POJ 3348 Cows(凸包+多边形面积)
Description
Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with his overseas adventure, they are forced to save money on buying fence posts by using trees as fence posts wherever possible. Given the locations of some trees, you are to help farmers try to create the largest pasture that is possible. Not all the trees will need to be used.
However, because you will oversee the construction of the pasture yourself, all the farmers want to know is how many cows they can put in the pasture. It is well known that a cow needs at least 50 square metres of pasture to survive.
Input
The first line of input contains a single integer, n (1 ≤ n ≤ 10000), containing the number of trees that grow on the available land. The next n lines contain the integer coordinates of each tree given as two integers x and y separated by one space (where -1000 ≤ x, y ≤ 1000). The integer coordinates correlate exactly to distance in metres (e.g., the distance between coordinate (10; 11) and (11; 11) is one metre).
Output
You are to output a single integer value, the number of cows that can survive on the largest field you can construct using the available trees.
题目大意:给n个点,求凸包,然后求这个凸包的面积。
思路:跟题目大意一样……
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 #include <cmath> 6 using namespace std; 7 8 const int MAXN = 10010; 9 const double EPS = 1e-8; 10 const double PI = acos(-1.0);//3.14159265358979323846 11 12 inline int sgn(double x) { 13 return (x > EPS) - (x < -EPS); 14 } 15 16 struct Point { 17 double x, y; 18 Point() {} 19 Point(double x, double y): x(x), y(y) {} 20 void read() { 21 scanf("%lf%lf", &x, &y); 22 } 23 bool operator == (const Point &rhs) const { 24 return sgn(x - rhs.x) == 0 && sgn(y - rhs.y) == 0; 25 } 26 bool operator < (const Point &rhs) const { 27 if(y != rhs.y) return y < rhs.y; 28 return x < rhs.x; 29 } 30 Point operator + (const Point &rhs) const { 31 return Point(x + rhs.x, y + rhs.y); 32 } 33 Point operator - (const Point &rhs) const { 34 return Point(x - rhs.x, y - rhs.y); 35 } 36 Point operator * (const int &b) const { 37 return Point(x * b, y * b); 38 } 39 Point operator / (const int &b) const { 40 return Point(x / b, y / b); 41 } 42 double length() const { 43 return sqrt(x * x + y * y); 44 } 45 Point unit() const { 46 return *this / length(); 47 } 48 }; 49 typedef Point Vector; 50 51 double dist(const Point &a, const Point &b) { 52 return (a - b).length(); 53 } 54 55 double cross(const Point &a, const Point &b) { 56 return a.x * b.y - a.y * b.x; 57 } 58 //ret >= 0 means turn left 59 double cross(const Point &sp, const Point &ed, const Point &op) { 60 return sgn(cross(sp - op, ed - op)); 61 } 62 63 double area(const Point& a, const Point &b, const Point &c) { 64 return fabs(cross(a - c, b - c)) / 2; 65 } 66 67 struct Seg { 68 Point st, ed; 69 Seg() {} 70 Seg(Point st, Point ed): st(st), ed(ed) {} 71 void read() { 72 st.read(); ed.read(); 73 } 74 }; 75 typedef Seg Line; 76 77 bool isOnSeg(const Seg &s, const Point &p) { 78 return (p == s.st || p == s.ed) || 79 (((p.x - s.st.x) * (p.x - s.ed.x) < 0 || 80 (p.y - s.st.y) * (p.y - s.ed.y) < 0) && 81 sgn(cross(s.ed, p, s.st) == 0)); 82 } 83 84 bool isIntersected(const Point &s1, const Point &e1, const Point &s2, const Point &e2) { 85 return (max(s1.x, e1.x) >= min(s2.x, e2.x)) && 86 (max(s2.x, e2.x) >= min(s1.x, e1.x)) && 87 (max(s1.y, e1.y) >= min(s2.y, e2.y)) && 88 (max(s2.y, e2.y) >= min(s1.y, e1.y)) && 89 (cross(s2, e1, s1) * cross(e1, e2, s1) >= 0) && 90 (cross(s1, e2, s2) * cross(e2, e1, s2) >= 0); 91 } 92 93 bool isIntersected(const Seg &a, const Seg &b) { 94 return isIntersected(a.st, a.ed, b.st, b.ed); 95 } 96 97 bool isParallel(const Seg &a, const Seg &b) { 98 return sgn(cross(a.ed - a.st, b.ed - b.st)) == 0; 99 } 100 101 //return Ax + By + C =0 's A, B, C 102 void Coefficient(const Line &L, double &A, double &B, double &C) { 103 A = L.ed.y - L.st.y; 104 B = L.st.x - L.ed.x; 105 C = L.ed.x * L.st.y - L.st.x * L.ed.y; 106 } 107 108 Point intersection(const Line &a, const Line &b) { 109 double A1, B1, C1; 110 double A2, B2, C2; 111 Coefficient(a, A1, B1, C1); 112 Coefficient(b, A2, B2, C2); 113 Point I; 114 I.x = - (B2 * C1 - B1 * C2) / (A1 * B2 - A2 * B1); 115 I.y = (A2 * C1 - A1 * C2) / (A1 * B2 - A2 * B1); 116 return I; 117 } 118 119 bool isEqual(const Line &a, const Line &b) { 120 double A1, B1, C1; 121 double A2, B2, C2; 122 Coefficient(a, A1, B1, C1); 123 Coefficient(b, A2, B2, C2); 124 return sgn(A1 * B2 - A2 * B1) == 0 && sgn(A1 * C2 - A2 * C1) == 0 && sgn(B1 * C2 - B2 * C1) == 0; 125 } 126 127 struct Poly { 128 int n; 129 Point p[MAXN];//p[n] = p[0] 130 void init(Point *pp, int nn) { 131 n = nn; 132 for(int i = 0; i < n; ++i) p[i] = pp[i]; 133 p[n] = p[0]; 134 } 135 double area() { 136 if(n < 3) return 0; 137 double s = p[0].y * (p[n - 1].x - p[1].x); 138 for(int i = 1; i < n; ++i) 139 s += p[i].y * (p[i - 1].x - p[i + 1].x); 140 return s / 2; 141 } 142 }; 143 144 void Graham_scan(Point *p, int n, int *stk, int &top) {//stk[0] = stk[top] 145 sort(p, p + n); 146 top = 1; 147 stk[0] = 0; stk[1] = 1; 148 for(int i = 2; i < n; ++i) { 149 while(top && cross(p[i], p[stk[top]], p[stk[top - 1]]) >= 0) --top; 150 stk[++top] = i; 151 } 152 int len = top; 153 stk[++top] = n - 2; 154 for(int i = n - 3; i >= 0; --i) { 155 while(top != len && cross(p[i], p[stk[top]], p[stk[top - 1]]) >= 0) --top; 156 stk[++top] = i; 157 } 158 } 159 160 /*******************************************************************************************/ 161 162 Point p[MAXN]; 163 Poly poly; 164 int stk[MAXN], top; 165 int n, T; 166 167 int solve() { 168 poly.n = top; 169 for(int i = 0; i <= top; ++i) poly.p[i] = p[stk[i]]; 170 double ret = poly.area() + EPS; 171 return int(ret / 50); 172 } 173 174 int main() { 175 scanf("%d", &n); 176 for(int i = 0; i < n; ++i) p[i].read(); 177 Graham_scan(p, n, stk, top); 178 printf("%d\n", solve()); 179 }