hdu 1392（凸包）

传送门
题意：几乎是模板题，求凸包的周长

用Graham扫描法解决（还有一个叫jarvis步行法，目测也差不多）。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>using namespace std;const int maxn=102;struct NODE {    int x,y;}a[maxn],s[maxn];int top,n;inline int read() {    int x=0;char c=getchar();    while (c<'0'||c>'9') c=getchar();    while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();    return x;}inline double dis(NODE a,NODE b) {    return sqrt((double)(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}inline int cross(NODE a,NODE b,NODE c) {    return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);}inline int cmp1(NODE a,NODE b) {    return a.y<b.y||(a.y==b.y&&a.x<b.x);}inline int cmp2(NODE c,NODE d) {    int temp=cross(a[0],c,d);    if (temp==0) return dis(a[0],c)<dis(a[0],d);//three points collinear    return temp>0;}int main() {//  freopen("hdu 1392.in","r",stdin);    while (n=read()) {        for (int i=0;i<n;++i) a[i].x=read(),a[i].y=read();        if (n==1||n==2) {printf("%.2lf\n",n^1?dis(a[0],a[1]):0.0);continue;}//special judge        sort(a,a+n,cmp1),sort(a+1,a+n,cmp2),        top=-1,        s[++top]=a[0],s[++top]=a[1],s[++top]=a[2];        for (int i=3;i<n;++i) {            while (top&&cross(s[top-1],s[top],a[i])<=0) --top;            //strictly '<=', if three points are collinear, choose the first ans the last             s[++top]=a[i];        }        double c=0.0;        for (int i=1;i<=top;++i) c+=dis(s[i-1],s[i]);        c+=dis(s[top],s[0]);        printf("%.2lf\n",c);    }    return 0;}