POJ1113 Wall 凸包

Posted 2020-08-17 声声醉如兰

题目大意：建立围墙将城堡围起来，要求围墙至少距离城堡L，拐角处用圆弧取代，求围墙的长度。

题目思路：围墙长度=凸包周长+（2*PI*L），另外不知道为什么C++poj会RE，G++就没问题。

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#define INF 0x3f3f3f3f
#define MAX 100005
#define PI acos(-1)
using namespace std;

struct node
{
    int x,y;
}point[MAX];

int Stuck[MAX],n,l,top;

bool cmp(struct node A,struct node B)
{
    if(A.y < B.y)
        return true;
    if(A.y==B.y && A.x<B.x)
        return true;
    return false;
}

int Cross(int x1,int y1,int x2,int y2,int x3,int y3)
{
    return (x1-x2)*(y1-y3)-(x1-x3)*(y1-y2);
}

void Graham()
{
    memset(Stuck,0,sizeof(Stuck));
    Stuck[0]=0;
    Stuck[1]=1;
    top=1;
    for(int i=2;i<n;i++)
    {
        while(top>0 && Cross(point[Stuck[top]].x,point[Stuck[top]].y,point[Stuck[top-1]].x,point[Stuck[top-1]].y,point[i].x,point[i].y)<=0)
            top--;
         Stuck[++top]=i;
    }
    int len=top;
    Stuck[++top]=n-2;
    for(int i=n-3;i>=0;i--)
    {
        while(top!=len && Cross(point[Stuck[top]].x,point[Stuck[top]].y,point[Stuck[top-1]].x,point[Stuck[top-1]].y,point[i].x,point[i].y)<=0)
            top--;
        Stuck[++top]=i;
    }
}

double Dist(int pos1,int pos2)
{
    return sqrt((point[pos1].x-point[pos2].x)*(point[pos1].x-point[pos2].x)*1.0+(point[pos1].y-point[pos2].y)*(point[pos1].y-point[pos2].y));
}

int main()
{
    int k;
    while(scanf("%d%d",&n,&l)!=EOF)
    {
        k=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&point[i].x,&point[i].y);
            if(point[k].y > point[i].y || (point[k].x>point[i].x && point[k].y==point[i].y))
                k=i;
        }
        swap(point[0],point[k]);
        sort(point+1,point+n,cmp);
        Graham();
        double sum=0;
        sum+=Dist(Stuck[0],Stuck[top])+2*PI*l;
        for(int i=1;i<=top;i++)
            sum+=Dist(Stuck[i],Stuck[i-1]);
        int ans=(int)(sum+0.5);
        printf("%d\\n",ans);
    }
    return 0;
}

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