链接：http://acm.hdu.edu.cn/showproblem.php?pid=1348
Wall
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3229 Accepted Submission(s): 919
Your task is to help poor Architect to save his head, by writing a program that will find the minimum possible length of the wall that he could build around the castle to satisfy King's requirements.
The task is somewhat simplified by the fact, that the King's castle has a polygonal shape and is situated on a flat ground. The Architect has already established a Cartesian coordinate system and has precisely measured the coordinates of all castle's vertices in feet.
Next N lines describe coordinates of castle's vertices in a clockwise order. Each line contains two integer numbers Xi and Yi separated by a space (-10000 <= Xi, Yi <= 10000) that represent the coordinates of ith vertex. All vertices are different and the sides of the castle do not intersect anywhere except for vertices.
This problem contains multiple test cases!
The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.
The output format consists of N output blocks. There is a blank line between output blocks.
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <algorithm> using namespace std; const int MAX=;
const double eps=1e-; typedef struct point
{
double x,y;
}point; point c[MAX];
int top; bool dy(double x,double y)
{
return x>y+eps;
}
bool xy(double x,double y)
{
return x<y-eps;
}
bool xyd(double x,double y)
{
return x<y+eps;
}
bool dyd(double x,double y)
{
return x>y-eps;
}
bool dd(double x,double y)
{
return fabs(x-y)<eps;
} double crossProduct(point a,point b,point c)
{
return (c.x-a.x)*(b.y-a.y)-(c.y-a.y)*(b.x-a.x);
}
double dist(point a,point b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
} bool cmp(point a,point b)
{
if(dd(a.y,b.y))
{
return xy(a.x,b.x);
}
return xy(a.y,b.y);
}
bool cmp1(point a,point b)
{
double len=crossProduct(c[],a,b);
if(dd(len,0.0))
{
return xy(dist(c[],a),dist(c[],b));
}
return xy(len,0.0);
} point stk[MAX]; double dis()
{
double sum=0.0;
for(int i=;i<top;i++)
{
sum+=dist(stk[i],stk[i+]);
}
sum+=dist(stk[top],stk[]);
return sum;
} double Graham(int n)
{
sort(c,c+n,cmp);
sort(c+,c+n,cmp1);
top=;
stk[top++]=c[];
stk[top++]=c[];
stk[top++]=c[];
top--;
for(int i=;i<n;i++)
{
while()
{
point a,b;
a=stk[top];
b=stk[top-];
if(xyd(crossProduct(a,b,c[i]),0.0))
{
top--;
}
else
break;
}
stk[++top]=c[i];
}
return dis();
} int main()
{
int i,j,k,t;
int n,m;
int cas=;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
for(i=;i<n;i++)
{
scanf("%lf%lf",&c[i].x,&c[i].y);
}
if(cas++)
{
printf("\n");
}
double re=*acos(0.0)*m;
double sum=Graham(n);
printf("%.0lf\n",sum+re);
}
}