POJ 1113:Wall(凸包)
http://poj.org/problem?id=1113
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 34616 | Accepted: 11821 |
Description

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.
Input
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.
Output
Sample Input
9 100
200 400
300 400
300 300
400 300
400 400
500 400
500 200
350 200
200 200
Sample Output
1628
Hint
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <iostream>
using namespace std;
#define N 1005
const double PI = acos(-1.0); struct point
{
int x,y;
}p[N]; int stak[N],top, n, L; int cross(point p0, point p1, point p2)
{
return (p1.x - p0.x) * (p2.y - p0.y) - (p1.y - p0.y) * (p2.x - p0.x);
}//计算叉积 p0p1 x p0p2 double dis(point p1, point p2)
{
return sqrt( (double)(p2.x - p1.x) * (p2.x - p1.x) + (double)(p2.y - p1.y) * (p2.y - p1.y) );
}//计算距离 p1p2的距离 bool cmp(point p1, point p2)
{
int tmp = cross(p[], p1, p2);
if(tmp > ) return true;
else if( tmp == && dis(p[], p1) < dis(p[], p2) ) return true;
return false;
}//极角排序函数,角度相同则距离小的在前面 void init()
{
point pp;
scanf("%d%d", &p[].x, &p[].y);
pp.x = p[].x, pp.y = p[].y;
int tmp = ;
for(int i = ; i < n; i++) {
scanf("%d%d", &p[i].x, &p[i].y);
if( (pp.y > p[i].y) || (pp.y == p[i].y) && (pp.x > p[i].x) ) {
tmp = i;
pp = p[i];
}
}
p[tmp] = p[];
p[] = pp;
sort(p+, p+n, cmp);
}//输入函数并进行预处理,将所有的点输入, 在最左下方的点放到p[0], 并进行极角排序 void Graham()
{
if(n <= ) {
top = ;
stak[] = ;
}
else if(n == ) {
top = ;
stak[] = ;
stak[] = ;
}
else{
stak[] = ;
stak[] = ;
top = ;
for(int i = ; i < n; i++) {
while( top > && cross(p[stak[top-]], p[stak[top]], p[i]) <= )
top--;
top++;
stak[top] = i;
}
}
}//凸包Graham模板 int main()
{
while(~scanf("%d%d",&n,&L)) {
init();
Graham(); //本题求的是凸包的周长 + 一个圆的周长
double res = ;
for(int i = ; i < top; i++) {
res += dis(p[stak[i]], p[stak[i+]]);
}
res += dis(p[stak[]], p[stak[top]]);
res += * L * PI; printf("%.0f\n", res);
}
return ;
}
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