POJ 1113 解题报告

为了以后搜索方便，注：这道题是凸包问题。用的是convex hull的扫描算法。

这道题实际上是个convex hull的问题。因为无论有些点凹进去多少，实际上是由那些convex hull上的凸点决定了最短的边界的形状。所以最终的围栏周长是convex hull的周长加上一个完整的园的周长（半径就是L）。

代码用的是我之前的代码：http://blog.csdn.net/thestoryofsnow/article/details/38516855

1113

Accepted

180K

16MS

C++

2488B

/*
ID: thestor1
LANG: C++
TASK: poj1113
*/
#include <iostream>
#include <fstream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <limits>
#include <string>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#include <cassert>

using namespace std;

class Point{
public:
int x, y;

Point()
{
Point(0, 0);
}

Point (int x, int y)
{
this->x = x;
this->y = y;
}

bool operator <(const Point &p) const {
return x < p.x || (x == p.x && y < p.y);
}

friend ostream& operator<<(ostream& os, const Point& p);
};

ostream& operator<<(ostream& os, const Point& p)
{
os << p.x << "\t" << p.y;
return os;
}

int cross(const Point &left, Point &middle, Point &right)
{
return (left.x - middle.x) * (right.y - middle.y) - (right.x - middle.x) * (left.y - middle.y);
}

int crossProduct(int x1, int y1, int x2, int y2)
{
return x1 * y2 - x2 * y1;
}

double dis(Point p1, Point p2)
{
return sqrt((double)((p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y)));
}

double convexhull(std::vector<Point> &points)
{
sort(points.begin(), points.end());

// for (int i = 0; i < points.size(); ++i)
// {
// 	cout<<points[i]<<endl;
// }

std::vector<int> convexhull(2 * points.size());
int k = 0;
for (int i = 0; i < points.size(); ++i)
{
while (k >= 2 && cross(points[convexhull[k - 2]], points[convexhull[k - 1]], points[i]) <= 0)
{
k--;
}
convexhull[k] = i;
k++;
}

for (int i = points.size() - 2, t = k + 1; i >= 0; --i)
{
while (k >= t && cross(points[convexhull[k - 2]], points[convexhull[k - 1]], points[i]) <= 0)
{
k--;
}
convexhull[k] = i;
k++;
}
convexhull.resize(k);

// cout<<"convexhull:"<<endl;
// for (int i = 0; i < convexhull.size(); ++i)
// {
// 	cout<<points[convexhull[i]]<<endl;
// }

double perimeter = 0.00;
for (int i = 0; i < convexhull.size() - 1; ++i)
{
perimeter += dis(points[convexhull[i]], points[convexhull[i + 1]]);
}

// cout << "perimeter:" << perimeter << endl;
return perimeter;
}

int main()
{
int N, L;
scanf("%d%d", &N, &L);
std::vector<Point> points(N);
for (int i = 0; i < N; ++i)
{
scanf("%d%d", &points[i].x, &points[i].y);
}
double perimeter = convexhull(points);
double circle = 2 * 3.141592653 * L;
printf("%.0f\n", perimeter + circle);
return 0;
}