先求出凸包，然后利用凸包求出面积，除以50就是答案
代码：
#include<cstdio>#include<cmath>#include<algorithm>using namespace std;const int MAXN=10005;struct Point {double x, y;Point() {}Point(double x, double y) {this->x = x;this->y = y;}void read() {scanf("%lf%lf", &x, &y);}} list[MAXN], p[MAXN];typedef Point Vector;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);}double Cross(Vector A, Vector B) {return A.x * B.y – A.y * B.x;} //叉积int stack[MAXN], top;int xmult(Point p0, Point p1, Point p2) {return (p1.x – p0.x) * (p2.y – p0.y) – (p1.y – p0.y) * (p2.x – p0.x);}double dist(Point p1,Point p2) {return sqrt((double)(p2.x-p1.x)*(p2.x-p1.x)+(p2.y-p1.y)*(p2.y-p1.y));}//极角排序函数 ， 角度相同则距离小的在前面bool cmp(Point p1,Point p2) {int tmp= xmult(list[0], p1, p2);if(tmp > 0) return true;else if(tmp == 0 && dist(list[0], p1) < dist(list[0], p2)) return true;else return false;}//输入并把最左下方的点放在list[0]并且进行极角排序void init(int n) {int i, k;Point p0;scanf("%lf%lf", &list[0].x, &list[0].y);p0.x = list[0].x;p0.y = list[0].y;k = 0;for(i = 1; i < n; i++) {scanf("%lf%lf", &list[i].x, &list[i].y);if((p0.y > list[i].y) || ((p0.y == list[i].y) && (p0.x > list[i].x))) {p0.x = list[i].x;p0.y = list[i].y;k = i;}}list[k] = list[0];list[0] = p0;sort(list + 1, list + n, cmp);}void graham(int n) {int i;if(n == 1) {top = 0; stack[0] = 0;}if(n == 2) {top = 1;stack[0] = 0;stack[1] = 1;}if(n > 2) {for(i = 0; i <= 1; i++) stack[i] = i;top = 1;for(i = 2; i < n; i++) {while(top > 0 && xmult(list[stack[top – 1]], list[stack[top]], list[i]) <= 0) top–;top++;stack[top]=i;}}for (int i = 0; i <= top; i++)p[i] = list[stack[i]];}double PolygonArea(Point *p, int n) {double area = 0;for (int i = 1; i < n – 1; i++)area += Cross(p[i] – p[0], p[i + 1] – p[0]);return fabs(area) / 2;}int n;int main() {while (~scanf("%d", &n)) {init(n);graham(n);printf("%d\n", (int)(PolygonArea(p, top +1) / 50.0));}return 0;}
，自己打败自己的远远多于比别人打败的。
