SCU4445——模拟——Right turn

Right turn

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

frog is initially in grid (0,0), heading grid (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≤103). Each of the following n lines contains 2 integers xi,yi. (|xi|,|yi|≤109,(xi,yi)≠(0,0), all (xi,yi) are distinct)

Output

For each test, write 1 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

/*

   模拟题 机器人碰了就往右拐

   每次判断要与所有点判断并且是靠自己最近的点

*/

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<cmath>

using namespace std;



const int inf = 0x3f3f3f3f;

int dx[] = { 1, 0, -1, 0};

int dy[] = { 0, -1, 0, 1};

int vis[1100][4];

int obx[1100], oby[1100];

int n;



void work()

{

    memset(vis, 0, sizeof(vis));

    int dir = 0;

    int nx , ny, dis;

    nx = ny = 0;

    for(int step = 0; ; step++){

        int mx = inf;

        dis = inf;

        int index = -1;

        for(int i = 1; i <= n ; i++){

            if(dir == 0 ){

                if(ny == oby[i] && nx < obx[i]){

                    dis = obx[i] - nx - 1;

                }

            }        

            else if(dir == 1){

                if(nx == obx[i] && ny > oby[i]){

                //   printf("%d ",i);

                    dis = ny - oby[i] - 1;

                }

            }

            else if(dir == 2){

                if(ny == oby[i] && nx > obx[i]){

                    dis = nx - obx[i] - 1;

                }

            }

            else {

                if(nx == obx[i] && ny < oby[i]){

                    dis = oby[i] - ny - 1;

                }

            }

            if(dis < mx){

                mx = dis;

                index = i;

            }

        }

       // printf("%d %d %d\n",index, step,dir);

    

      if(index == -1) {

            printf("%d\n",step);

            return ;

        }

      if(vis[index][dir]){

          printf("-1\n");

          return ;

      }



        vis[index][dir] = 1;

            if(dir == 0){

                nx = obx[index] - 1; ny = oby[index];

            }

            else if(dir == 1){

                nx = obx[index]; ny = oby[index] + 1;

            }

            else if(dir == 2){

                nx = obx[index] + 1; ny = oby[index];

            }

            else {

                nx = obx[index]; ny = oby[index] - 1;

            }

            //  printf("%d %d %d \n",nx,ny, dir);

        dir = (dir + 1) % 4;

    }

}



int main()

{

    while(~scanf("%d", &n)){

        for(int i = 1; i <= n; i++){

            scanf("%d%d", &obx[i], &oby[i]);

        }

        work();

    }

    return 0;

}