定义一个二维数组：
int maze[5][5] = {
0, 1, 0, 0, 0,
0, 1, 0, 1, 0,
0, 0, 0, 0, 0,
0, 1, 1, 1, 0,
0, 0, 0, 1, 0,
};
它表示一个迷宫，其中的1表示墙壁，0表示可以走的路，只能横着走或竖着走，不能斜着走，要求编程序找出从左上角到右下角的最短路线。
Input
一个5 × 5的二维数组，表示一个迷宫。数据保证有唯一解。
Output
左上角到右下角的最短路径，格式如样例所示。
Sample Input
0 1 0 0 0
0 1 0 1 0
0 0 0 0 0
0 1 1 1 0
0 0 0 1 0
Sample Output
(0, 0)
(1, 0)
(2, 0)
(2, 1)
(2, 2)
(2, 3)
(2, 4)
(3, 4)
(4, 4)
典型的模板题
dfs代码：
#include<string.h> #include<iostream> using namespace std; const int dir[][2] = {1,0,0,1,-1,0,0,-1}; char mp[10][10]; struct node{int x;int y;}; node l[30],m[30]; int minv = 0x3f3f3f3f; int vis[10][10]; void dfs(int x,int y,int cnt) { if(x == 4 && y == 4) { if(cnt < minv) { minv = cnt; for(int i = 0 ; i < cnt ; i++) m[i] = l[i]; } return; } l[cnt].x = x; l[cnt].y = y; for(int i = 0 ; i < 4 ; i++) { int nx = x + dir[i][0]; int ny = y + dir[i][1]; if(nx >= 0 && ny >= 0 && nx < 5 && ny < 5 && mp[nx][ny] == '0' && !vis[nx][ny]) { vis[nx][ny] = 1; dfs(nx,ny,cnt+1); } } } int main(void) { memset(vis,0,sizeof vis); for(int i = 0 ; i < 5 ; i++) for(int j = 0 ; j < 5 ; j++) cin >> mp[i][j]; dfs(0,0,0); for(int i = 0 ; i < minv ; i++) cout << "(" << m[i].x << ", " << m[i].y << ")" << endl; cout << "(4, 4)" << endl; } bfs代码：
#include<stack> #include<queue> #include<string.h> #include<iostream> using namespace std; const int dir[][2] = {1,0,0,1,-1,0,0,-1}; struct node{int x,y;}; char mp[10][10]; node pro[10][10]; int vis[10][10]; void print() { stack<node>s; node now = pro[5][5]; node next; while(1) { if(now.x == 1 && now.y == 1) break; s.push(now); next = pro[now.x][now.y]; now = next; } while(s.size()) { now = s.top(); s.pop(); cout << "(" << now.x - 1 << ", " << now.y - 1 << ")" << endl; } } void bfs() { queue<node>q; node now,next; now.x = 1; now.y = 1; q.push(now); while(q.size()) { now = q.front(); q.pop(); if(now.x == 5 && now.y == 5) break; for(int i = 0 ; i < 4 ; i++) { next.x = now.x + dir[i][0]; next.y = now.y + dir[i][1]; if(next.x >= 1 && next.y >= 1 && next.x <= 5 && next.y <= 5 && !vis[next.x][next.y] && mp[next.x][next.y] != '1') { vis[next.x][next.y] = 1; pro[next.x][next.y] = now; q.push(next); } } } } int main(void) { memset(vis,0,sizeof vis); for(int i = 1 ; i <= 5 ; i++) for(int j = 1 ; j <= 5 ; j++) cin >> mp[i][j]; bfs(); cout << "(0, 0)" << endl; print(); cout << "(4, 4)" << endl; return 0; } 转载请标明出处:dfs-bfs-迷宫问题
文章来源: https://blog.csdn.net/qq_42891420/article/details/89682639