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此题为凸包问题模板题，题目中所给点均为整点，考虑到数据范围问题求norm()时先转换成double了，把norm()那句改成<vector>压栈即可求得凸包。
初次提交被坑得很惨，在GDB中可以完美运行A掉，到OJ上就频频RE（此处应有黑人问号）
后来发现了问题，原因是玩杂耍写了这样的代码
struct point {
int x, y;
point (){
scanf("%d%d", &x, &y);
}
...
}pt[MAXN];
于是乎，在swap
void swap(point &a, point &b){
point t = a;
a = b;
b = t;
}
的时候临时变量把数据读进去了，GG
//Ps. 我感觉这个代码高亮比上次的好看
//POJ1113
//凸包
//AC 2016.10.13
#include "cstdio"
#include "cstdlib"
#include "cmath"
#include "iostream"
#define MAXN 1010
const double pi = acos(-1.0);
double sqr(double x){
return x * x;
}
struct point {
int x, y;
point (){}
point (int X, int Y): x(X), y(Y) {}
double norm(){
return sqrt(sqr(x) + sqr(y));
}
friend bool operator < (const point &p1, const point &p2){
return (p1.x < p2.x)||(p1.x == p2.x)&&(p1.y < p2.y);
}
friend bool operator > (const point &p1, const point &p2){
return (p1.x > p2.x)||(p1.x == p2.x)&&(p1.y > p2.y);
}
friend point operator >> (const point &p1, const point &p2){
return point(p2.x - p1.x, p2.y - p1.y);
}
friend int operator ^ (const point &p1, const point &p2){
return p1.x * p2.y - p1.y * p2.x;
}
}pt[MAXN];
void swap(point &a, point &b){
point t = a;
a = b;
b = t;
}
int main(){
int n, l;
freopen("fin.c", "r", stdin);
scanf("%d%d", &n, &l);
for (int i = 0; i < n; i++){
scanf("%d%d", &pt[i].x, &pt[i].y);
for (int j = i; j && (pt[j - 1] > pt[j]); j--)
swap(pt[j - 1], pt[j]);
}
int cur = 0;
double res = 0;
while (1){
int tmp = - 1;
for (int i = 0; i < n; i++)
if (i != cur){
if (tmp == - 1)
tmp = i;
else if (((pt[cur] >> pt[i]) ^ (pt[cur] >> pt[tmp])) > 0)
tmp = i;
}
res += (pt[cur] >> pt[tmp]).norm();
cur = tmp;
if ((tmp == - 1)||(!tmp)) break;
}
printf("%d\n", (int)(res + 2 * pi * l + 0.5));
return 0;
}
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原文地址：http://www.cnblogs.com/xlnx/p/5961809.html
