POJ 1279 Art Gallery 半平面交 + 多边形面积

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
#define eps 1e-8

struct point
{
double x, y;
}p[1505], tmp[1505], q[1505];

double a, b, c;

void get_line(point p1, point p2)
{
a = p2.y - p1.y;
b = p1.x - p2.x;
c = p2.x * p1.y - p2.y * p1.x;
}

point intersect(point p1, point p2)
{
double u = fabs(a * p1.x + b * p1.y + c);
double v = fabs(a * p2.x + b * p2.y + c);
point ret;
ret.x = (p1.x * v + p2.x * u) / (u + v);
ret.y = (p1.y * v + p2.y * u) / (u + v);
return ret;
}

int n, m;

void cut()
{
int tm = 0;
int i;
for(i = 1; i <= m; i++)
{
if(a * q[i].x + b * q[i].y + c >= 0)
tmp[++tm] = q[i];
else
{
if(a * q[i-1].x + b * q[i-1].y + c > 0)
tmp[++tm] = intersect(q[i-1], q[i]);

if(a * q[i+1].x + b * q[i+1].y + c > 0)
tmp[++tm] = intersect(q[i], q[i+1]);
}
}
for(i = 1; i <= tm; i++)
q[i] = tmp[i];
q[0] = tmp[tm];
q[tm + 1] = tmp[1];
m = tm;
}

void solve()
{
int i;
for(i = 1; i <= n; i++)
q[i] = p[i];
p[n+1] = p[1];
q[0] = q
;
q[n+1] = q[1];
m = n;
for(i = 1; i <= n; i++)
{
get_line(p[i], p[i+1]);
cut();
}
}

double cal(int n, point *p)
{
int i;
double s = 0;
p[n+1] = p[1];
for(i = 1; i <= n; i++)
s += p[i].x * p[i+1].y - p[i].y * p[i+1].x;
return fabs(s) / 2;
}

int main()
{
int i, j, cas;
scanf("%d", &cas);
while(cas--)
{
scanf("%d", &n);
for(i = 1;i <= n; i++)
scanf("%lf%lf", &p[i].x, &p[i].y);
solve();
if(m >= 3)printf("%.2f\n", cal(m, q) );
else printf("0.00\n");
}
return 0;
}