HDU4740

九野的博客，转载请注明出处：http://blog.csdn.net/acmmmm/article/details/11711743

题意：驴和老虎在方格中跑，跑的方式:径直跑，若遇到边界或之前走过的点则转向，驴向右转，虎向左转，若转向后还不能跑则一直呆着不动，

问：他们是否会相遇，会输出相遇坐标，不会输出-1

#include <iostream>#include <algorithm>#include <cstdlib>#include <cstring>#include <string>#include <vector>#include <cstdio>#include <queue>#include <cmath>#include <string.h>#include <assert.h>#include <stack>#include <sstream>#include <map>#include <set>#define M 1020#define LL __int64using namespace std;bool vis1[M][M];bool vis2[M][M];int dx[4]={0,1,0,-1};int dy[4]={1,0,-1,0};int main(){    int i,j,x1,y1,z1,x2,y2,z2,xx1,yy1,xx2,yy2,n;    bool flag,ok1,ok2;    while(scanf("%d",&n),n)    {        scanf("%d%d%d",&x1,&y1,&z1);        scanf("%d%d%d",&x2,&y2,&z2);        memset(vis1,false,sizeof(vis1));        memset(vis2,false,sizeof(vis2));        ok1=true;        ok2=true;        flag=false;        while(1)        {            if(x1==x2&&y1==y2)            {                flag=true;                break;            }            if(!ok1&&!ok2)                break;            vis1[x1][y1]=true;            vis2[x2][y2]=true;            if(ok1)            {                xx1=x1+dx[z1];                yy1=y1+dy[z1];                if(xx1>=0&&xx1<n&&yy1>=0&&yy1<n&&!vis1[xx1][yy1])                {                    x1=xx1;                    y1=yy1;                    z1=z1;                }                else                {                    xx1=x1+dx[(z1+1)%4];                    yy1=y1+dy[(z1+1)%4];                    if(xx1>=0&&xx1<n&&yy1>=0&&yy1<n&&!vis1[xx1][yy1])                    {                        x1=xx1;                        y1=yy1;                        z1=(z1+1)%4;                    }                    else                    {                        x1=x1;                        y1=y1;                        z1=z1;                        ok1=false;                    }                }            }            if(ok2)            {                xx2=x2+dx[z2];                yy2=y2+dy[z2];                if(xx2>=0&&xx2<n&&yy2>=0&&yy2<n&&!vis2[xx2][yy2])                {                    x2=xx2;                    y2=yy2;                    z2=z2;                }                else                {                    xx2=x2+dx[((z2-1)%4+4)%4];                    yy2=y2+dy[((z2-1)%4+4)%4];                    if(xx2>=0&&xx2<n&&yy2>=0&&yy2<n&&!vis2[xx2][yy2])                    {                        x2=xx2;                        y2=yy2;                        z2=((z2-1)%4+4)%4;                    }                    else                    {                        x2=x2;                        y2=y2;                        z2=z2;                        ok2=false;                    }                }            }        }        if(flag)        {            printf("%d %d\n",x1,y1);        }        else        {            printf("-1\n");        }    }    return 0;}