【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

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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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