sgu124
| 124. Broken line time limit per test: 0.5 sec. 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 xi1,yi1,xi2,yi2; 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 |
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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;
}