poj 3083 dfs+bfs

题意：给一张地图，分别求向左优先、向右优先、最短路三种走法从起点到终点的步数；

思路：向左向右采用dfs，最短路采用bfs；

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char mm[550][505],flag;
int n,m,s1,s2,e1,e2,i,j,k,step1,step2,step3;
int vis[550][550];
int shortest[550][550];
int x[4]={1,0,-1,0};
int y[4]={0,1,0,-1};
int xx[4]={1,0,-1,0};
int yy[4]={0,-1,0,1};
struct node
{
    int x,y;
    node(){};
    node(int x,int y){
        this->x=x;
        this->y=y;
      }
}q[100005];
void dfs1(int k1,int k2,int dir,int step)
{

    if(mm[k1][k2]=='E')
    {
        printf("%d",step);
        flag=1;return;
    }
    for(i=dir;i<4;i++)
    {
        if(flag) break;
        int x1=k1+xx[i];
        int y1=k2+yy[i];
        if(mm[x1][y1]!='#'&&x1>0&&x1<=n&&y1>0&&y1<=m)
        {
            dir=(i+3)%4;
            dfs1(x1,y1,dir,step+1);
        }
        if(i==3) i-=4;
    }
}
void dfs2(int k1,int k2,int dir,int step)
{
    if(mm[k1][k2]=='E')
    {
        printf(" %d",step);
        flag=1;return;
    }
    for(i=dir;i<4;i++)
    {
        if(flag) break;
        int x1=k1+x[i];
        int y1=k2+y[i];
        if(mm[x1][y1]!='#'&&x1>0&&x1<=n&&y1>0&&y1<=m)
        {
            dir=(i+3)%4;
            dfs2(x1,y1,dir,step+1);
        }
        if(i==3) i-=4;
    }
}
void bfs(int sx,int sy)
{
    int b=0,e=0,i;
    q[e++]=node(sx,sy);
    vis[sx][sy]=1;
    shortest[sx][sy]=1;
    while(b<e)
    {
        if(q[b].x==e1&&q[b].y==e2)
        {
            printf(" %d\n",shortest[e1][e2]);
            break;
        }
        node now=q[b++];
        for(i=0;i<4;i++)
        {
            int x1=now.x+xx[i];
            int y1=now.y+yy[i];
            if(x1>0&&x1<=n&&y1>0&&y1<=m&&!vis[x1][y1]&&mm[x1][y1]!='#')
            {
                vis[x1][y1]=1;
                shortest[x1][y1]=shortest[now.x][now.y]+1;
                q[e++]=node(x1,y1);
            }
        }
    }

}
int main()
{
    int t,len;
    scanf("%d",&t);
    while(t--)
    {
        memset(mm,0,sizeof(mm));
        scanf("%d%d",&m,&n);
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            {
                scanf(" %c",&mm[i][j]);
                if(mm[i][j]=='S') s1=i,s2=j;
                if(mm[i][j]=='E') e1=i,e2=j;
            }
        }
        flag=0;
        dfs1(s1,s2,0,1);
        flag=0;
        dfs2(s1,s2,0,1);
        memset(vis,0,sizeof(vis));
        bfs(s1,s2);
    }
    return 0;
}