SGU 124. Broken line（射線法判斷一個點是否在一個多邊形內）

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

題意：給你一些平行於座標軸的線段，這些線段連成一個多邊形，讓你判斷一個點是在這個多邊形上，多邊形內還是多邊形外。

因爲給的邊是亂序的，所以不能用叉積判斷。

所以用射線法，以要求的點往一個方向作射線（當然不能過多邊形的端點），判斷與多邊形的幾條邊相交，如果是奇數條，那麼就是在多邊形內，否則就是在多邊形外。

判斷在不在多邊形上就不說了。

當然這題還有個坑就是這個多邊形的邊有可能是有多條線段組成的，你作的射線如果穿過兩條平行的且相連的兩條線段，就有可能會算成兩條，實際上只能算一條，所以我們判斷的時候把這些線段當成左閉右開的就好了。

（這題裏面的線段都是平行於座標軸的，所以就不用叉積判斷相交了）

#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;
}