我的想法是：先把點點按照順or逆時針排序(直接看code比較快)，然後取一個定點，算出那個定點和另外兩個連續的點構成的三角形面積。
http://codepad.org/WBmoELb5
#include <iostream> #include <stdio.h> #include <algorithm> #include <cmath> #include <iomanip> using namespace std; const int MAX_N = 12; struct Point { double x; double y; } point[MAX_N]; double cross(Point o,Point a,Point b) { return (a.x - o.x) * (b.y - o.y) - (a.y - o.y) * (b.x - o.x); } double dis(Point a,Point b) { return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y)); } double get_area(Point a,Point b,Point c) { double p = dis(a,b); double q = dis(b,c); double r = dis(a,c); double s = (p+q+r)/2; return sqrt(s*(s-p)*(s-q)*(s-r)); } bool cmp(const Point &p1,const Point &p2) { return cross(point[0],p1,p2) < 0.0; } int main () { int n; while (scanf("%d",&n) != EOF) { for (int X=0;n>X;X++) { double i,j; cin >> i >> j; point[X].x=i; point[X].y=j; } sort(point+1,point+n,cmp); double ans = 0.0; for (int x=1;n-1>x;x++) { ans+=get_area(point[0],point[x],point[x+1]); } cout<<fixed<<setprecision(2)<<ans<<endl; } }
