sgu124

124. Broken line

time limit per test: 0.5 sec.
memory limit per test: 4096 KB

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 (4ЈK 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 integerx_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, Y₀in 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

40 0 0 33 3 3 00 3 3 33 0 0 02 2

Sample Output

INSIDE

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Author: Alex Y. Suslov, Sergey V. MironovResource: 5th Southern Subregional Contest. Saratov 2002Date: 2002-10-10

题目大意，给你一个多边形以及一个点（保证多边形每条边都平行于坐标轴），判断该点是否在多边形内部

方法：可以考虑从该点做一条射线，判断与每一条边是否相交即可。

如果相交次数不是2的倍数，那么说明该点在多边形内部，否则在多边形外部。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>using namespace std;struct point{    double x,y;};struct edge{    point a,b;}e[10101];int n;point p1;int main(){    //init    std::ios::sync_with_stdio(false);    cin >> n;    for (int i=1;i<=n;++i){cin >> e[i].a.x >> e[i].a.y    >> e[i].b.x >> e[i].b.y;if (e[i].a.x==e[i].b.x &&     e[i].a.y>e[i].b.y) swap(e[i].a.y,e[i].b.y);if (e[i].a.y==e[i].b.y &&    e[i].a.x>e[i].b.x) swap(e[i].a.x,e[i].b.x);    }    cin >> p1.x >> p1.y;    //solve    for (int i=1;i<=n;++i){if (e[i].a.x==e[i].b.x && e[i].a.x==p1.x &&    e[i].a.y<=p1.y && e[i].b.y>=p1.y){    cout << "BORDER" << endl;    return 0;}else if (e[i].a.y==e[i].b.y && e[i].a.y==p1.y && e[i].a.x<=p1.x && e[i].b.x >=p1.x){    cout << "BORDER" << endl;    return 0;}    }    int k=0;    for (int i=1;i<=n;++i){if ((e[i].a.y == e[i].b.y) && (e[i].a.y>p1.y) &&    (e[i].a.x<p1.x) && (e[i].b.x>=p1.x))    k++;    }    //print    if (k & 1==1) cout << "INSIDE" << endl;    else cout << "OUTSIDE" << endl;    return 0;}