sgu 124 Broken line

最新推荐文章于 2015-06-04 13:51:53 发布

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最新推荐文章于 2015-06-04 13:51:53 发布

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分类专栏： sgu 计算几何

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本文介绍了一种简单的方法来判断给定点是否位于由直线段构成的多边形内部。通过从该点引出一条特定方向的射线并与多边形的所有边进行相交检查，我们可以得出点的位置信息。

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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, 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

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;
}