搜索 dfs

基础题：

input:

n m

下面是n行m列的矩阵，由‘.’和‘x’组成，求由(1,1)到(n,m)的最短路

 

#include<iostream>using namespace std;#define M 100#define INF 10000000int dir[4][2]={0,1,0,-1,1,0,-1,0};char map[M][M];int ans,step=0,a=0;int n,m;//suppose '.' means can walk// 'x' means cannot walkbool judge(int x,int y){if(x>=0&&x<n&&y>=0&&y<n&&map[x][y]=='.')return 1;return 0;}void dfs(int x,int y,int step)//cal the min step{if(x==n-1&&y==m-1){if(step<ans)ans=step;}for(int i=0;i<4;i++){int tx=x+dir[i][0];int ty=y+dir[i][1];if(judge(tx,ty)){step++;map[tx][ty]='x';dfs(tx,ty,step);step--;map[tx][ty]='.';}}}int main(){while(cin>>n>>m&&m&&n){ans=INF;step=0;a=0;for(int i=0;i<n;i++){getchar();for(int j=0;j<m;j++){scanf("%c",&map[i][j]);if(map[i][j]=='.')a++;}}dfs(0,0,0);if(ans!=INF)cout<<ans<<endl;elsecout<<"Cannot achieve"<<endl;}}