SGU 124. Broken line（射线法判断一个点是否在一个多边形内）

There is a closed broken line on a plane with sides parallel to coordinate axes, without self-crossings and self-contacts. The broken line consists of K segments. You have to determine, whether a given point with coordinates (X₀,Y₀) is inside this closed broken line, outside or belongs to the broken line.

Input

The first line contains integer K (4 Ј K Ј 10000) - the number of broken line segments. Each of the following N lines contains coordinates of the beginning and end points of the segments (4 integer x_i1,y_i1,x_i2,y_i2; all numbers in a range from -10000 up to 10000 inclusive). Number separate by a space. The segments are given in random order. Last line contains 2 integers X₀ and Y₀ - the coordinates of the given point delimited by a space. (Numbers X₀, Y₀ in a range from -10000 up to 10000 inclusive).

Output

The first line should contain:

INSIDE - if the point is inside closed broken line,

OUTSIDE - if the point is outside,

BORDER - if the point belongs to broken line.

Sample Input

4
0 0 0 3
3 3 3 0
0 3 3 3
3 0 0 0
2 2

Sample Output

INSIDE

题意：给你一些平行于座标轴的线段，这些线段连成一个多边形，让你判断一个点是在这个多边形上，多边形内还是多边形外。

因为给的边是乱序的，所以不能用叉积判断。

所以用射线法，以要求的点往一个方向作射线（当然不能过多边形的端点），判断与多边形的几条边相交，如果是奇数条，那么就是在多边形内，否则就是在多边形外。

判断在不在多边形上就不说了。

当然这题还有个坑就是这个多边形的边有可能是有多条线段组成的，你作的射线如果穿过两条平行的且相连的两条线段，就有可能会算成两条，实际上只能算一条，所以我们判断的时候把这些线段当成左闭右开的就好了。

（这题里面的线段都是平行于座标轴的，所以就不用叉积判断相交了）

#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long

using namespace std;

int x[10010],y[10010],xx[10010],yy[10010];

int main(void)
{
    int n,i,j;
    while(scanf("%d",&n)==1)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d%d%d%d",&x[i],&y[i],&xx[i],&yy[i]);
            if(x[i] > xx[i])
                swap(x[i],xx[i]);
            if(y[i] > yy[i])
                swap(y[i],yy[i]);
        }
        int tx,ty;
        scanf("%d%d",&tx,&ty);
        int flag = 0;
        for(i=1;i<=n;i++)
        {
            if(x[i] == xx[i] && x[i] == tx && y[i] <= ty && ty <= yy[i])
                flag = 1;
            if(y[i] == yy[i] && y[i] == ty && x[i] <= tx && tx <= xx[i])
                flag = 1;
        }
        if(flag == 1)
            printf("BORDER\n");
        else
        {
            int cnt = 0;
            for(i=1;i<=n;i++)
            {
                if(x[i] == xx[i])
                    continue;
                if(x[i] <= tx && tx < xx[i] && y[i] > ty)
                    cnt++;
            }
            if(cnt % 2 == 0)
                printf("OUTSIDE\n");
            else
                printf("INSIDE\n");
        }
    }

    return 0;
}