链接：http://poj.org/problem?id=2284

题意：一个自动画图的机器在纸上（无限大）画图，笔尖从不离开纸，有n个指令，每个指令是一个坐标，因为笔尖不离开纸，所以相邻的坐标会连有一条直线，最后画笔再回到起始点。所以这个图是一个连通图，并且画笔走过的路径是一个欧拉回路。现在问题来了，这个图形将平面分成了几部分。

思路：题目说明白一些就是告诉你一些几何信息问平面被分成了几部分。可以用欧拉公式来做

欧拉公式：假设图的顶点个数为n，边数为m，区域数位r，则有 n - m + r = 2，前提必须是连通图

知道任意两个就能求第三个

#include<cstring>
#include<string>
#include<fstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cctype>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<stack>
#include<ctime>
#include<cstdlib>
#include<functional>
#include<cmath>
using namespace std;
#define PI acos(-1.0)
#define MAXN 90010
#define eps 1e-7
#define INF 0x3F3F3F3F      //0x7FFFFFFF
#define LLINF 0x7FFFFFFFFFFFFFFF
#define seed 1313131
#define MOD 1000000007
#define ll long long
#define ull unsigned ll
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

struct Point{   //点
    double x, y;
    Point(double a = 0, double b = 0){
        x = a;
        y = b;
    }
};
struct LineSegment{ //线段
    Point s, e;
    LineSegment(Point a, Point b){
        s = a;
        e = b;
    }
};
struct Line{    //直线
    double a, b, c;
};
bool operator < (Point p1, Point p2){
    return (p1.x < p2.x || p1.x == p2.x) && p1.y < p2.y;
}
bool operator == (Point p1, Point p2){
    return fabs(p1.x - p2.x) < eps && fabs(p1.y - p2.y) < eps;
}
bool Online(LineSegment l, Point p){    //判断点是否在线段上
    return fabs((l.e.x - l.s.x) * (p.y - l.s.y) - (p.x - l.s.x) * (l.e.y - l.s.y)) < eps
        && (p.x - l.s.x) * (p.x - l.e.x) < eps && (p.y - l.s.y) * (p.y - l.e.y) < eps;
}
Line MakeLine(Point p1, Point p2){  //将线段延长为直线
    Line l;
    if(p2.y > p1.y){
        l.a = p2.y - p1.y;
        l.b = p1.x - p2.x;
        l.c = p1.y * p2.x - p1.x * p2.y;
    }
    else{
        l.a = p1.y - p2.y;
        l.b = p2.x - p1.x;
        l.c = p1.x * p2.y - p1.y * p2.x;
    }
    return l;   //返回直线
}
bool LineIntersect(Line l1, Line l2, Point &p){ //判断直线是否相交，并求出交点p
    double d = l1.a * l2.b - l2.a * l1.b;
    if(fabs(d) < eps) return false;
    //求交点
    p.x = (l2.c * l1.b - l1.c * l2.b) / d;
    p.y = (l2.a * l1.c - l1.a * l2.c) / d;
    return true;
}
bool LineSegmentIntersect(LineSegment l1, LineSegment l2, Point &p){    //判断线段是否相交
    Line a, b;
    a = MakeLine(l1.s, l1.e), b = MakeLine(l2.s, l2.e); //将线段延长为直线
    if(LineIntersect(a, b, p))  //如果直线相交
        return Online(l1, p) && Online(l2, p);  //判断直线交点是否在线段上，是则线段相交
    else
        return false;
}

bool cmp(Point a, Point b){
    if(fabs(a.x - b.x) < eps)   return a.y < b.y;
    else    return a.x < b.x;
}
Point p[MAXN], Intersection[MAXN];
int N, m, n;
int main(){
    int i, j, cas = 1;
    while(scanf("%d", &N), N){
        m = n = 0;
        for(i = 0; i < N; i++){
            scanf("%lf%lf", &p[i].x, &p[i].y);
        }
        for(i = 0; i < N; i++){
            for(j = 0; j < N; j++){
                LineSegment l1(p[i], p[(i + 1) % N]), l2(p[j], p[(j + 1) % N]);
                Point p;
                if(LineSegmentIntersect(l1, l2, p))
                    Intersection[n++] = p;  //记录交点
            }
        }
        sort(Intersection, Intersection + n, cmp);
        n = unique(Intersection, Intersection + n) - Intersection;
        for(i = 0; i < n; i++){
            for(j = 0; j < N; j++){
                LineSegment t(p[j], p[(j + 1) % N]);
                if(Online(t, Intersection[i]) && !(t.s == Intersection[i]))  //若有交点落在边上，则该边分裂成两条边
                    m++;
            }
        }
        printf("Case %d: There are %d pieces.\n", cas++, 2 - n + m);
    }
    return 0;
}

POJ--2284--That Nice Euler Circuit【平面图欧拉公式】