模拟+贪心 SCU 4445 Right turn

题目传送门

/*
    题意；从原点出发，四个方向，碰到一个点向右转，问多少次才能走出，若不能输出-1
    模拟：碰到的点横坐标相等或纵坐标相等，然而要先满足碰到点最近，
        当没有转向或走到之前走过的点结束循环。dir数组使得代码精简巧妙
    对点离原点排序竟然submit failed，别人的代码有毒！
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <cmath>
using namespace std;

const int MAXN = 1e3 + 10;
const int INF = 0x3f3f3f3f;
int x[MAXN], y[MAXN];
int dir[4][2] = {1, 0, 0, -1, -1, 0, 0, 1};
int n;
int vis[4][MAXN];

int work(void)
{
    int nx = 0, ny = 0;    int d = 0;    int res = 0;
    memset (vis, 0, sizeof (vis));

    while (true)
    {
        int mx = INF, p = -1;
        for (int i=1; i<=n; ++i)
        {
            if (d == 0 || d == 2)
            {
                if (ny != y[i])    continue;
                if ((x[i] - nx) * dir[d][0] < 0)    continue;
            }
            else
            {
                if (nx != x[i])    continue;
                if ((y[i] - ny) * dir[d][1] < 0)    continue;
            }

            int tmp = abs (x[i] - nx) + abs (y[i] - ny);
            if (tmp < mx)    {mx = tmp; p = i;}
        }

        if (p == -1)    return res;
        if (vis[d][p])    return -1;
        vis[d][p] = 1;
        nx = x[p] - dir[d][0];    ny = y[p] - dir[d][1];
        d = (d + 1) % 4;    res++;
    }

}

int main(void)        //SCU 4445 Right turn
{
//    freopen ("E.in", "r", stdin);

    while (scanf ("%d", &n) == 1)
    {
        for (int i=1; i<=n; ++i)
        {
            scanf ("%d%d", &x[i], &y[i]);
        }
        printf ("%d
", work ());
    }

    return 0;
}