ECNU 1624 求交集多边形面积
其实我是想做hdu3060的。。。写完了发现。。。这题坑姐，还有凹的情况。。。无奈先找个OJ把俩凸的代码给交了。。
求俩凸多边形面积。水题啊。本来我想了一种方法，想着能水过去呢。一直WA。后来用半平面交算了。。。好久没写半平面交的题了。。。
刚改了点BUG，我的那个方法也水过去啦！！！(*^__^*) 嘻嘻……
我的是M*N的算法，半平面交是log(N+M)*(N+M)的。
我这个就很好想啦，既然是求面积的交，因为是凸多边形，所以两个的交集一定也是凸的（如果存在的话）。
而且稍微想一下就很清楚，交得的凸多边形的顶点要么是交点，要么是在另一个凸多边形里的点。所以把这些个点弄到一个集合里，极角排序，然后求面积即可~~~
需要特判是否一个完全包含另一个，还有不想交的情况~
中间一直WA是因为，求得的点集如果有重点，那么用极角排序（用atan2）是很不靠谱的！！！因为两个相同点的atan2的结果是0！而极角排序是从负值到正值，如果俩点相同，经过极角排序后，它俩不在一起了，哭。所以，在极角排序前去重下就没问题啦。
有很大改进空间滴。。
我的N*M做法
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <limits.h>
#include <string.h>
#include <string>
#include <algorithm>
#define MID(x,y) ( ( x + y ) >> 1 )
#define L(x) ( x << 1 )
#define R(x) ( x << 1 | 1 )
#define FOR(i,s,t) for(int i=(s); i<(t); i++)
#define BUG puts("here!!!")
#define STOP system("pause")
#define file_r(x) freopen(x, "r", stdin)
#define file_w(x) freopen(x, "w", stdout)
using namespace std;
const int MAX = 510;
const double eps = 1e-6;
bool dy(double x,double y) { return x > y + eps;} // x > y
bool xy(double x,double y) { return x < y - eps;} // x < y
bool dyd(double x,double y) { return x > y - eps;} // x >= y
bool xyd(double x,double y) { return x < y + eps;} // x <= y
bool dd(double x,double y) { return fabs( x - y ) < eps;} // x == y
struct point{
double x, y;
void get()
{
scanf("%lf%lf", &x, &y);
}
bool operator==(const point &a)
{
return dd(a.x, x) && dd(a.y, y);
}
};
double crossProduct(point a,point b,point c)//向量 ac 在 ab 的方向 顺时针是正
{
return (c.x - a.x)*(b.y - a.y) - (b.x - a.x)*(c.y - a.y);
}
point l2l_inst_p(point u1,point u2,point v1,point v2)
{
point ans = u1;
double t = ((u1.x - v1.x)*(v1.y - v2.y) - (u1.y - v1.y)*(v1.x - v2.x))/
((u1.x - u2.x)*(v1.y - v2.y) - (u1.y - u2.y)*(v1.x - v2.x));
ans.x += (u2.x - u1.x)*t;
ans.y += (u2.y - u1.y)*t;
return ans;
}
bool onSegment(point a, point b, point c)
{
if( dd(crossProduct(a,b,c),0.0) && dyd(c.x,min(a.x,b.x)) &&
xyd(c.x,max(a.x,b.x)) && dyd(c.y,min(a.y,b.y)) && xyd(c.y,max(a.y,b.y)) )
return true;
return false;
}
bool s2s_inst(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;
return false;
}
point p1[MAX], p2[MAX];
bool pin_convexh(point *p,int n,point a)
{
p[n] = p[0]; p[n+1] = p[1];
for(int i=0; i<n; i++)
if( xy(crossProduct(p[i],p[i+1],a)*
crossProduct(p[i+1],p[i+2],a),0.0) )
return false;
return true;
}
double area_polygon(point p[],int n)
{
if( n < 3 ) return 0.0;
double s = 0.0;
for(int i=0; i<n; i++)
s += p[(i+1)%n].y * p[i].x - p[(i+1)%n].x * p[i].y;
return fabs(s)/2.0;
}
bool ainb(point *a, int n, point *b, int m)
{
FOR(i, 0, n)
if( !pin_convexh(b, m, a[i]) )
return false;
return true;
}
bool inst(point *a, int n, point *b, int m)
{
FOR(i, 0, n)
FOR(k, 0, m)
if( s2s_inst(a[i], a[i+1], b[k], b[k+1]) )
return true;
return false;
}
point t[MAX*2], C;
bool cmp(point a,point b)
{
double t1 = atan2(a.y - C.y, a.x - C.x);
double t2 = atan2(b.y - C.y, b.x - C.x);
if( dd(t1, t2) ) return xy(fabs(a.x),fabs(b.x));
return xy(t1, t2);
}
bool cmp_p(point a, point b)
{
if( dd(a.x, b.x) )
return xy(a.y, b.y);
return xy(a.x, b.x);
}
bool cmp_equal(point a, point b)
{
return a == b;
}
double solve(int n, int m)
{
double area1 = area_polygon(p1, n);
double area2 = area_polygon(p2, m);
if( ainb(p1, n, p2, m) || ainb(p2, m, p1, n) ) // 如果一个在另一个里面，答案是面积最小的
return min(area1, area2);
if( !inst(p1, n, p2, m) ) // 如果两个不相交
return 0;
// 如果两个相交，纠结啊
int cnt = 0;
FOR(i, 0, n)
FOR(k, 0, m)
if( s2s_inst(p1[i], p1[i+1], p2[k], p2[k+1]) )
t[cnt++] = l2l_inst_p(p1[i], p1[i+1], p2[k], p2[k+1]);
FOR(i, 0, n)
if( pin_convexh(p2, m, p1[i]) )
t[cnt++] = p1[i];
FOR(i, 0, m)
if( pin_convexh(p1, n, p2[i]) )
t[cnt++] = p2[i];
sort(t, t+cnt, cmp_p);
cnt = unique(t, t+cnt, cmp_equal) - t;
C = t[0];
sort(t+1, t+cnt, cmp);
cnt = unique(t, t+cnt, cmp_equal) - t;
double area = area_polygon(t, cnt);
return area;
}
int main()
{
int n, m;
while( ~scanf("%d", &n) )
{
FOR(i, 0, n)
p1[i].get();
scanf("%d", &m);
FOR(i, 0, m)
p2[i].get();
p1[n] = p1[0]; p2[m] = p2[0];
double ans = solve(n, m);
printf("%.2lf\n", ans);
}
return 0;
}
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <limits.h>
#include <string.h>
#include <string>
#include <algorithm>
#define MID(x,y) ( ( x + y ) >> 1 )
#define L(x) ( x << 1 )
#define R(x) ( x << 1 | 1 )
#define FOR(i,s,t) for(int i=(s); i<(t); i++)
#define BUG puts("here!!!")
#define STOP system("pause")
#define file_r(x) freopen(x, "r", stdin)
#define file_w(x) freopen(x, "w", stdout)
using namespace std;
const int MAX = 510;
const double eps = 1e-8;
bool dy(double x,double y) { return x > y + eps;} // x > y
bool xy(double x,double y) { return x < y - eps;} // x < y
bool dyd(double x,double y) { return x > y - eps;} // x >= y
bool xyd(double x,double y) { return x < y + eps;} // x <= y
bool dd(double x,double y) { return fabs( x - y ) < eps;} // x == y
struct point{
double x, y;
void get()
{
scanf("%lf%lf", &x, &y);
}
bool operator==(const point &a)
{
return dd(a.x, x) && dd(a.y, y);
}
};
double crossProduct(point a,point b,point c)//向量 ac 在 ab 的方向 顺时针是正
{
return (c.x - a.x)*(b.y - a.y) - (b.x - a.x)*(c.y - a.y);
}
double area_polygon(point p[],int n)
{
if( n < 3 ) return 0.0;
double s = 0.0;
for(int i=0; i<n; i++)
s += p[(i+1)%n].y * p[i].x - p[(i+1)%n].x * p[i].y;
return fabs(s)/2.0;
}
point t[MAX*2], p1[MAX], p2[MAX];
struct line{
point a, b;
double ang;
};
line ln[MAX*2], deq[MAX*2];
bool cmp_equal(point a, point b)
{
return a == b;
}
bool equal_ang(line a,line b) // 第一次unique的比较函数
{
return dd(a.ang, b.ang);
}
bool cmphp(line a,line b) // 排序的比较函数
{
if( dd(a.ang,b.ang) ) return xy(crossProduct(b.a,b.b,a.a),0.0);
return xy(a.ang,b.ang);
}
bool equal_p(point a,point b)//第二次unique的比较函数
{
return dd(a.x,b.x) && dd(a.y,b.y);
}
void makeline_hp(double x1,double y1,double x2,double y2,line &l)
{
l.a.x = x1; l.a.y = y1; l.b.x = x2; l.b.y = y2;
l.ang = atan2(y2 - y1,x2 - x1);
}
void makeline_hp(point a,point b,line &l) // 线段(向量ab)左侧侧区域有效
{
l.a = a; l.b = b;
l.ang = atan2(b.y - a.y,b.x - a.x); // 如果是右侧区域，改成a.y - b.y,a.x - b.x
}
bool parallel(line u,line v)
{
return dd( (u.a.x - u.b.x)*(v.a.y - v.b.y) - (v.a.x - v.b.x)*(u.a.y - u.b.y) , 0.0 );
}
point l2l_inst_p(line l1,line l2)
{
point ans = l1.a;
double t = ((l1.a.x - l2.a.x)*(l2.a.y - l2.b.y) - (l1.a.y - l2.a.y)*(l2.a.x - l2.b.x))/
((l1.a.x - l1.b.x)*(l2.a.y - l2.b.y) - (l1.a.y - l1.b.y)*(l2.a.x - l2.b.x));
ans.x += (l1.b.x - l1.a.x)*t;
ans.y += (l1.b.y - l1.a.y)*t;
return ans;
}
void inst_hp_nlogn(line *ln,int n,point *s,int &len)
{
len = 0;
sort(ln,ln+n,cmphp);
n = unique(ln,ln+n,equal_ang) - ln;
int bot = 0,top = 1;
deq[0] = ln[0]; deq[1] = ln[1];
for(int i=2; i<n; i++)
{
if( parallel(deq[top],deq[top-1]) || parallel(deq[bot],deq[bot+1]) )
return ;
while( bot < top && dy(crossProduct(ln[i].a,ln[i].b,
l2l_inst_p(deq[top],deq[top-1])),0.0) )
top--;
while( bot < top && dy(crossProduct(ln[i].a,ln[i].b,
l2l_inst_p(deq[bot],deq[bot+1])),0.0) )
bot++;
deq[++top] = ln[i];
}
while( bot < top && dy(crossProduct(deq[bot].a,deq[bot].b,
l2l_inst_p(deq[top],deq[top-1])),0.0) ) top--;
while( bot < top && dy(crossProduct(deq[top].a,deq[top].b,
l2l_inst_p(deq[bot],deq[bot+1])),0.0) ) bot++;
if( top <= bot + 1 ) return ;
for(int i=bot; i<top; i++)
s[len++] = l2l_inst_p(deq[i],deq[i+1]);
if( bot < top + 1 ) s[len++] = l2l_inst_p(deq[bot],deq[top]);
len = unique(s,s+len,equal_p) - s;
}
double solve(int n, int m)
{
int cnt = 0;
FOR(i, 0, n)
makeline_hp(p1[i], p1[i+1], ln[cnt++]);
FOR(i, 0, m)
makeline_hp(p2[i], p2[i+1], ln[cnt++]);
int len;
inst_hp_nlogn(ln, cnt, t, len);
if( len == 0 ) return 0;
double area = area_polygon(t, len);
return area;
}
int main()
{
int n, m;
while( ~scanf("%d", &n) )
{
for(int i=n-1; i>=0; i--)
p1[i].get();
scanf("%d", &m);
for(int i=m-1; i>=0; i--)
p2[i].get();
p1[n] = p1[0]; p2[m] = p2[0];
double ans = solve(n, m);
printf("%.2lf\n", ans);
}
return 0;
}