首页 > 代码库 > LightOj1203 - Guarding Bananas(凸包求多边形中的最小角)

LightOj1203 - Guarding Bananas(凸包求多边形中的最小角)

题目链接:http://lightoj.com/volume_showproblem.php?problem=1203

题意:给你一个点集,求凸包中最小的角;模板题,但是刚开始的时候模板带错了,错的我都想吐了;

技术分享
#include <stdio.h>#include <algorithm>#include <cstring>#include <cmath>using namespace std;#define met(a, b) memset(a, b, sizeof(a))const double eps = 1e-10;const double PI = acos(-1);const int N = 150010;struct point{    double x, y;    point(double x=0, double y=0) : x(x), y(y){}    friend point operator - (const point& p1, const point& p2)    {        return point(p1.x-p2.x, p1.y-p2.y);    }    friend double operator ^ (const point& p1, const point& p2)    {        return p1.x*p2.y - p1.y*p2.x;    }}p[N], res[N];double Dist(point p1, point p2){    double dx = p1.x - p2.x, dy = p1.y - p2.y;    return sqrt(dx*dx + dy*dy);}bool cmp1(point p1, point p2){    if(p1.y == p2.y)        return p1.x < p2.x;    return p1.y < p2.y;}bool cmp2(point p1, point p2)///极角排序;若极角相同,距离近的在前面;{    double k = (p1-p[0])^(p2-p[0]);    if( k>eps || (fabs(k)<eps && Dist(p1, p[0]) < Dist(p2, p[0]) ))        return 1;    return 0;}int Graham(int n){    res[0] = p[0];if(n == 1) return 1;    res[1] = p[1];if(n == 2) return 2;    int top = 1;    for(int i=2; i<n; i++)    {        while(top && ((res[top]-res[top-1])^(p[i]-res[top-1])) <= 0) top--;        res[++top] = p[i];    }    return top+1;}int main(){    int n, T, tCase = 1;    scanf("%d", &T);    while(T--)    {        scanf("%d", &n);        for(int i=0; i<n; i++)            scanf("%lf %lf", &p[i].x, &p[i].y);        sort(p, p+n, cmp1);///p[0]为最下方靠左的点;        sort(p+1, p+n, cmp2);///以p[0]为基点,按叉积进行排序;        int cnt = Graham(n);///求凸包的顶点个数cnt,保存在res中,下标从0开始;        if(cnt < 3)        {            printf("Case %d: 0\n", tCase++);            continue;        }        double ans = 1000;        res[cnt] = res[0], res[cnt+1] = res[1];        for(int i=1; i<=cnt; i++)        {            double a = Dist(res[i-1], res[i+1]);            double b = Dist(res[i-1], res[i]);            double c = Dist(res[i], res[i+1]);            ans = min(ans, acos((b*b+c*c-a*a)/(2*b*c)));        }        printf("Case %d: %.6f\n", tCase++, ans*180/PI);    }    return 0;}
View Code

 

LightOj1203 - Guarding Bananas(凸包求多边形中的最小角)