Poj 1113 Hoj 1605 Wall

来源：互联网发布：java初级工程师考证编辑：程序博客网时间：2024/05/13 17:44

题目连接:http://poj.org/problem?id=1113

http://acm.hit.edu.cn/hoj/problem/view?id=1605

本题是凸包的典型应用。求凸包用建议水平序，因为极角序要处理共线的问题。

参考：http://www.cnblogs.com/Booble/archive/2011/03/10/1980089.html

有一个问题是水平序的常用模板中。左链是否有机会回溯到右链。本题不加判断也是可以的。不知其它题目如何。

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std;struct Point{    int x;    int y;    Point() {}    Point(int _x,int _y):x(_x),y(_y) {}    friend Point operator + (Point a,Point b)    {        return Point(a.x+b.x , a.y+b.y);    }    friend Point operator - (Point a,Point b)    {        return Point(a.x-b.x , a.y-b.y);    }};Point p[1002];Point stack[10002];double det(Point a,Point b){    return a.x*b.y-a.y*b.x;}double det(Point a,Point b,Point o){    return det(a-o,b-o);}double det(Point a,Point b,Point c,Point d){    return det(b-a,d-c);}bool cmp(Point a,Point b){    return a.y<b.y || (a.y == b.y && a.x<b.x);}double dis(Point a,Point b){    return sqrt( (a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y) );}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);#endif    int n,l;    while(scanf(" %d %d",&n,&l)!=EOF)    {        for(int i=0; i<n; i++)        {            scanf(" %d %d",&p[i].x,&p[i].y);        }        sort(p,p+n,cmp);        int top = -1;        stack[++top] = p[0];        stack[++top] = p[1];        for(int i=2; i<n; i++)        {            while(top && det(stack[top],p[i],stack[top-1]) < 0)            {                top--;            }            stack[++top] = p[i];        }        //int midtop = top;        for(int i=n-2; i>=0; i--)        {            //左链有可能会回溯到右链么？top>midtop是否不加也可以？            //while(top>midtop && det(stack[top],p[i],stack[top-1]) < 0)            while(det(stack[top],p[i],stack[top-1]) < 0)            {                top--;            }            stack[++top] = p[i];        }        double ans = 2 * l * acos(-1.0);        for(int i=0; i<top; i++)        {            ans += dis(stack[i],stack[i+1]);        }        ans += 0.5;        printf("%d\n",(int)ans);    }    return 0;}