【凸包构造】poj 1113 Wall

poj 1113 Wall

http://poj.org/problem?id=1113

问题描述：凸包长度

问题答案是凸包的长度+以l为半径的圆周长（围墙是一个圆角多边形，圆周的那部分之和为一个圆）

http://blog.csdn.net/zhengnanlee/article/details/9633357

思路

凸包构造+遍历求周长+圆周长

参考代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<vector>
#include<algorithm>
#include<set>
#include<sstream>
#define eps 1e-9
#define pi acos(-1)
using namespace std;

typedef long long ll;

const int _max = 1e3 + 10;

int n,l;

int dcmp(double x){
if(fabs(x)<eps) return 0;else return x < 0?-1:1;
}

struct point{
double x,y;
}p[_max],res[_max];

bool mult(point sp,point ep,point op){
return (sp.x - op.x) * (ep.y - op.y)
>= (ep.x - op.x) * (sp.y - op.y);
}

bool operator < (const point &l, const point &r){
return l.y < r.y ||(l.y == r.y && l.x < r.x);
}

double len(point a,point b){
return hypot(b.x-a.x,b.y-a.y);
}

int graham(point pnt[],int n, point res[]){//构造凸包
int i,len ,k = 0,top = 1;
sort(pnt,pnt+n);
if(n == 0)return 0; res[0] = pnt[0];
if(n == 1)return 1; res[1] = pnt[1];
if(n == 2)return 2; res[2] = pnt[2];
for(int i =2; i < n; ++ i){
while(top && mult(pnt[i],res[top],res[top-1]))
top--;
res[++top] = pnt[i];
}
len = top; res[++top] = pnt[n - 2];
for(i = n - 3; i >= 0; -- i){
while(top!=len && mult(pnt[i],res[top],res[top-1]))
top--;
res[++top] = pnt[i];
}
return top;//返回凸包中点的个数
}

int main(){
#ifndef ONLINE_JUDGE
freopen("input.txt","r",stdin);
#endif // ONLINE_JUDGE
while(scanf("%d%d",&n,&l) == 2){
for(int i = 0; i < n; ++ i)
scanf("%lf%lf",&p[i].x,&p[i].y);
n = graham(p,n,res);
double tar = 2 * pi * l; //圆的周长+凸包的周长
for(int i = 0; i < n; ++ i)
tar+=len(res[i],res[(i+1)%n]);
printf("%.0f\n",tar);//四舍五入
}
return 0;
}

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