zoj 1010

#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<iostream>
using namespace std;

const int maxn = 1000+10;
int n;

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);
}
}p[maxn];

double Cross(Point a, Point b) { //叉积
return a.x*b.y - a.y*b.x;
}

// = 0 表示,相交于端点也是认定为相交
bool SegmentProperIntersection(Point s1, Point e1, Point s2, Point e2) {
double c1 = Cross(s2-s1, e1-s1), c2 = Cross(e1-s1, e2-s1),
c3 = Cross(s1-s2, e2-s2), c4 = Cross(e2-s2, e1-s2);
if(c1*c2 >= 0 && c3*c4 >= 0) return true;
return false;
}

bool haveCross() {

//从第 三 条线段开始, 一直到第 N-1 条线段
for(int i = 2; i < n-1; i++) { //检查 Lin(i,i+1) 与前面每一条不直接相连的线段
for(int j = 1; j < i; j++) {
if(SegmentProperIntersection(p[i], p[i+1], p[j-1], p[j])) {
return true;
}
}
}

//判断最后一条线段p[0]_p[n-1]
//从第二条线段一直比较到第 n-2 条线段
for(int i = 1; i < n-2 ; i++) { //p[i]__p[i+1]
if(SegmentProperIntersection(p[0],p[n-1],p[i],p[i+1])) {
return true;
}
}
return false;
}

//叉积求面积
double Area() {
double area = 0;
for(int i = 1; i < n-1; i++)
area += Cross(p[i]-p[0], p[i+1]-p[0]);
return fabs(area) / 2.0;
}

int main()
{
int test = 0;
while(scanf("%d", &n) != EOF) {
if(n == 0) break;

double x,y;
for(int i = 0; i < n; i++) {
scanf("%lf%lf", &x,&y);
p[i] = Point(x,y);
}

if(n < 3) { //如果小于三个点, 肯定不能求面积
printf("Figure %d: Impossible\n", ++test);
printf("\n");
continue;
}

if(haveCross()) { //如果有线段相交的情况
printf("Figure %d: Impossible\n", ++test);
printf("\n");
continue;
}
else printf("Figure %d: %.2lf", ++test,Area());
printf("\n");
}
return 0;
}