学步园

题目描述：

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 (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, 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

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

Sample Output

INSIDE 

一道比较简单的计算几何，就是求给定点是否在一个多边形内部，我们可以从这个点引出一条射线，计算其与所有边的交点，如果是奇数个，说明在内部，否则在外部。

那么我们取怎样的射线比较合适？

因为题目告诉我们边平行与坐标轴，我们可以取与Y正方向平行大致向左偏一点的射线，就可以避免其和平行于Y轴的边相交，这就是代码中    if(p[i].a.x<P.x&&p[i].b.x>=P.x)k++;  左边取小于号的原因。

附上AC代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#include<algorithm>
#include<vector>
#include<cstdlib>

#define inf 0xfffffff
#define CLR(a,b) memset((a),(b),sizeof((a)))
#define FOR(a,b) for(int a=1;a<=(b);(a)++)

using namespace std;
int const nMax = 10010;
int const base = 10;
typedef int LL;
typedef pair<LL,LL> pij;

struct point{
    int x,y;
    bool operator <(const point& b)const {
        if(x==b.x)return y<b.y;
        else return x<b.x;
    }
} P;

struct seg{
    point a,b;
};
seg p[nMax];

int n;
int main(){
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d%d%d%d",&p[i].a.x,&p[i].a.y,&p[i].b.x,&p[i].b.y);
        if(p[i].b<p[i].a)swap(p[i].a,p[i].b);
    }
    scanf("%d%d",&P.x,&P.y);
    int k=0;
    for(int i=0;i<n;i++){
        if(p[i].a.x==p[i].b.x){
            if(P.x==p[i].a.x&&P.y>=p[i].a.y&&P.y<=p[i].b.y){
                printf("BORDER\n");
                return 0;
            }
        }else {
            if(p[i].a.y==P.y&&p[i].a.x<=P.x&&p[i].b.x>=P.x){
                printf("BORDER\n");
                return 0;
            }
        }
    }
    for(int i=0;i<n;i++){
        if(p[i].a.y==p[i].b.y&&p[i].a.y>P.y){
            if(p[i].a.x<P.x&&p[i].b.x>=P.x)k++;
        }
    }
    if(k&1)printf("INSIDE\n");
    else   printf("OUTSIDE\n");
    return 0;
}