After knowing the basic principles of DFS&BFS, let's look at a few examples.
Example 1: Looking for a girlfriend
Topic description:
X, as a loyal fan of outdoor sports, always doesn't want to stay at home. Now, he wants to pull the dead house Y out of the house. Ask what is the shortest time from X's home to Y's home.
To simplify the problem, we abstract the map as an n*m matrix with row numbers 1 to n from top to bottom and columns 1 to m from left to right. In the matrix, 'X' indicates the initial coordinate of X, 'Y' indicates the position of Y, '#' indicates that the current position cannot be walked, and '*' indicates that the current position is passable. X can only move up, down, left, and right adjacent '*' each time, and each move takes 1 second.
Enter :
Multiple sets of inputs. Each set of test data first input two integers n, m (1<= n , m<=15 ) represents the map size. The next n lines, m characters each. Ensure that the entered data is legal.
output:
If X can reach Y's home, output the minimum time, otherwise output -1.
example input
The DFS code is as follows:
#include<stdio.h> #include<string.h> int mv[16][16]; char map[16][16]; int mx[4]={-1,0,1,0}; int my[4]={0,-1,0,1}; int min; struct B { int x,y,years; }; /*DFS开始*/ void dfs(int x,int y,int n,int m,int ans) { int i; if(ans>=min) return ; if(ans<min&&map[x][y]=='Y'){ min = ans; return ; } struct B t; for(i=0;i<4;i++) { t.x=x+mx[i]; t.y=y+my[i]; if(0<=t.x&&t.x<n&&0<=t.y&&t.y<m&&mv[t.x][t.y]==0&&map[t.x][t.y]!='#'){ mv[t.x][t.y]=1; dfs(t.x,t.y,n,m,ans+1); mv[t.x][t.y]=0; } } } intmain () { int i,j,n,m; while(~scanf("%d %d",&n,&m)){
#include<stdio.h> #include<string.h> int mv[16][16]; char map[16][16]; int mx[4]={-1,0,1,0}; int my[4]={0,-1,0,1}; struct B { int x,y,years; }q[300],t,f; int main() { int i,j,n,m,flag,s,e; while(~scanf("%d %d",&n,&m)){ for(i=0;i<n;i++) scanf("%s",map[i]); memset(mv,0,sizeof(mv)); for(i=0;i<n;i++) { for(j=0;j<m;j++) if(map[i][j]=='X') break; if(j!=m)break; } // BFS start s=flag=e= 0 ; tx =i,ty=j,t.ans= 0 ; q[e++]=t;//进栈 mv[t.x][t.y]=1; while(s<e){ t=q[s++];//出栈 if(map[t.x][t.y]=='Y'){ printf("%d\n",t.ans); flag=1; break; } for(i=0;i<4;i++) { f.x=t.x+mx[i]; f.y=t.y+my[i]; if(0<=f.x&&f.x<n&&0<=f.y&&f.y<m&&mv[f.x][f.y]==0&&map[f.x][f.y]!='#'){ f.years =t.years+ 1 ; q[e++]=f; mv[f.x][f.y]=1; } } } if(flag==0)printf("-1\n"); } return 0; }
for(i=0;i<n;i++) scanf("%s",map[i]); memset(mv,0,sizeof(mv)); min=300; for(i=0;i<n;i++) { for(j=0;j<m;j++) if(map[i][j]=='X') break; if(j!=m)break; } dfs(i,j,n,m,0); min==300?printf("-1\n"):printf("%d\n",min); } return 0; }
3 3
X#Y
***
#*#
3 3
X#Y
*#*
#*#
Sample output