poj 1113 Wall(Graham扫描法求凸包)

【题目大意】:给出一些点,求凸包周长,并加上给定圆周长


【解题思路】:极角排序,然后求凸包。模版testing


【代码】:

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
                   
using namespace std;
                   
#define eps 1e-8
#define PI acos(-1.0)
#define inf 1<<30
#define linf 1LL<<60
#define pb push_back
#define lc(x) (x << 1)
#define rc(x) (x << 1 | 1)
#define lowbit(x) (x & (-x))
#define ll long long
#define INF 999999999.9

struct Point{
    double x, y, dis;
}pt[1005],stack[1005],p0;
int top,tot;
int n;
double l;

//计算几何距离
double get_dis(double x1, double y1, double x2, double y2){
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}

//极角比较, 返回-1: p0p1在p0p2的右侧,返回0:p0,p1,p2共线
int Cmp_PolarAngel(struct Point p1, struct Point p2, struct Point pb){
    double delta=(p1.x-pb.x)*(p2.y-pb.y)-(p2.x-pb.x)*(p1.y-pb.y);
    if (delta<0.0) return 1;
    else if (delta==0.0) return 0;
    else return -1;
}

// 判断向量p2p3是否对p1p2构成左旋
bool Is_LeftTurn(struct Point p3, struct Point p2, struct Point p1){
    int type=Cmp_PolarAngel(p3, p1, p2);
    if (type<0) return true;
    return false;
}

//先按极角排,再按距离由小到大排
int Cmp(const void*p1, const void*p2){
    struct Point*a1=(struct Point*)p1;
    struct Point*a2=(struct Point*)p2;
    int type=Cmp_PolarAngel(*a1, *a2, p0);
    if (type<0) return -1;
    else if (type==0){
        if (a1->disdis) return -1;
        else if (a1->dis==a2->dis) return 0;
        else return 1;
    }
    else return 1;
}

//求凸包
void Solve(int n){
    int k;
    p0.x=p0.y=INF;
    for (int i=0; i=1 && Is_LeftTurn(pt[i],stack[top],stack[top-1])==false)
            top--;
        stack[++top]=pt[i];
    }
    stack[top+1]=stack[0];
    double ans=0;
    ans=2.0*PI*(double)l;
    for (int i=1; i<=top+1; i++){
        ans+=get_dis(stack[i-1].x,stack[i-1].y,stack[i].x,stack[i].y);
    }
    printf("%.0f\n",ans);
}

int main(){
    while (~scanf("%d%lf",&n,&l)){
        Solve(n);
    }
    return 0;
}


你可能感兴趣的:(凸包,计算几何,POJ)