zoj 1081 判断点在多边形内

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=81Points Within

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Statement of the Problem

Several drawing applications allow us to draw polygons and almost all of them allow us to fill them with some color. The task of filling a polygon reduces to knowing which points are inside it, so programmers have to colour only those points.

You're expected to write a program which tells us if a given point lies inside a given polygon described by the coordinates of its vertices. You can assume that if a point is in the border of the polygon, then it is in fact inside the polygon.

Input Format

The input file may contain several instances of the problem. Each instance consists of: (i) one line containing integers N, 0 < N < 100 and M, respectively the number of vertices of the polygon and the number of points to be tested. (ii) N lines, each containing a pair of integers describing the coordinates of the polygon's vertices; (iii) M lines, each containing a pair of integer coordinates of the points which will be tested for "withinness" in the polygon.

You may assume that: the vertices are all distinct; consecutive vertices in the input are adjacent in the polygon; the last vertex is adjacent to the first one; and the resulting polygon is simple, that is, every vertex is incident with exactly two edges and two edges only intersect at their common endpoint. The last instance is followed by a line with a 0 (zero).

Output Format

For the ith instance in the input, you have to write one line in the output with the phrase "Problem i:", followed by several lines, one for each point tested, in the order they appear in the input. Each of these lines should read "Within" or "Outside", depending on the outcome of the test. The output of two consecutive instances should be separated by a blank line.

Sample Input

3 1
0 0
0 5
5 0
10 2
3 2
4 4
3 1
1 2
1 3
2 2
0

Sample Output

Problem 1:
Outside

Problem 2:
Outside
Within

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Source: South America 2001

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题意就是判断是否在多边形内，是个的话就输出Within……

看书上敲的，不是太懂，没用到射线求交点稀里糊涂的就求出来了

先贴个好点的代码：转至：http://blog.csdn.net/zxy_snow/article/details/6339621

#include <queue>
#include <stack>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <limits.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int MAX = ;
const double eps = 1e-;
struct point
{
    double x,y;
};
struct beeline
{
    point a,b;
};
point p[MAX];
int n;

bool dy(double x,double y)  // x > y
{
    return x > y + eps;
}
bool xy(double x,double y) // x < y
{
    return x < y - eps;
}
bool dyd(double x,double y) // x >= y
{
    return x > y - eps;
}
bool xyd(double x,double y) // x <= y
{
    return x < y + eps;
}
bool dd(double x,double y)  // x == y
{
    return fabs( x - y ) < eps;
}
double crossProduct(point a,point b,point c)//向量 ac 在 ab 的方向
{
    return (c.x - a.x)*(b.y - a.y) - (b.x - a.x)*(c.y - a.y);
}
bool onSegment(point a, point b, point c)
{
    double maxx = max(a.x,b.x);
    double maxy = max(a.y,b.y);
    double minx = min(a.x,b.x);
    double miny = min(a.y,b.y);
    if( dd(crossProduct(a,b,c),0.0) && dyd(c.x,minx) && xyd(c.x,maxx) && dyd(c.y,miny) && xyd(c.y,maxy) )
        return true;
    return false;
}
bool segIntersect(point p1,point p2, point p3, point p4)
{
    double d1 = crossProduct(p3,p4,p1);
    double d2 = crossProduct(p3,p4,p2);
    double d3 = crossProduct(p1,p2,p3);
    double d4 = crossProduct(p1,p2,p4);
    if( xy(d1 * d2,0.0) && xy( d3*d4,0.0 ) )
        return true;
    if( dd(d1,0.0) && onSegment(p3,p4,p1) )
        return true;
    if( dd(d2,0.0) && onSegment(p3,p4,p2) )
        return true;
    if( dd(d3,0.0) && onSegment(p1,p2,p3) )
        return true;
    if( dd(d4,0.0) && onSegment(p1,p2,p4) )
        return true;
    return false;
}
bool inPolygon(point pot)
{
    int count = ;
    beeline l;
    l.a = pot;
    l.b.x = 1e10;
    l.b.y = pot.y;
    p[n] = p[];
    for(int i=; i<n; i++)
    {
        if( onSegment(p[i],p[i+],pot) )
            return true;
        if( !dd(p[i].y,p[i+].y) )//水平边不考虑
        {
            int tmp = -;
            if( onSegment(l.a,l.b,p[i]) )//对于顶点与射线相交，该顶点应是所属边上纵坐标上较大的
                tmp = i;
            else if( onSegment(l.a,l.b,p[i+]) )
                tmp = i+;
            if( tmp != - && dd(p[tmp].y,max(p[i].y,p[i+].y)) )
                count++;
            else if( tmp == - && segIntersect(p[i],p[i+],l.a,l.b) )//相交
                count++;
        }
    }
    if( count %  ==  )
        return true;
    return false;
}
int main()
{
    int m;
    int ind = ;
    point pot;
    while( ~scanf("%d",&n) && n )
    {
        if( ind !=  )
            printf("\n");
        scanf("%d",&m);
        for(int i=; i<n; i++)
            scanf("%lf %lf",&p[i].x,&p[i].y);
        printf("Problem %d:\n",ind++);
        while( m-- )
        {
            scanf("%lf %lf",&pot.x,&pot.y);
            if( inPolygon(pot) )
                printf("Within\n");
            else
                printf("Outside\n");
        }
    }
    return ;
}

下面是我的代码，写的自己不是很懂

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <iostream>
#include <stdlib.h>
#include <map>

#define eps 1.0e-5
inline double max(double a,double b){return a>b?a:b;}
inline double min(double a,double b){return a<b?a:b;}
inline double dabs(double a ){return a<?-a:a;}

struct point
{
    double x,y;
};

point poly[];
int n,m;

bool online(const point &p1,const point &p2,const point &p3)
{
    if(p2.x>=min(p1.x,p3.x)&&p2.x<=max(p1.x,p3.x)&&
        p2.y>=min(p1.y,p3.y)&&p2.y<=max(p1.y,p3.y))
        {
            if(dabs((p2.x-p1.x)*(p3.y-p1.y)-(p3.x-p1.x)*(p2.y-p1.y))<=eps)
            return true;
        }
        return false;
}

bool insidepolygon(point p)
{
    int  count = ;
    double xinters;
    point p1,p2;
    p1=poly[];

    for(int i=;i<=n;i++)
    {
        p2=poly[i%n];
        if(online(p1,p,p2))return true;
        if(p.y>min(p1.y,p2.y))
        {
            if(p.y<=max(p1.y,p2.y))
            {
                if(p.x<=max(p1.x,p2.x))
                {
                    if(p1.y!=p2.y)
                    {
                        xinters=(p.y-p1.y)*(p2.x-p1.x)/(p2.y-p1.y)+p1.x;
                        //(p.y-p1.y)/(xinter-p1.x)=(p2.y-p1.y)/(p2.x-p1.x)斜率
                        if(p1.x == p2.x||p.x<=xinters)count++;
                    }
                }
            }
        }
        p1=p2;
    }
    if(count% == )
        return false;
    return true;
}

int main()
{
    int i,j;
    point p;
    int cas=;
    while(scanf("%d",&n)!=EOF)
    {
        if(n == )break;
        if(cas>)printf("\n");
        printf("Problem %d:\n",cas++);

        scanf("%d",&m);
        for(i=;i<n;i++)
        {
            scanf("%lf%lf",&poly[i].x,&poly[i].y);
        }

        for(j=;j<m;j++)
        {
            scanf("%lf%lf",&p.x,&p.y);
            if(insidepolygon(p))
            {
                printf("Within\n");
            }
            else
            {
                printf("Outside\n");
            }
        }
    }
    return ;
}