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hdu 2440 Watch out the Animal(凸包+费马点)

读懂了题的话就不难发现是要求费马点,不过输入点的顺序是不一定的,而且求的是凸包的费马点,不是所有点的费马点。。。
 
#include<algorithm>  
#include<iostream>  
#include<cstring>  
#include<cstdlib>  
#include<fstream>  
#include<sstream>  
#include<bitset>  
#include<vector>  
#include<string>  
#include<cstdio>  
#include<cmath>  
#include<stack>  
#include<queue>  
#include<stack>  
#include<map>  
#include<set>  
#define FF(i, a, b) for(int i=a; i<b; i++)  
#define FD(i, a, b) for(int i=a; i>=b; i--)  
#define REP(i, n) for(int i=0; i<n; i++)  
#define CLR(a, b) memset(a, b, sizeof(a))  
#define debug puts("**debug**")  
#define LL long long  
#define PB push_back  
#define eps 1e-10  
using namespace std;  
  
//hdu 2440  
const int maxn = 111;  
int n, T;  
struct Point  
{  
    double x, y;  
    Point (double x=0, double y=0):x(x), y(y) {}  
}p[maxn], pt[maxn], ft;  
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); }  
  
bool operator < (const Point& a, const Point& b)  
{  
    return a.x < b.x || (a.x == b.x && a.y < b.y);  
}  
  
int dcmp(double x)  
{  
    if(fabs(x) < eps) return 0;  
    return x < 0 ? -1 : 1;  
}  
  
bool operator == (const Point& a, const Point& b)  
{  
    return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;  
}  
  
double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }  
double Length(Vector A) { return sqrt(Dot(A, A)); }  
double Angel(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }  
double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }  
  
int ConvexHull(Point *p, int n, Point* ch)//凸包  
{  
    sort(p, p+n);  
    int m = 0;  
    REP(i, n)  
    {  
        while(m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;  
        ch[m++] = p[i];  
    }  
    int k = m;  
    FD(i, n-2, 0)  
    {  
        while(m > k && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;  
        ch[m++] = p[i];  
    }  
    if(n > 1) m--;  
    return m;  
}  
  
inline double dist(Point a, Point b)  
{  
    return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y));  
}  
  
double fermat_point(Point *pt, int n, Point& ptres)  
{  
    Point u, v;  
    double step = 0.0, curlen, explen, minlen;  
    int i, j, k, idx;  
    bool flag;  
    u.x = u.y = v.x = v.y = 0.0;  
    REP(i, n)  
    {  
        step += fabs(pt[i].x) + fabs(pt[i].y);  
        u.x += pt[i].x;  
        u.y += pt[i].y;  
    }  
    u.x /= n;  
    u.y /= n;  
    flag = 0;  
    while(step > eps)  
    {  
        for(k = 0; k < 10; step /= 2, ++k)  
            for(i = -1; i <= 1; ++i)  
                for(j = -1; j <= 1; ++j)  
                {  
                    v.x = u.x + step*i;  
                    v.y = u.y + step*j;  
                    curlen = explen = 0.0;  
                        REP(idx, n)  
                    {  
                        curlen += dist(u, pt[idx]);  
                        explen += dist(v, pt[idx]);  
                    }  
                    if(curlen > explen)  
                    {  
                        u = v;  
                        minlen = explen;  
                        flag = 1;  
                    }  
                }  
    }  
    ptres = u;  
    return flag ? minlen : curlen;  
}  
  
int main()  
{  
    scanf("%d", &T);  
    while(T--)  
    {  
        scanf("%d", &n);  
        REP(i, n) scanf("%lf%lf", &p[i].x, &p[i].y);  
        int m = ConvexHull(p, n, pt);  
        double ans = fermat_point(pt, m, ft);  
        printf("%.0f\n", ans);  
        if(T > 0) puts("");  
    }  
    return 0;  
}  

 


补充:软件开发 , C++ ,
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