SCU 4445 Right turn

模拟。

每次找一下即将要遇到的那个点，这个数据范围可以暴力找，自己的写的时候二分了一下。如果步数大于$4*n$一定是$-1$。

#include<bits/stdc++.h>
using namespace std;

const int INF = 0x7FFFFFFF;
const int mod = 1e9 + ;
const int N = 5e6 + ;
const int M = 1e4 + ;
const double eps = 1e-;
int T,n,m;

struct P
{
    int x,y;
    P(int X=,int Y=)
    {
        x=X;
        y=Y;
    }
}p[],q[];

bool cmp1(P a,P b)
{
    if(a.x!=b.x) return a.x<b.x;
    return a.y<b.y;
}

bool cmp2(P a,P b)
{
    if(a.y!=b.y) return a.y<b.y;
    return a.x<b.x;
}

int nowx,nowy,nowd;

int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=;i<=n;i++)
        {
            scanf("%d%d",&p[i].x,&p[i].y);
            q[i].x=p[i].x;
            q[i].y=p[i].y;
        }

        sort(p+,p++n,cmp1);
        sort(q+,q++n,cmp2);

        nowx=nowy=; nowd=;

        int ans=;

        while()
        {
            if(nowd==)
            {
                int k = lower_bound(q+,q+n+,P(nowx,nowy),cmp2)-q;
                if(k==n+||q[k].y!=nowy)
                {
                    printf("%d\n",ans);
                    break;
                }
                else ans++,nowx = q[k].x-,nowd=(nowd+)%;
            }
            else if(nowd==)
            {
                int k = lower_bound(q+,q+n+,P(nowx,nowy),cmp2)-q-;
                if(k==||q[k].y!=nowy)
                {
                    printf("%d\n",ans);
                    break;
                }
                else ans++,nowx = q[k].x+,nowd=(nowd+)%;
            }

            else if(nowd==)
            {
                int k = lower_bound(p+,p+n+,P(nowx,nowy),cmp1)-p-;
                if(k==||p[k].x!=nowx)
                {
                    printf("%d\n",ans);
                    break;
                }
                else ans++,nowy = p[k].y+,nowd=(nowd+)%;
            }

            else
            {
                int k = lower_bound(p+,p+n+,P(nowx,nowy),cmp1)-p;
                if(k==n+||p[k].x!=nowx)
                {
                    printf("%d\n",ans);
                    break;
                }
                else ans++,nowy = p[k].y-,nowd=(nowd+)%;
            }

            if(ans>*n)
            {
                printf("-1\n");
                break;
            }
        }
    }
    return ;
}