(模板)poj1113(graham扫描法求凸包)

题目链接:https://vjudge.net/problem/POJ-1113

题意:简化下题意即求凸包的周长+2×PI×r。

思路:用graham求凸包,模板是kuangbin的。

AC code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;

const int maxn=1005;
const double PI=acos(-1.0);

struct Point{
    int x,y;
    Point():x(0),y(0){}
    Point(int x,int y):x(x),y(y){}
}list[maxn];
int stack[maxn],top;

//计算叉积p0p1×p0p2
int cross(Point p0,Point p1,Point p2){
    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
//计算p1p2的距离
double dis(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=cross(list[0],p1,p2);
    if(tmp>0) return true;
    else if(tmp==0&&dis(list[0],p1)<dis(list[0],p2)) return true;
    else return false;
}
//输入,把最左下角放在list[0],并且进行极角排序 
void init(int n){
    Point p0;
    scanf("%d%d",&list[0].x,&list[0].y);
    p0.x=list[0].x;
    p0.y=list[0].y;
    int k=0;
    for(int i=1;i<n;++i){
        scanf("%d%d",&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);
}
//graham扫描法求凸包,凸包顶点存在stack栈中
//从栈底到栈顶一次是逆时针方向排列的
void graham(int n){
    if(n==1){
        top=0;
        stack[0]=0;
        return;
    }
    top=1;
    stack[0]=0;
    stack[1]=1;
    for(int i=2;i<n;++i){
        while(top>0&&cross(list[stack[top-1]],list[stack[top]],list[i])<=0) --top;
        stack[++top]=i;
    }
}
int main(){
    int n,r;
    scanf("%d%d",&n,&r);
    init(n);
    graham(n);
    double res=2*PI*r;
    for(int i=0;i<top;++i)
        res+=dis(list[stack[i]],list[stack[i+1]]);
    res+=dis(list[stack[0]],list[stack[top]]);
    printf("%d\n",(int)(res+0.5));
    return 0;
}

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转载自www.cnblogs.com/FrankChen831X/p/11824907.html