题意:给出点集,求凸包的面积
思路:主要是求面积的考察,固定一个点顺序枚举两个点叉积求三角形面积和除2即可
/** @Date : 2017-07-19 16:07:11 * @FileName: POJ 3348 凸包面积 叉积.cpp * @Platform: Windows * @Author : Lweleth ([email protected]) * @Link : https://github.com/ * @Version : $Id$ */ #include <stdio.h> #include <iostream> #include <string.h> #include <algorithm> #include <utility> #include <vector> #include <map> #include <set> #include <string> #include <stack> #include <queue> #include <math.h> //#include <bits/stdc++.h> #define LL long long #define PII pair<int ,int> #define MP(x, y) make_pair((x),(y)) #define fi first #define se second #define PB(x) push_back((x)) #define MMG(x) memset((x), -1,sizeof(x)) #define MMF(x) memset((x),0,sizeof(x)) #define MMI(x) memset((x), INF, sizeof(x)) using namespace std; const int INF = 0x3f3f3f3f; const int N = 1e5+20; const double eps = 1e-8; struct point { double x, y; point(){} point(double _x, double _y){x = _x, y = _y;} point operator -(const point &b) const { return point(x - b.x, y - b.y); } double operator *(const point &b) const { return x * b.x + y * b.y; } double operator ^(const point &b) const { return x * b.y - y * b.x; } }; double xmult(point p1, point p2, point p0) { return (p1 - p0) ^ (p2 - p0); } double distc(point a, point b) { return sqrt((double)((b - a) * (b - a))); } int sign(double x) { if(fabs(x) < eps) return 0; if(x < 0) return -1; else return 1; } int cmpC(point a, point b)//水平序排序 { return sign(a.x - b.x) < 0 || (sign(a.x - b.x) == 0 && sign(a.y - b.y) < 0); } point stk[10100]; int Graham(point *p, int n)//水平序 { sort(p, p + n, cmpC); int top = 0; for(int i = 0; i < n; i++) { while(top >= 2 && sign(xmult(stk[top - 2], stk[top - 1], p[i])) < 0) top--; stk[top++] = p[i]; } int tmp = top; for(int i = n - 2; i >= 0; i--) { while(top > tmp && sign(xmult(stk[top - 2],stk[top - 1] ,p[i] )) < 0) top--; stk[top++] = p[i]; } if(n > 1) top--; return top; } double Area(point *p, int n) { double ans = 0; for(int i = 1; i < n - 1; i++) ans+= xmult(p[i], p[i + 1], p[0]); return ans / 2; } point p[10100]; double x, y; int n; int main() { while(cin >> n) { for(int i = 0; i < n; i++) { scanf("%lf%lf", &x, &y); p[i] = point(x, y); } int m = Graham(p, n); /*for(int i = 0; i < m; i++) printf("%lf %lf\n", stk[i].x, stk[i].y);*/ double ans = Area(stk, m); printf("%d\n", (int)(ans / 50)); } return 0; }
时间: 2024-10-05 18:03:56