Right turn(SCU-4445) (离散化+模拟)

Right turn(SCU-4445) (离散化+模拟)

frog is trapped in a maze. The maze is infinitely large and divided into grids. It also consists of nn obstacles, where the ii-th obstacle lies in grid (xi,yi)(xi,yi).

frog is initially in grid (0,0)(0,0), heading grid (1,0)(1,0). She moves according to The Law of Right Turn: she keeps moving forward, and turns right encountering a obstacle.

The maze is so large that frog has no chance to escape. Help her find out the number of turns she will make.

Input

The input consists of multiple tests. For each test:

The first line contains 1 integer n (0≤n≤1000). Each of the following n lines contains 2 integers xi,y(|xi|,|yi|≤109,(xi,yi)≠(0,0) all (xi,yi)are distinct)

Output

For each test, write 11 integer which denotes the number of turns, or -1 if she makes infinite turns.

Sample Input

    2
    1 0
    0 -1
    1
    0 1
    4
    1 0
    0 1
    0 -1
    -1 0

Sample Output

    2
    0
    -1

题意:

​ 一个迷宫有n块石头,一个人从(0,0)出发,一开始是面向右边,如果他遇到石头就要向右转,问这个人要转几次弯,如果会无限次转弯就输出-1

我看网上题解大部分都是二分找点走，而这里采用的是坐标离散化，然后自动转弯走。

思路：

​ 离散化上所有石头坐标，这样形成的地图最多为$2\ast n$ * $2\ast n$，然后while循环自动转完即可，走到边界离散化的行或列就算走出去了。

代码：

#include
#define mset(a,b) memset(a,b,sizeof(a))
typedef long long ll;
using namespace std;
int X[1100],Y[1100],sx[1100],sy[1100],XX[2100],YY[2100];
int dir[4][2]={
1,0,
0,-1,
-1,0,
0,1
};
int book[2100][2100][4];
int flies[2100][2100];
void getd(int dr,int &dx,int &dy){
    dx=dir[dr][0];
    dy=dir[dr][1];
}
int main()
{
    //1e9+1 1e9+1
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    while(cin>>n)
    {
        int cnt=0;
        X[0]=Y[0]=XX[0]=YY[0]=0;
        for(int k=-1;k<1;++k)
        {
            XX[cnt]=k;
            YY[cnt]=k;
            cnt++;
        }
        for(int i=1;i<=n;++i){
            cin>>X[i]>>Y[i];
            for(int k=-1;k<1;++k)
            {
                XX[cnt]=X[i]+k;
                YY[cnt]=Y[i]+k;
                cnt++;
            }
        }
        X[n+1]=Y[n+1]=XX[cnt]=YY[cnt]=1e9+10;
        cnt++;
        X[n+2]=Y[n+2]=XX[cnt]=YY[cnt]=-1e9-10;
        cnt++;
        int tx,ty;
        sort(XX,XX+cnt);
        sort(YY,YY+cnt);
        tx=unique(XX,XX+cnt)-XX;
        ty=unique(YY,YY+cnt)-YY;
        for(int i=0;i<n+3;++i){
            sx[i]=lower_bound(XX,XX+tx,X[i])-XX;
            sy[i]=lower_bound(YY,YY+ty,Y[i])-YY;
            if(i>=1&&i<=n){
                flies[sx[i]][sy[i]]=1;
            }
        }
        int lx=sx[n+2],rx=sx[n+1],ly=sy[n+2],ry=sy[n+1];
        int nx=sx[0],ny=sy[0],dr=0,tus=0,flag=0;
        book[nx][ny][dr]=1;
        while(true)
        {
            if(nx<=lx||nx>=rx||ny<=ly||ny>=ry){
                flag=1;
                break;
            }
            int dx,dy;
            getd(dr,dx,dy);
            if(flies[nx+dx][ny+dy]==0){
                nx+=dx;
                ny+=dy;
                book[nx][ny][dr]=1;
                continue;
            }
            if(book[nx][ny][(dr+1)%4]==0)
            {
                dr=(dr+1)%4;
                tus++;
                book[nx][ny][dr]=1;
                continue;
            }
            break;
        }
        if(!flag)
            cout<<-1<<endl;
        else
            cout<<tus<<endl;
            mset(book,0);
            mset(flies,0);
    }
    return 0;
}