HDOJ1728(限制转弯的迷宫问题)

用bfs进行深搜,求出每个可达点的最小转弯数

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int MAX_N=105;
char g[MAX_N][MAX_N];
int vis[MAX_N][MAX_N];
int n,m;
int k,sx,sy,ex,ey;
struct node{
    int x,y,turns;
    node(){}
    node(int cy,int cx,int cturns):x(cx),y(cy),turns(cturns){}
};
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
bool judge(int r,int c)
{
    if(1<=r&&r<=m&&1<=c&&c<=n&&g[r][c]==‘.‘)
    {
        return true;
    }
    return false;
}
void bfs()
{
    if(sy==ey&&sx==ex)
    {
        printf("yes\n");
        return;
    }
    queue<node> que;
    que.push(node(sy,sx,-1));
    vis[sy][sx]=1;
    while(!que.empty())
    {
        node now=que.front();que.pop();
        for(int i=0;i<4;i++)
        {
            int ny=now.y+dy[i];
            int nx=now.x+dx[i];
            while(judge(ny,nx))//按一条路深搜
            {
                if(!vis[ny][nx])
                {
                //    printf("%c (%d,%d) %d\n",g[ny][nx],ny,nx,now.turns+1);
                    que.push(node(ny,nx,now.turns+1));
                    vis[ny][nx]=1;
                    if(ny==ey&&nx==ex&&now.turns+1<=k)
                    {
                        printf("yes\n");
                        return ;
                    }
                }
                ny+=dy[i];
                nx+=dx[i];
            }
        }
    }

    printf("no\n");
}
int main()
{
    int t;scanf("%d",&t);
    while(t--)
    {
        memset(vis,0,sizeof(vis));
        scanf("%d %d",&m,&n);
        scanf("%*c");
        for(int i=1;i<=m;i++)
        {
            for(int j=1;j<=n;j++)
            {
                scanf("%c",&g[i][j]);
            }
            scanf("%*c");
        }
        scanf("%d %d %d %d %d",&k,&sx,&sy,&ex,&ey);
        bfs();
    }

    return 0;
}