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| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 6668 | Accepted: 2725 |
Description

Input
Output
Sample Input
1 7 0 0 4 4 4 7 9 7 13 -1 8 -6 4 -4
Sample Output
80.00
/*
poj 1279 半平面交核面积
给你一个多边形的图书馆.要求得到一块地方能看见墙上所有的点,并求出面积
在半平面模板上加个求面积公式即可.
而且输入并没有指定顺时针还是逆时针,可以通过求面积进行判断.
hhh-2016-05-11 21:01:47
*/
#include <iostream>
#include <vector>
#include <cstring>
#include <string>
#include <cstdio>
#include <queue>
#include <cmath>
#include <algorithm>
#include <functional>
#include <map>
using namespace std;
#define lson (i<<1)
#define rson ((i<<1)|1)
typedef long long ll;
using namespace std;
const int maxn = 1510;
const double PI = 3.1415926;
const double eps = 1e-8;
int sgn(double x)
{
if(fabs(x) < eps) return 0;
if(x < 0)
return -1;
else
return 1;
}
struct Point
{
double x,y;
Point() {}
Point(double _x,double _y)
{
x = _x,y = _y;
}
Point operator -(const Point &b)const
{
return Point(x-b.x,y-b.y);
}
double operator ^(const Point &b)const
{
return x*b.y-y*b.x;
}
double operator *(const Point &b)const
{
return x*b.x + y*b.y;
}
};
struct Line
{
Point s,t;
double k;
Line() {}
Line(Point _s,Point _t)
{
s = _s;
t = _t;
k = atan2(t.y-s.y,t.x-s.x);
}
Point operator &(const Line &b) const
{
Point res = s;
double ta = ((s-b.s)^(b.s-b.t))/((s-t)^(b.s-b.t));
res.x += (t.x-s.x)*ta;
res.y += (t.y-s.y)*ta;
return res;
}
};
bool HPIcmp(Line a,Line b)
{
if(fabs(a.k-b.k) > eps) return a.k<b.k;
return ((a.s-b.s)^(b.t-b.s)) < 0;
}
Line li[maxn];
double CalArea(Point p[],int n)
{
double ans = 0;
for(int i = 0;i < n;i++)
{
ans += (p[i]^p[(i+1)%n])/2;
}
return ans;
}
double HPI(Line line[],int n,Point res[],int &resn)
{
int tot =n;
sort(line,line+n,HPIcmp);
tot = 1;
for(int i = 1; i < n; i++)
{
if(fabs(line[i].k - line[i-1].k) > eps)
line[tot++] = line[i];
}
int head = 0,tail = 1;
li[0] = line[0];
li[1] = line[1];
resn = 0;
for(int i = 2; i < tot; i++)
{
if(fabs((li[tail].t-li[tail].s)^(li[tail-1].t-li[tail-1].s)) < eps||
fabs((li[head].t-li[head].s)^(li[head+1].t-li[head+1].s)) < eps)
return 0;
while(head < tail && (((li[tail] & li[tail-1]) - line[i].s) ^ (line[i].t-line[i].s)) > eps)
tail--;
while(head < tail && (((li[head] & li[head+1]) - line[i].s) ^ (line[i].t-line[i].s)) > eps)
head++;
li[++tail] = line[i];
}
while(head < tail && (((li[tail] & li[tail-1]) - li[head].s) ^ (li[head].t-li[head].s)) > eps)
tail--;
while(head < tail && (((li[head] & li[head-1]) - li[tail].s) ^ (li[tail].t-li[tail].t)) > eps)
head++;
if(tail <= head+1)
return 0;
for(int i = head; i < tail; i++)
res[resn++] = li[i]&li[i+1];
if(head < tail-1)
res[resn++] = li[head]&li[tail];
double tans = 0;
for(int i = 0;i < resn;i++)
{
tans += (res[i]^res[(i+1)%resn])/2;
}
return fabs(tans);
}
Point p0;
Point lis[maxn];
Line line[maxn];
double dist(Point a,Point b)
{
return sqrt((a-b)*(a-b));
}
bool cmp(Point a,Point b)
{
double t = (a-p0)^(b-p0);
if(sgn(t) > 0)return true;
else if(sgn(t) == 0 && sgn(dist(a,lis[0])-dist(b,lis[0])) <= 0)
return true;
else
return false;
}
int main()
{
//freopen("in.txt","r",stdin);
int n,T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i = 0; i < n; i++)
{
scanf("%lf%lf",&lis[i].x,&lis[i].y);
}
int ans;
if(CalArea(lis,n) < 0)
reverse(lis,lis+n);
for(int i = 0; i < n; i++)
{
line[i] = Line(lis[i],lis[(i+1)%n]);
}
printf("%.2f\n",HPI(line,n,lis,ans));
}
return 0;
}
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原文地址:http://www.cnblogs.com/Przz/p/5510568.html