您的位置:首页 > 其它

【poj2187】 Beauty Contest

2016-09-27 21:42 302 查看
http://poj.org/problem?id=2187 (题目链接)

题意

  求点集上两点间最长距离

Solution

  凸包+旋转卡壳。

  旋转卡壳是看起来很难,但是很好意会也很好实现的算法,但是要真正的搞懂搞透还是有点难度,有篇博客写得很好,也就不再赘述了。

代码

// poj2187
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<map>
#define inf 2147483640
#define LL long long
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout);
using namespace std;
inline LL getint() {
LL x=0,f=1;char ch=getchar();
while (ch>'9' || ch<'0') {if (ch=='-') f=-1;ch=getchar();}
while (ch>='0' && ch<='9') {x=x*10+ch-'0';ch=getchar();}
return x*f;
}

const int maxn=50010;
struct point {int x,y;}p[maxn];
int n,top,ans,s[maxn];

int dis(point a,point b) {
return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
int cross(point p0,point p1,point p2) {
return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
bool cmp(point a,point b) {
int t=cross(p[1],a,b);
if (t>0) return 1;
if (t<0) return 0;
return dis(p[1],a)<dis(p[1],b);
}
void Graham() {
top=0;
if (n==1) s[++top]=1;
else if (n==2) {s[++top]=1;s[++top]=2;}
else {
s[++top]=1;s[++top]=2;
for (int i=3;i<=n;i++) {
while (top>1 && cross(p[s[top-1]],p[s[top]],p[i])<=0) top--;
s[++top]=i;
}
}
}
void RC() {
int q=2;
ans=dis(p[s[1]],p[s[2]]);
for (int i=1;i<=top;i++) {
while (abs(cross(p[s[i%top+1]],p[s[i]],p[s[q%top+1]]))>abs(cross(p[s[i%top+1]],p[s[i]],p[s[(q-1)%top+1]]))) q=q%top+1;
ans=max(ans,max(dis(p[s[i%top+1]],p[s[q]]),dis(p[s[i]],p[s[q]])));
}
}
int main() {
while (scanf("%d",&n)!=EOF) {
int k=1;
for (int i=1;i<=n;i++) {
scanf("%d%d",&p[i].x,&p[i].y);
if (p[i].x==p[k].x ? p[i].y<p[k].y : p[i].x<p[k].x) k=i;
}
point p0=p[1];p[1]=p[k];p[k]=p0;
sort(p+2,p+1+n,cmp);
Graham();
RC();
printf("%d\n",ans);
}
return 0;
}


  
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: