Codeforces 50C Happy Farm 5 凸包

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== 难得的y出了一道计算几何。。

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <math.h>#include <iostream>using namespace std;#define INF 999999999.9#define PI acos(-1.0)#define ll long longstruct Point{ll x, y, dis;}pt[400005], stack[400005], p0;ll top, tot;//计算几何距离ll Dis(ll x1, ll y1, ll x2, ll y2){return (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);}//极角比较， 返回-1: p0p1在 p0p2的右侧，返回 0:p0,p1,p2共线int Cmp_PolarAngel(struct Point p1, struct Point p2, struct Point pb){double delta=(p1.x-pb.x)*(p2.y-pb.y)-(p2.x-pb.x)*(p1.y-pb.y);if (delta<0.0) return 1;else if (delta==0.0) return 0;else return -1;}// 判断向量p2p3是否对 p1p2构成左旋bool Is_LeftTurn(struct Point p3, struct Point p2, struct Point p1){int type=Cmp_PolarAngel(p3, p1, p2);if (type<0) return true;return false;}//先按极角排，再按距离由小到大排int Cmp(const void*p1, const void*p2){struct Point*a1=(struct Point*)p1;struct Point*a2=(struct Point*)p2;int type=Cmp_PolarAngel(*a1, *a2, p0);if (type<0) return -1;else if (type==0){if (a1->dis<a2->dis) return -1;else if (a1->dis==a2->dis) return 0;else return 1;}else return 1;}//求凸包ll step[4][2] = {0,1,0,-1,1,0,-1,0};void Solve(ll n){ll i, k;p0.x=p0.y=INF;ll x,y;for (i=0;i<n;i++){scanf("%I64d %I64d",&x, &y);for(ll j = 0; j < 4; j++){Point P = {x+step[j][0], y+step[j][1]};pt[i*4+j] = P;if (pt[i*4+j].y < p0.y){p0.y=pt[i*4+j].y;p0.x=pt[i*4+j].x;k=i*4+j;}else if (pt[i*4+j].y==p0.y){if (pt[i*4+j].x<p0.x){p0.x=pt[i*4+j].x;k=i*4+j;}}}}n *= 4;pt[k]=pt[0];pt[0]=p0;for (i=1;i<n;i++)pt[i].dis=Dis(pt[i].x,pt[i].y, p0.x,p0.y);qsort(pt+1, n-1, sizeof(struct Point), Cmp);//去掉极角相同的点tot=1;for (i=2;i<n;i++)if (Cmp_PolarAngel(pt[i], pt[i-1], p0))pt[tot++]=pt[i-1];pt[tot++]=pt[n-1];top=1;stack[0]=pt[0];stack[1]=pt[1];for (i=2;i<tot;i++){while (top>=1 && Is_LeftTurn(pt[i], stack[top], stack[top-1])==false)top--;stack[++top]=pt[i];}}inline ll Abs(ll x){return x>0?x:-x;}ll len(Point a, Point b){return Abs(a.x-b.x) + Abs(a.y-b.y)-min(Abs(a.x-b.x), Abs(a.y-b.y));}int main (){ll n, x, y;while (cin>>n){Solve(n);ll ans = 0;for(ll i = 0; i < top; i++)ans += len(stack[i], stack[i+1]);ans += len(stack[top], stack[0]);cout<<ans<<endl;}return 0;}