poj3348 Cows【凸包面积】

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题目链接：http://poj.org/problem?id=3348
题意：给你n个点，让你围城一个多边形，也就是求凸包，然后50平方米才能有一头牛，问你这些点构成的多边形能养几头牛
解析：求凸包面积/50，向下取整

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>using namespace std;const int maxn = 1e5+100;const double eps = 1e-10;const double pi = acos(-1.0);struct point{    double x,y;    point() {}    point(double _x,double _y)    {        x = _x;        y = _y;    }    bool operator < (const point &b) const    {        if(y==b.y)            return x<b.x;        return y<b.y;    }}a[maxn],ans[maxn];double x_mul(point p0,point p1,point p2){    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}double dis(point p1,point p2){    return sqrt((p2.x-p1.x)*(p2.x-p1.x)+(p2.y-p1.y)*(p2.y-p1.y));}bool cmp(point t1,point t2){    double tmp = x_mul(a[0],t1,t2);    if(tmp==0)        return dis(a[0],t1)<dis(a[0],t2);    else        return tmp>0;}double graham(int n){    sort(a,a+n);    sort(a+1,a+n,cmp);    ans[0] = a[0];    ans[1] = a[1];    ans[2] = a[2];    int top = 2;    for(int i=3;i<n;i++)    {        while(top>=2&&x_mul(ans[top-1],ans[top],a[i])<0)            top--;        ans[++top] = a[i];    }    double res = 0;    for(int i=0;i<top;i++)        res += x_mul(point(0.0,0.0),ans[i],ans[i+1]);    res += x_mul(point(0.0,0.0),ans[top],ans[0]);    return res/2.0;}int main(void){    int n;    scanf("%d",&n);    for(int i=0;i<n;i++)        scanf("%lf %lf",&a[i].x,&a[i].y);    double sum = graham(n)/50.0;    printf("%.0f\n",floor(sum));    return 0;}

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