题意：一笔画，问该图形将平面分成多少个区域
分析：训练指南P260，欧拉定理：平面图定点数V，边数E，面数F，则V + F - E = 2。那么找出新增的点和边就可以了。用到了判断线段相交，求交点，判断点在线上
/************************************************
* Author :Running_Time
* Created Time :2015/10/22 星期四 09:10:09
* File Name :LA_3263.cpp
************************************************/ #include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std; #define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 300 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
struct Point {
double x, y;
Point (double x=0, double y=0) : x (x), y (y) {}
};
typedef Point Vector;
double dot(Vector A, Vector B) {
return A.x * B.x + A.y * B.y;
}
double cross(Vector A, Vector B) {
return A.x * B.y - A.y * B.x;
}
int dcmp(double x) {
if (fabs (x) < EPS) return 0;
else return x < 0 ? -1 : 1;
}
Vector operator + (Vector A, Vector B) {
return Vector (A.x + B.x, A.y + B.y);
}
Vector operator - (Vector A, Vector B) {
return Vector (A.x - B.x, A.y - B.y);
}
Vector operator * (Vector A, double p) {
return Vector (A.x * p, A.y * p);
}
Vector operator / (Vector A, double p) {
return Vector (A.x / p, A.y / p);
}
bool operator < (const Point &a, const Point &b) {
return a.x < b.x || (a.x == b.x && a.y < b.y);
}
bool operator == (const Point &a, const Point &b) {
return dcmp (a.x - b.x) == 0 && dcmp (a.y - b.y) == 0;
}
double length(Vector A) {
return sqrt (dot (A, A));
}
Point point_inter(Point p, Vector V, Point q, Vector W) {
Vector U = p - q;
double t = cross (W, U) / cross (V, W);
return p + V * t;
}
Point point_proj(Point p, Point a, Point b) {
Vector V = b - a;
return a + V * (dot (V, p - a) / dot (V, V));
}
bool inter(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) * dcmp (c2) < 0 && dcmp (c3) * dcmp (c4) < 0;
}
bool on_seg(Point p, Point a1, Point a2) {
return dcmp (cross (a1 - p, a2 - p)) == 0 && dcmp (dot (a1 - p, a2 - p)) < 0;
}
Point P[N], V[N*N]; int main(void) {
int n, cas = 0;
while (scanf ("%d", &n) == 1) {
if (!n) break;
for (int i=0; i<n; ++i) {
scanf ("%lf%lf", &P[i].x, &P[i].y);
V[i] = P[i];
}
n--;
int v = n, e = n;
for (int i=0; i<n; ++i) {
for (int j=i+1; j<n; ++j) {
if (inter (P[i], P[i+1], P[j], P[j+1])) {
V[v++] = point_inter (P[i], P[i+1] - P[i], P[j], P[j+1] - P[j]);
}
}
}
sort (V, V+v);
v = unique (V, V+v) - V;
for (int i=0; i<v; ++i) {
for (int j=0; j<n; ++j) {
if (on_seg (V[i], P[j], P[j+1])) e++;
}
}
printf ("Case %d: There are %d pieces.\n", ++cas, e + 2 - v);
} return 0;
}