pku 2284 That Nice Euler Circuit
题意：
给你ｎ个点第ｎ个点保证与第０个点相交，然后求这ｎ个点组成的图形可以把整个平面分成几个面
思路：
这里的解题关键是知道关于多面体的欧拉定理
多面体：
设v为顶点数，e为棱数，f是面数，则
v-e+f=2-2p
p为欧拉示性数，例如
p=0 的多面体叫第零类多面体
p=1 的多面体叫第一类多面体
这里满足的是零类多面体，我们只要求出该图形的 点v,边e即可。 怎么求点v呢？ 两部分一部分是原来的n-1个顶点，然后是交出来的，我们只要判断线段相交求直线交点即可，然偶可能会摇头重复的交点去掉，求边的话我们只要求出一个规范相交的点肯定会增加一条边，枚举点然后判断 点是否在线段上（除了端点），然偶求解即可。
#include <iostream> #include <cstdio> #include <cmath> #include <vector> #include <cstring> #include <algorithm> #include <string> #include <set> #include <functional> #include <numeric> #include <sstream> #include <stack> #include <map> #include <queue> #define CL(arr, val) memset(arr, val, sizeof(arr)) #define lc l,m,rt<<1 #define rc m + 1,r,rt<<1|1 #define pi acos(-1.0) #define L(x) (x) << 1 #define R(x) (x) << 1 | 1 #define MID(l, r) (l + r) >> 1 #define Min(x, y) (x) < (y) ? (x) : (y) #define Max(x, y) (x) < (y) ? (y) : (x) #define E(x) (1 << (x)) #define iabs(x) (x) < 0 ? -(x) : (x) #define OUT(x) printf("%I64d\n", x) #define lowbit(x) (x)&(-x) #define Read() freopen("din.txt", "r", stdin) #define Write() freopen("d.out", "w", stdout) #define ll unsigned long long #define keyTree (chd[chd[root][1]][0]) #define M 100007 #define N 317 using namespace std; const int inf = 0x7f7f7f7f; const int mod = 1000000007; struct Point { double x,y; Point(double tx = 0,double ty = 0) : x(tx),y(ty){} }; typedef Point Vtor; Vtor operator + (Vtor A,Vtor B) { return Vtor(A.x + B.x,A.y + B.y); } Vtor operator - (Point A,Point B) { return Vtor(A.x - B.x,A.y - B.y); } Vtor operator * (Vtor A,double p) { return Vtor(A.x*p,A.y*p); } Vtor operator / (Vtor A,double p) { return Vtor(A.x/p,A.y/p); } const double eps = 1e-8; int dcmp(double x) { if (fabs(x) < eps) return 0; else return x > 0? 1 : -1; } bool operator < (Point A,Point B) { return A.x < B.x || (A.x == B.x && A.y < B.y); } bool operator == (Point A,Point B) { return dcmp(A.x - B.x) == 0 && dcmp(A.y - B.y) == 0 ; } double Dot(Vtor A,Vtor B) { return A.x*B.x + A.y*B.y; } double Length(Vtor A) { return sqrt(Dot(A,A)); } double Angle(Vtor A,Vtor B) { return acos(Dot(A,B)/Length(A)/Length(B)); } double Cross(Vtor A,Vtor B) { return A.x*B.y - A.y*B.x; } double Area2(Point A,Point B,Point C) { return Cross(B - A,C - A); } Vtor Rotate(Vtor A,double rad) { return Vtor(A.x*cos(rad) - A.y*sin(rad),A.x*sin(rad) + A.y*cos(rad)); } /*>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>*/ bool SegmentPI(Point a1,Point a2,Point b1,Point b2) { double c1 = Cross(a2 - a1,b1 - a1), c2 = Cross(a2 - a1, b2 - a1), c3 = Cross(b2 - b1,a1 - b1), c4 = Cross(b2 - b1, a2 - b1); return dcmp(c1*c2) < 0 && dcmp(c3*c4) < 0; } bool OnSegment(Point p,Point a1,Point a2) { return dcmp(Cross(a2 - p, a1 - p)) == 0 && dcmp(Dot(a2 - p,a1 - p)) < 0; } Point GetLineIn(Point P,Vtor v,Point Q,Vtor w) { Vtor u = P - Q; double t = Cross(w,u)/Cross(v,w); return P + v*t; } Point P[N],v[N*N]; int n; int main() { Read(); int Ei,Vi; int cas = 1; while (~scanf("%d",&n)) { if (!n) break; Vi = 0; for (int i = 0; i < n; ++i) scanf("%lf%lf",&P[i].x,&P[i].y),v[Vi++] = P[i]; Ei = --n; for (int i = 0; i < n; ++i) { for (int j = i + 1; j < n; ++j) { if (SegmentPI(P[i],P[i + 1],P[j],P[j + 1])) { v[Vi++] = GetLineIn(P[i],P[i + 1] - P[i],P[j],P[j + 1] - P[j]); } } } sort(v, v + Vi); Vi = unique(v,v + Vi) - v; for (int i = 0; i < Vi; ++i) { for (int j = 0; j < n; ++j) { if (OnSegment(v[i],P[j],P[j + 1])) Ei++; } } // printf("%d %d\n",Vi,Ei); printf("Case %d: There are %d pieces.\n",cas++,Ei - Vi + 2); } return 0; }