基础算法（蓝桥杯）--DFS经典题型1

1、B站视频链接：B04 DFS 迷宫 方案数_哔哩哔哩_bilibili

题index.php/tags-41973.html" >���链接：迷宫 - 洛谷

 

#include using namespace std;const int N=10;int m,n,t,sx,sy,fx,fy,ans;int a,b;//记录偏移坐标int g[N][N];int dx[4]={-1,0,1,0},dy[4]={0,1,0,-1};void dfs(int x,int y){if(x==fx&&y==fy){ans++;return;}for(int i=0;i>n>>m>>t>>sx>>sy>>fx>>fy;for(int i=0;i>a>>b,g[a][b]=1;//标记障碍物g[sx][sy]=1;//标记起点dfs(sx,sy);cout

pimg height="964" src="https://m.xiaogongju.com.cn/uploads/allimg/20250423/-250423023P5259.png" width="1012" //p h32、B站视频链接：B05 DFS 跳马 方案数_哔哩哔哩_bilibili/h3 p题目链接：跳马问题 - 洛谷 /p pimg height="1074" src="https://m.xiaogongju.com.cn/uploads/allimg/20250423/-250423023PC50.png" width="1200" //p pimg height="1009" src="https://m.xiaogongju.com.cn/uploads/allimg/20250423/-250423023PHS.png" width="1200" //p pre #include bits/stdc++.husing namespace std;int m,n,ans;//探照灯偏移量 int dx[4]={2,1,-1,-2};int dy[4]={1,2, 2, 1};void dfs(int x,int y){if(x==n&&y==m){ans++;return;}for(int i=0;i4;i++){int a=x+dx[i],b=y+dy[i];if(a0||an||bm)continue;//越界dfs(a,b); }}int main(){cin>>n>>m;dfs(0,0);cout