广搜+深搜（搜索最短路径+最短步数）

#include <iostream>
#include <string.h>
#include <stack>
#include <queue>
#include <algorithm>
using namespace std;
const int maxn=10;

struct P
{
	int x,y;
}point[maxn];

struct PP
{
	int fx,fy;
}path[maxn][maxn];

int num[maxn][maxn];
int vis[maxn][maxn];
int dx[4]={0,-1,1,0};
int dy[4]={-1,0,0,1};
int total=0;

int sx=1,sy=1;
int ex=4,ey=4;

char map[6][6]=
{
    {‘#‘,‘#‘,‘#‘,‘#‘,‘#‘,‘#‘},
    {‘#‘,‘.‘,‘.‘,‘.‘,‘#‘,‘#‘},
    {‘#‘,‘.‘,‘#‘,‘.‘,‘.‘,‘#‘},
    {‘#‘,‘.‘,‘.‘,‘.‘,‘#‘,‘#‘},
    {‘#‘,‘#‘,‘.‘,‘.‘,‘.‘,‘#‘},
    {‘#‘,‘#‘,‘#‘,‘#‘,‘#‘,‘#‘}
};   

int ok(int x,int y)
{
	 if(x<0||x>5||y<0||y>5)
        return 0;
    if(map[x][y]==‘#‘)
        return 0;
    if(vis[x][y]==1)
        return 0;
    return 1;
}

void dfs(int x,int y,int step)
{
	if(x==ex&&y==ey)
	{
		total++;
		for(int i=0;i<step;i++)
		{
			cout<<"("<<point[i].x<<","<<point[i].y<<")";
		}
		cout<<endl;
		return;
	}

	for(int i=0;i<4;i++)
	{
		int curx=x+dx[i];
		int cury=y+dy[i];
		if(ok(curx,cury))
		{
			vis[curx][cury]=1;
			point[step].x=curx;
			point[step].y=cury;
			dfs(curx,cury,step+1);
			vis[curx][cury]=0;
		}
	}
}

void bfs(int x,int y)
{
	num[x][y]=0;
	queue<P>q;
	P a,b;
	a.x=x;
	a.y=y;
	path[a.x][a.y].fx=0;
	path[a.x][a.y].fy=0;
	q.push(a);
	while(!q.empty())
	{
		b=q.front();
		q.pop();
		for(int i=0;i<4;i++)
		{
			a.x=b.x+dx[i];
			a.y=b.y+dy[i];
			while(ok(a.x,a.y))
			{
				vis[a.x][a.y]=1;
				q.push(a);
				path[a.x][a.y].fx=dx[i];
				path[a.x][a.y].fy=dy[i];
				num[a.x][a.y]=num[b.x][b.y]+1;
			}
		}
	}
}

void print(int x,int y)
{
	if(x==sx&&y==sy)
	{
		cout<<"("<<x<<","<<y<<")";
		return;
	}
	print(x-path[x][y].fx,y-path[x][y].fy);
	cout<<"("<<x<<","<<y<<")";
} 

int main()
{
	memset(vis,0,sizeof(vis));
    memset(num,0,sizeof(num)); 

	point[0].x=sx;
	point[0].y=sy;
	vis[sx][sy]=1;
	dfs(sx,sy,1);
	cout<<"There are "<<total<<" paths to arrive."<<endl;

	memset(vis,0,sizeof(vis));
	vis[sx][sy]=1;
	bfs(sx,sy);

    for(int i=0;i<6;i++)
    {
        for(int j=0;j<6;j++)
            cout<<num[i][j]<<" ";
        cout<<endl;
    }  

    cout<<endl;
    cout<<num[ex][ey]<<endl<<endl;;
    print(ex,ey);
	cout<<endl;
    return 0;
}

原文地址：https://www.cnblogs.com/KasenBob/p/10624186.html