矩形面积
Problem Description
小度熊有一个桌面，小度熊剪了很多矩形放在桌面上，小度熊想知道能把这些矩形包围起来的面积最小的矩形的面积是多少。
Input
第一行一个正整数 T，代表测试数据组数($1 \leq T \leq 20$)，接下来 T 组测试数据。
每组测试数据占若干行，第一行一个正整数 $N(1 \leq N < \leq 1000)$，代表矩形的数量。接下来 N 行，每行 8 个整数$x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4$，代表矩形的四个点坐标，坐标绝对值不会超过10000。
Output
对于每组测试数据，输出两行：
第一行输出"Case #i:"，i 代表第 i 组测试数据。
第二行包含1 个数字，代表面积最小的矩形的面积，结果保留到整数位。
Sample Input
2
2
5 10 5 8 3 10 3 8
8 8 8 6 7 8 7 6
1
0 0 2 2 2 0 0 2
Sample Output
Case #1:
17
Case #2:
4
Problem‘s Link: http://acm.hdu.edu.cn/showproblem.php?pid=5251
Mean:
略
analyse:
把n个矩形的点输入多边形寻找最小覆盖面积矩形的模版代码中就OK，网上套的模板。
Time complexity: O(n)
Source code:
/*
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-05-31-23.18
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#define LL long long
#define ULL unsigned long long
using namespace std;
const double eps = 1e-8;
const int N = 4004;
int sign(double d)
{
return d < -eps ? -1 : (d > eps);
}
struct point
{
double x, y;
point operator-(point d){
point dd;
dd.x = this->x - d.x;
dd.y = this->y - d.y;
return dd;
}
point operator+(point d){
point dd;
dd.x = this->x + d.x;
dd.y = this->y + d.y;
return dd;
}
void read(){ scanf("%lf%lf", &x, &y); }
}ps[N];
int n, cn;
double dist(point d1, point d2)
{
return sqrt(pow(d1.x - d2.x, 2.0) + pow(d1.y - d2.y, 2.0));
}
double dist2(point d1, point d2)
{
return pow(d1.x - d2.x, 2.0) + pow(d1.y - d2.y, 2.0);
}
bool cmp(point d1, point d2)
{
return d1.y < d2.y || (d1.y == d2.y && d1.x < d2.x);
}
//st1-->ed1叉乘st2-->ed2的值
double xmul(point st1, point ed1, point st2, point ed2)
{
return (ed1.x - st1.x) * (ed2.y - st2.y) - (ed1.y - st1.y) * (ed2.x - st2.x);
}
double dmul(point st1, point ed1, point st2, point ed2)
{
return (ed1.x - st1.x) * (ed2.x - st2.x) + (ed1.y - st1.y) * (ed2.y - st2.y);
}
//多边形类
struct poly
{
static const int N = 50005; //点数的最大值
point ps[N+5]; //逆时针存储多边形的点,[0,pn-1]存储点
int pn; //点数
poly() { pn = 0; }
//加进一个点
void push(point tp)
{
ps[pn++] = tp;
}
//第k个位置
int trim(int k)
{
return (k+pn)%pn;
}
void clear(){ pn = 0; }
};
//返回含有n个点的点集ps的凸包
poly graham(point* ps, int n)
{
std::sort(ps, ps + n, cmp);
poly ans;
if(n <= 2){
for(int i = 0; i < n; i++)
{
ans.push(ps[i]);
}
return ans;
}
ans.push(ps[0]);
ans.push(ps[1]);
point* tps = ans.ps;
int top = -1;
tps[++top] = ps[0];
tps[++top] = ps[1];
for(int i = 2; i < n; i++)
{
while(top > 0 && xmul(tps[top - 1], tps[top], tps[top - 1], ps[i]) <= 0) top--;
tps[++top] = ps[i];
}
int tmp = top; //注意要赋值给tmp！
for(int i = n - 2; i >= 0; i--)
{
while(top > tmp && xmul(tps[top - 1], tps[top], tps[top - 1], ps[i]) <= 0) top--;
tps[++top] = ps[i];
}
ans.pn = top;
return ans;
}
//求点p到st->ed的垂足，列参数方程
point getRoot(point p, point st, point ed)
{
point ans;
double u=((ed.x-st.x)*(ed.x-st.x)+(ed.y-st.y)*(ed.y-st.y));
u = ((ed.x-st.x)*(ed.x-p.x)+(ed.y-st.y)*(ed.y-p.y))/u;
ans.x = u*st.x+(1-u)*ed.x;
ans.y = u*st.y+(1-u)*ed.y;
return ans;
}
//next为直线(st,ed)上的点，返回next沿(st,ed)右手垂直方向延伸l之后的点
point change(point st, point ed, point next, double l)
{
point dd;
dd.x = -(ed - st).y;
dd.y = (ed - st).x;
double len = sqrt(dd.x * dd.x + dd.y * dd.y);
dd.x /= len, dd.y /= len;
dd.x *= l, dd.y *= l;
dd = dd + next;
return dd;
}
//求含n个点的点集ps的最小面积矩形，并把结果放在ds(ds为一个长度是4的数组即可,ds中的点是逆时针的)中，并返回这个最小面积。
double getMinAreaRect(point* ps, int n, point* ds)
{
int cn, i;
double ans;
point* con;
poly tpoly = graham(ps, n);
con = tpoly.ps;
cn = tpoly.pn;
if(cn <= 2)
{
ds[0] = con[0]; ds[1] = con[1];
ds[2] = con[1]; ds[3] = con[0];
ans=0;
}
else
{
int l, r, u;
double tmp, len;
con[cn] = con[0];
ans = 1e40;
l = i = 0;
while(dmul(con[i], con[i+1], con[i], con[l])
>= dmul(con[i], con[i+1], con[i], con[(l-1+cn)%cn]))
{
l = (l-1+cn)%cn;
}
for(r=u=i = 0; i < cn; i++){
while(xmul(con[i], con[i+1], con[i], con[u])
<= xmul(con[i], con[i+1], con[i], con[(u+1)%cn]))
{
u = (u+1)%cn;
}
while(dmul(con[i], con[i+1], con[i], con[r])
<= dmul(con[i], con[i+1], con[i], con[(r+1)%cn]))
{
r = (r+1)%cn;
}
while(dmul(con[i], con[i+1], con[i], con[l])
>= dmul(con[i], con[i+1], con[i], con[(l+1)%cn]))
{
l = (l+1)%cn;
}
tmp = dmul(con[i], con[i+1], con[i], con[r]) - dmul(con[i], con[i+1], con[i], con[l]);
tmp *= xmul(con[i], con[i+1], con[i], con[u]);
tmp /= dist2(con[i], con[i+1]);
len = xmul(con[i], con[i+1], con[i], con[u])/dist(con[i], con[i+1]);
if(sign(tmp - ans) < 0)
{
ans = tmp;
ds[0] = getRoot(con[l], con[i], con[i+1]);
ds[1] = getRoot(con[r], con[i+1], con[i]);
ds[2] = change(con[i], con[i+1], ds[1], len);
ds[3] = change(con[i], con[i+1], ds[0], len);
}
}
}
return ans+eps;
}
int main()
{
int t;
scanf("%d",&t);
for(int Cas = 1; Cas <=t; ++Cas)
{
printf("Case #%d:\n",Cas);
int num = 0;
scanf("%d",&num);
for(int i = 0; i < num ; ++i)
{
ps[i*4+0].read();
ps[i*4+1].read();
ps[i*4+2].read();
ps[i*4+3].read();
}
struct point ds[4];
double ans = getMinAreaRect(ps,num*4,ds);
printf("%.0lf\n",ans);
}
return 0;
}
时间： 2024-10-13 11:09:16