Description

Being the only living descendant of his grandfather, Kamran the Believer inherited all of the grandpa's belongings. The most valuable one was a piece of convex polygon shaped farm in the grandpa's birth village. The farm was originally separated from the neighboring farms by a thick rope hooked to some spikes (big nails) placed on the boundary of the polygon. But, when Kamran went to visit his farm, he noticed that the rope and some spikes are missing. Your task is to write a program to help Kamran decide whether the boundary of his farm can be exactly determined only by the remaining spikes.

Input

The first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains an integer n (1 <= n <= 1000) which is the number of remaining spikes. Next, there are n lines, one line per spike, each containing a pair of integers which are x and y coordinates of the spike.

Output

There should be one output line per test case containing YES or NO depending on whether the boundary of the farm can be uniquely determined from the input.
 
题目大意:原本有一个凸多边形,现在我们不知道它的边,它的一部分点也有可能消失了,问用剩下的点能不能确定那个多边形。
思路:先求个凸包,然后判断是否凸包相邻的两点之间都存在一个题目给出的点使得这个点在这条边上。
因为有一个点在这个边上呢,这条边就确定下来了。
嘛其实没有这么麻烦,因为我在做模板所以……
啊,对了,全部点是一条直线的时候输出NO,我上了求多边形面积的模板←_←
 
代码(0MS):
 #include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std; const int MAXN = ;
const double EPS = 1e-;
const double PI = acos(-1.0);//3.14159265358979323846 inline int sgn(double x) {
return (x > EPS) - (x < -EPS);
} struct Point {
double x, y;
Point() {}
Point(double x, double y): x(x), y(y) {}
void read() {
scanf("%lf%lf", &x, &y);
}
bool operator == (const Point &rhs) const {
return sgn(x - rhs.x) == && sgn(y - rhs.y) == ;
}
bool operator < (const Point &rhs) const {
if(y != rhs.y) return y < rhs.y;
return x < rhs.x;
}
Point operator + (const Point &rhs) const {
return Point(x + rhs.x, y + rhs.y);
}
Point operator - (const Point &rhs) const {
return Point(x - rhs.x, y - rhs.y);
}
Point operator * (const int &b) const {
return Point(x * b, y * b);
}
Point operator / (const int &b) const {
return Point(x / b, y / b);
}
double length() const {
return sqrt(x * x + y * y);
}
Point unit() const {
return *this / length();
}
};
typedef Point Vector; double dist(const Point &a, const Point &b) {
return (a - b).length();
} double cross(const Point &a, const Point &b) {
return a.x * b.y - a.y * b.x;
}
//ret >= 0 means turn left
double cross(const Point &sp, const Point &ed, const Point &op) {
return sgn(cross(sp - op, ed - op));
} double area(const Point& a, const Point &b, const Point &c) {
return fabs(cross(a - c, b - c)) / ;
} struct Seg {
Point st, ed;
Seg() {}
Seg(Point st, Point ed): st(st), ed(ed) {}
void read() {
st.read(); ed.read();
}
};
typedef Seg Line; bool isOnSeg(const Seg &s, const Point &p) {
return (p == s.st || p == s.ed) ||
(((p.x - s.st.x) * (p.x - s.ed.x) < ||
(p.y - s.st.y) * (p.y - s.ed.y) < ) &&
sgn(cross(s.ed, p, s.st) == ));
} bool isIntersected(const Point &s1, const Point &e1, const Point &s2, const Point &e2) {
return (max(s1.x, e1.x) >= min(s2.x, e2.x)) &&
(max(s2.x, e2.x) >= min(s1.x, e1.x)) &&
(max(s1.y, e1.y) >= min(s2.y, e2.y)) &&
(max(s2.y, e2.y) >= min(s1.y, e1.y)) &&
(cross(s2, e1, s1) * cross(e1, e2, s1) >= ) &&
(cross(s1, e2, s2) * cross(e2, e1, s2) >= );
} bool isIntersected(const Seg &a, const Seg &b) {
return isIntersected(a.st, a.ed, b.st, b.ed);
} bool isParallel(const Seg &a, const Seg &b) {
return sgn(cross(a.ed - a.st, b.ed - b.st)) == ;
} //return Ax + By + C =0 's A, B, C
void Coefficient(const Line &L, double &A, double &B, double &C) {
A = L.ed.y - L.st.y;
B = L.st.x - L.ed.x;
C = L.ed.x * L.st.y - L.st.x * L.ed.y;
} Point intersection(const Line &a, const Line &b) {
double A1, B1, C1;
double A2, B2, C2;
Coefficient(a, A1, B1, C1);
Coefficient(b, A2, B2, C2);
Point I;
I.x = - (B2 * C1 - B1 * C2) / (A1 * B2 - A2 * B1);
I.y = (A2 * C1 - A1 * C2) / (A1 * B2 - A2 * B1);
return I;
} bool isEqual(const Line &a, const Line &b) {
double A1, B1, C1;
double A2, B2, C2;
Coefficient(a, A1, B1, C1);
Coefficient(b, A2, B2, C2);
return sgn(A1 * B2 - A2 * B1) == && sgn(A1 * C2 - A2 * C1) == && sgn(B1 * C2 - B2 * C1) == ;
} struct Poly {
int n;
Point p[MAXN];//p[n] = p[0]
void init(Point *pp, int nn) {
n = nn;
for(int i = ; i < n; ++i) p[i] = pp[i];
p[n] = p[];
}
double area() {
if(n < ) return ;
double s = p[].y * (p[n - ].x - p[].x);
for(int i = ; i < n; ++i)
s += p[i].y * (p[i - ].x - p[i + ].x);
return s / ;
}
}; void Graham_scan(Point *p, int n, int *stk, int &top) {
sort(p, p + n);
top = ;
stk[] = ; stk[] = ;
for(int i = ; i < n; ++i) {
while(top && cross(p[i], p[stk[top]], p[stk[top - ]]) >= ) --top;
stk[++top] = i;
}
int len = top;
stk[++top] = n - ;
for(int i = n - ; i >= ; --i) {
while(top != len && cross(p[i], p[stk[top]], p[stk[top - ]]) >= ) --top;
stk[++top] = i;
}
} /*******************************************************************************************/ Point p[MAXN];
Poly poly;
int stk[MAXN], top;
int n, T; bool check() {
poly.n = top;
for(int i = ; i < top; ++i) poly.p[i] = p[stk[i]];
poly.p[poly.n] = poly.p[];
if(sgn(poly.area()) == ) return false;
for(int i = ; i < poly.n; ++i) {
bool flag = false;
for(int j = ; j < n; ++j) {
if(p[j] == poly.p[i] || p[j] == poly.p[i + ]) continue;
if(isOnSeg(Seg(poly.p[i], poly.p[i + ]), p[j])) {
flag = true;
break;
}
}
if(!flag) return false;
}
return true;
} int main() {
scanf("%d", &T);
while(T--) {
scanf("%d", &n);
for(int i = ; i < n; ++i) p[i].read();
Graham_scan(p, n, stk, top);
if(check()) puts("YES");
else puts("NO");
}
}

POJ 1228 Grandpa's Estate(凸包唯一性判断)的更多相关文章

  1. POJ 1228 Grandpa's Estate 凸包 唯一性

    LINK 题意:给出一个点集,问能否够构成一个稳定凸包,即加入新点后仍然不变. 思路:对凸包的唯一性判断,对任意边判断是否存在三点及三点以上共线,如果有边不满足条件则NO,注意使用水平序,这样一来共线 ...

  2. POJ 1228 Grandpa's Estate(凸包)

    Grandpa's Estate Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 11289   Accepted: 3117 ...

  3. POJ 1228 - Grandpa's Estate 稳定凸包

    稳定凸包问题 要求每条边上至少有三个点,且对凸包上点数为1,2时要特判 巨坑无比,调了很长时间= = //POJ 1228 //稳定凸包问题,等价于每条边上至少有三个点,但对m = 1(点)和m = ...

  4. POJ 1228 Grandpa's Estate --深入理解凸包

    题意: 判断凸包是否稳定. 解法: 稳定凸包每条边上至少有三个点. 这题就在于求凸包的细节了,求凸包有两种算法: 1.基于水平序的Andrew算法 2.基于极角序的Graham算法 两种算法都有一个类 ...

  5. 简单几何(求凸包点数) POJ 1228 Grandpa's Estate

    题目传送门 题意:判断一些点的凸包能否唯一确定 分析:如果凸包边上没有其他点,那么边想象成橡皮筋,可以往外拖动,这不是唯一确定的.还有求凸包的点数<=2的情况一定不能确定. /********* ...

  6. poj - 1228 - Grandpa's Estate

    题意:原来一个凸多边形删去一些点后剩n个点,问这个n个点能否确定原来的凸包(1 <= 测试组数t <= 10,1 <= n <= 1000). 题目链接:http://poj. ...

  7. 【POJ】1228 Grandpa's Estate(凸包)

    http://poj.org/problem?id=1228 随便看看就能发现,凸包上的每条边必须满足,有相邻的边和它斜率相同(即共线或凸包上每个点必须一定在三点共线上) 然后愉快敲完凸包+斜率判定, ...

  8. 【POJ 1228】Grandpa's Estate 凸包

    找到凸包后暴力枚举边进行$check$,注意凸包是一条线(或者说两条线)的情况要输出$NO$ #include<cmath> #include<cstdio> #include ...

  9. 凸包稳定性判断:每条边上是否至少有三点 POJ 1228

    //凸包稳定性判断:每条边上是否至少有三点 // POJ 1228 #include <iostream> #include <cstdio> #include <cst ...

随机推荐

  1. 键盘录入6个int类型的数据存入数组arr中,将arr数组中的内容反转...

    一.有一道很有意思的数组操作相关编程题,闲来无事用JS解决了一下,问题描述如下: (1) 键盘录入6个int类型的数据存入数组arr中: (2) 将arr数组中的内容反转: (3) 将反转后的数组角标 ...

  2. Zabbix——设置报警阈值

    前提条件: 1. Zabbix-server 版本为4.0 2.邮件告警正常使用 3. 阈值改为1分钟进行邮件发送 点击: 找到agent,点击触发器: 设置网络ping包进行检测

  3. mysql中对于时间的处理,时间的滚动,求时间间隔,切换时区等等

    mysql中对于时间的处理,时间的滚动,求时间间隔,切换时区等等 在mysql的使用过程中,有时候会出现对时间进行的操作,比如时间向前向后滚动,求2个时间的间隔,或者切换时区的操作 1,时间向前滚动( ...

  4. vue实现多级弹窗

    webpack + vue 实现 弹窗功能 对于刚入门webpack + vue 不久的新人来说,这技术,确实有些不太友好,相比较于直接操纵dom元素的jQuery,直接操纵数据的 vue 在webp ...

  5. QWebView 与Js 交互

    我本愚钝,在网上搜了一下没找到可以运行的栗子,遂在这记录一下吧. 环境:win10 64位系统  qt 4.8.7 (mingw32) qtcreator(4.5.0) 1. 建立一个 Widgets ...

  6. Spark运行模式_spark自带cluster manager的standalone cluster模式(集群)

    这种运行模式和"Spark自带Cluster Manager的Standalone Client模式(集群)"还是有很大的区别的.使用如下命令执行应用程序(前提是已经启动了spar ...

  7. 学习/linux/list.h_双链表实现

    linux-3.5/include/linux/list.h 使用只含指针域的双向循环链表进行链表的操作. 下面是我选取部分list.h中代码: #ifndef _LINUX_LIST_H #defi ...

  8. 通过samba服务将centos7指定文件挂载到window下

    做嵌入式开发,windows下编辑代码,虚拟机上编译,为了方便打算在虚拟机下搭一个samba服务器,将文件夹映射到windows下,搜索网上的方法,内容大同小异,试了半天终于成功了.特此记录一下步骤, ...

  9. 利用 Python 插件 xlwings 读写 Excel

    Python 通过 xlwings 读取 Excel 数据 去年底公司让我做设备管理,多次委婉拒绝,最终还是做了.其实我比较喜欢技术.做管理后发现现场没有停机率统计,而原始数据有,每次要自己在Exce ...

  10. Java学习笔记二十四:Java中的Object类

    Java中的Object类 一:什么是Object类: Object类是所有类的父类,相当于所有类的老祖宗,如果一个类没有使用extends关键字明确标识继承另外一个类,那么这个类默认继承Object ...