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题意:平面上有一个包含n个端点的一笔画,第n个端点总是和第一个端点重合,因此团史一条闭合曲线。组成一笔画的线段可以相交,但是不会部分重叠。求这些线段将平面分成多少部分(包括封闭区域和无限大区域)。
分析:若是直接找出所有区域,或非常麻烦,而且容易出错。但用欧拉定理可以将问题进行转化,使解法变容易。
欧拉定理:设平面图的顶点数、边数和面数分别为V,E,F,则V+F-E=2。
这样,只需求出顶点数V和边数E,就可以求出F=E+2-V。
设平面图的结点由两部分组成,即原来的结点和新增的结点。由于可能出现三线共点,需要删除重复的点。
1 #include<cstdio> 2 #include<cmath> 3 #include<cstring> 4 #include<algorithm> 5 #include<iostream> 6 #include<memory.h> 7 #include<cstdlib> 8 #include<vector> 9 #define clc(a,b) memset(a,b,sizeof(a)) 10 #define LL long long int 11 using namespace std; 12 const int inf=0x3f3f3f3f; 13 const double eps = 1e-10; 14 const int N = 300 + 5; 15 16 struct Point 17 { 18 double x, y; 19 Point(double x = 0, double y = 0) : x(x), y(y) { } 20 }; 21 22 typedef Point Vector; 23 24 Vector operator + (Vector A, Vector B) 25 { 26 return Vector(A.x + B.x, A.y + B.y); 27 } 28 29 Vector operator - (Point A, Point B) 30 { 31 return Vector(A.x - B.x, A.y - B.y); 32 } 33 34 Vector operator * (Vector A, double p) 35 { 36 return Vector(A.x * p, A.y * p); 37 } 38 39 Vector operator / (Vector A, double p) 40 { 41 return Vector(A.x/p, A.y/p); 42 } 43 44 bool operator < (const Point& a, const Point& b) 45 { 46 return a.x < b.x || (a.x == b.x && a.y < b.y); 47 } 48 49 int dcmp(double x) 50 { 51 if(fabs(x) < eps) 52 return 0; 53 else 54 return x < 0 ? -1 : 1; 55 } 56 57 bool operator == (const Point& a, const Point& b) 58 { 59 return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0; 60 } 61 62 double Dot(Vector A, Vector B) 63 { 64 return A.x * B.x + A.y * B.y; 65 } 66 67 double Cross(Vector A, Vector B) 68 { 69 return A.x * B.y - A.y * B.x; 70 } 71 72 Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) 73 { 74 Vector u = P - Q; 75 double t = Cross(w, u) / Cross(v, w); 76 return P + v * t; 77 } 78 79 bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) 80 { 81 double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1), 82 c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1); 83 return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0; 84 } 85 86 bool OnSegment(Point p, Point a1, Point a2) 87 { 88 return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0; 89 } 90 91 Point P[N], V[N*N]; 92 93 int main() 94 { 95 int n, cas = 0; 96 while(~scanf("%d",&n) && n) 97 { 98 for(int i = 0; i < n; i++) 99 { 100 scanf("%lf%lf", &P[i].x, &P[i].y); 101 V[i] = P[i]; 102 } 103 n--; 104 int vcnt = n, ecnt = n; 105 for(int i = 0; i < n; i++) 106 for(int j = i + 1; j < n; j++) 107 { 108 if(SegmentProperIntersection(P[i], P[i+1], P[j], P[j+1])) 109 V[vcnt++] = GetLineIntersection(P[i], P[i+1]-P[i], P[j], P[j+1]-P[j]); 110 } 111 sort(V, V+vcnt); 112 vcnt = unique(V, V+vcnt) - V;//去掉相邻元素中重复的,使用前先排序 113 for(int i = 0; i < vcnt; i++) 114 for(int j = 0; j < n; j++) 115 if(OnSegment(V[i], P[j], P[j+1])) 116 ecnt++; 117 int ans = ecnt + 2 - vcnt; 118 printf("Case %d: There are %d pieces.\n", ++cas, ans); 119 } 120 return 0; 121 }
uvalive 3263 That Nice Euler Circuit
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原文地址:http://www.cnblogs.com/ITUPC/p/4868608.html