数组
#include<iostream> #include<vector> #include<cmath> #include<algorithm> using namespace std; int ans; struct direction{//也代表点 int row, col; direction(){ ; } direction(int row, int col):row(row), col(col){ ; } struct direction operator +(const struct direction &s){ direction ans; ans.row=s.row+row; ans.col=col+s.col; return ans; } }; vector<direction> direct; vector<vector<bool> > isvisit; int n; bool judge(const pair<direction ,int> a,const pair<direction ,int> b){ return a.second<b.second; } bool isvalid(direction &d){ if(d.col<0 || d.row<0 || d.col>=n || d.row>=n) return false; return true; } void BFS(direction start, direction end, int step, vector<vector<char>> &bitmap) { if(start.row==end.row && start.col == end.col){ if(step<ans){ ans=step; } return ; } if(ans!=(1<<30) && (abs(end.row-start.row)+abs(end.col-start.col)+step)>=ans) return; vector<pair<direction ,int>> tmp; for(int i=0; i<direct.size(); i++){ direction dest=direct[i]+start; if(isvalid(dest) && isvisit[dest.row][dest.col]==false && bitmap[dest.row][dest.col]!='#' && bitmap[dest.row][dest.col]!='@'){ tmp.push_back(make_pair(dest, abs(dest.row-end.row)+abs(dest.col-end.col))); } } sort(tmp.begin(), tmp.end(), judge); for(int i=0; i<tmp.size(); ++i){ isvisit[tmp[i].first.row][tmp[i].first.col]=true; if(isdigit(bitmap[tmp[i].first.row][tmp[i].first.col])) BFS(tmp[i].first, end, step+1, bitmap); else BFS(tmp[i].first, end, step+1, bitmap); isvisit[tmp[i].first.row][tmp[i].first.col]=false; } } int main(){ direction start,end; cin>>n; //cin>>start.row>>start.col>>end.row>>end.col; cin>>start.col>>start.row>>end.col>>end.row;//我真的无语了。。。 vector<vector<char>> bitmap(n, vector<char>(n, '\0')); //isvisit=vector<vector<bool>>(n, vector<bool>(n, false)); vector<bool> ttt(n, false); for(int i=0; i<n; i++){ string str; cin>>str; isvisit.push_back(ttt); //cout<<str<<endl<<endl; for(int j=0; j<n; j++){ //cin>>bitmap[i][j]; bitmap[i][j]=str[j]; } } direct.push_back(direction(1,0)); direct.push_back(direction(-1,0)); direct.push_back(direction(0,-1)); direct.push_back(direction(0,1)); ans=1<<30; isvisit[start.row][start.col]=true; BFS(start, end, 0, bitmap); if(ans==(1<<30)) cout<<-1; else cout<<ans<<endl; } 这输入,我人傻了。行列反过来的