HDU 5738 Eureka 统计共线的子集个数
Eureka
题目连接:
http://acm.hdu.edu.cn/showproblem.php?pid=5738
Description
Professor Zhang draws n points on the plane, which are conveniently labeled by 1,2,...,n. The i-th point is at (xi,yi). Professor Zhang wants to know the number of best sets. As the value could be very large, print it modulo 109+7.
A set P (P contains the label of the points) is called best set if and only if there are at least one best pair in P. Two numbers u and v (u,v∈P,u≠v) are called best pair, if for every w∈P, f(u,v)≥g(u,v,w), where f(u,v)=(xu−xv)2+(yu−yv)2−−−−−−−−−−−−−−−−−−−√ and g(u,v,w)=f(u,v)+f(v,w)+f(w,u)2.
Input
There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
The first line contains an integer n (1≤n≤1000) -- then number of points.
Each of the following n lines contains two integers xi and yi (−109≤xi,yi≤109) -- coordinates of the i-th point.
Output
For each test case, output an integer denoting the answer.
Sample Input
3
3
1 1
1 1
1 1
3
0 0
0 1
1 0
1
0 0
Sample Output
4
3
0
Hint
题意
问你有多少个子集是共线的。
题解:
其实就是枚举直线,然后统计一下就好了。
但是这道题会卡常数,所以得非常优越才行,我们队使用gcd去搞的,然后就搞过去了。。。。
cmp的极角排序要么很慢,要么就被卡精度了,非常烦的一道题……
代码
#include<bits/stdc++.h>
#define mp make_pair
using namespace std;
namespace fastIO{
#define BUF_SIZE 100000
#define OUT_SIZE 100000
#define ll long long
//fread->read
bool IOerror=0;
inline char nc(){
static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
if (p1==pend){
p1=buf; pend=buf+fread(buf,1,BUF_SIZE,stdin);
if (pend==p1){IOerror=1;return -1;}
//{printf("IO error!\n");system("pause");for (;;);exit(0);}
}
return *p1++;
}
inline bool blank(char ch){return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';}
inline void read(int &x){
bool sign=0; char ch=nc(); x=0;
for (;blank(ch);ch=nc());
if (IOerror)return;
if (ch=='-')sign=1,ch=nc();
for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
if (sign)x=-x;
}
inline void read(ll &x){
bool sign=0; char ch=nc(); x=0;
for (;blank(ch);ch=nc());
if (IOerror)return;
if (ch=='-')sign=1,ch=nc();
for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
if (sign)x=-x;
}
inline void read(double &x){
bool sign=0; char ch=nc(); x=0;
for (;blank(ch);ch=nc());
if (IOerror)return;
if (ch=='-')sign=1,ch=nc();
for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
if (ch=='.'){
double tmp=1; ch=nc();
for (;ch>='0'&&ch<='9';ch=nc())tmp/=10.0,x+=tmp*(ch-'0');
}
if (sign)x=-x;
}
inline void read(char *s){
char ch=nc();
for (;blank(ch);ch=nc());
if (IOerror)return;
for (;!blank(ch)&&!IOerror;ch=nc())*s++=ch;
*s=0;
}
inline void read(char &c){
for (c=nc();blank(c);c=nc());
if (IOerror){c=-1;return;}
}
//getchar->read
inline void read1(int &x){
char ch;int bo=0;x=0;
for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1;
for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());
if (bo)x=-x;
}
inline void read1(ll &x){
char ch;int bo=0;x=0;
for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1;
for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());
if (bo)x=-x;
}
inline void read1(double &x){
char ch;int bo=0;x=0;
for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1;
for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());
if (ch=='.'){
double tmp=1;
for (ch=getchar();ch>='0'&&ch<='9';tmp/=10.0,x+=tmp*(ch-'0'),ch=getchar());
}
if (bo)x=-x;
}
inline void read1(char *s){
char ch=getchar();
for (;blank(ch);ch=getchar());
for (;!blank(ch);ch=getchar())*s++=ch;
*s=0;
}
inline void read1(char &c){for (c=getchar();blank(c);c=getchar());}
//scanf->read
inline void read2(int &x){scanf("%d",&x);}
inline void read2(ll &x){
#ifdef _WIN32
scanf("%I64d",&x);
#else
#ifdef __linux
scanf("%lld",&x);
#else
puts("error:can't recognize the system!");
#endif
#endif
}
inline void read2(double &x){scanf("%lf",&x);}
inline void read2(char *s){scanf("%s",s);}
inline void read2(char &c){scanf(" %c",&c);}
inline void readln2(char *s){gets(s);}
//fwrite->write
struct Ostream_fwrite{
char *buf,*p1,*pend;
Ostream_fwrite(){buf=new char[BUF_SIZE];p1=buf;pend=buf+BUF_SIZE;}
void out(char ch){
if (p1==pend){
fwrite(buf,1,BUF_SIZE,stdout);p1=buf;
}
*p1++=ch;
}
void print(int x){
static char s[15],*s1;s1=s;
if (!x)*s1++='0';if (x<0)out('-'),x=-x;
while(x)*s1++=x%10+'0',x/=10;
while(s1--!=s)out(*s1);
}
void println(int x){
static char s[15],*s1;s1=s;
if (!x)*s1++='0';if (x<0)out('-'),x=-x;
while(x)*s1++=x%10+'0',x/=10;
while(s1--!=s)out(*s1); out('\n');
}
void print(ll x){
static char s[25],*s1;s1=s;
if (!x)*s1++='0';if (x<0)out('-'),x=-x;
while(x)*s1++=x%10+'0',x/=10;
while(s1--!=s)out(*s1);
}
void println(ll x){
static char s[25],*s1;s1=s;
if (!x)*s1++='0';if (x<0)out('-'),x=-x;
while(x)*s1++=x%10+'0',x/=10;
while(s1--!=s)out(*s1); out('\n');
}
void print(double x,int y){
static ll mul[]={1,10,100,1000,10000,100000,1000000,10000000,100000000,
1000000000,10000000000LL,100000000000LL,1000000000000LL,10000000000000LL,
100000000000000LL,1000000000000000LL,10000000000000000LL,100000000000000000LL};
if (x<-1e-12)out('-'),x=-x;x*=mul[y];
ll x1=(ll)floor(x); if (x-floor(x)>=0.5)++x1;
ll x2=x1/mul[y],x3=x1-x2*mul[y]; print(x2);
if (y>0){out('.'); for (size_t i=1;i<y&&x3*mul[i]<mul[y];out('0'),++i); print(x3);}
}
void println(double x,int y){print(x,y);out('\n');}
void print(char *s){while (*s)out(*s++);}
void println(char *s){while (*s)out(*s++);out('\n');}
void flush(){if (p1!=buf){fwrite(buf,1,p1-buf,stdout);p1=buf;}}
~Ostream_fwrite(){flush();}
}Ostream;
inline void print(int x){Ostream.print(x);}
inline void println(int x){Ostream.println(x);}
inline void print(char x){Ostream.out(x);}
inline void println(char x){Ostream.out(x);Ostream.out('\n');}
inline void print(ll x){Ostream.print(x);}
inline void println(ll x){Ostream.println(x);}
inline void print(double x,int y){Ostream.print(x,y);}
inline void println(double x,int y){Ostream.println(x,y);}
inline void print(char *s){Ostream.print(s);}
inline void println(char *s){Ostream.println(s);}
inline void println(){Ostream.out('\n');}
inline void flush(){Ostream.flush();}
//puts->write
char Out[OUT_SIZE],*o=Out;
inline void print1(int x){
static char buf[15];
char *p1=buf;if (!x)*p1++='0';if (x<0)*o++='-',x=-x;
while(x)*p1++=x%10+'0',x/=10;
while(p1--!=buf)*o++=*p1;
}
inline void println1(int x){print1(x);*o++='\n';}
inline void print1(ll x){
static char buf[25];
char *p1=buf;if (!x)*p1++='0';if (x<0)*o++='-',x=-x;
while(x)*p1++=x%10+'0',x/=10;
while(p1--!=buf)*o++=*p1;
}
inline void println1(ll x){print1(x);*o++='\n';}
inline void print1(char c){*o++=c;}
inline void println1(char c){*o++=c;*o++='\n';}
inline void print1(char *s){while (*s)*o++=*s++;}
inline void println1(char *s){print1(s);*o++='\n';}
inline void println1(){*o++='\n';}
inline void flush1(){if (o!=Out){if (*(o-1)=='\n')*--o=0;puts(Out);}}
struct puts_write{
~puts_write(){flush1();}
}_puts;
inline void print2(int x){printf("%d",x);}
inline void println2(int x){printf("%d\n",x);}
inline void print2(char x){printf("%c",x);}
inline void println2(char x){printf("%c\n",x);}
inline void print2(ll x){
#ifdef _WIN32
printf("%I64d",x);
#else
#ifdef __linux
printf("%lld",x);
#else
puts("error:can't recognize the system!");
#endif
#endif
}
inline void println2(ll x){print2(x);printf("\n");}
inline void println2(){printf("\n");}
#undef ll
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace fastIO;
const int maxn = 1e3+15;
const int mod = 1e9 + 7;
const double PI = acos(-1.0);
int vis[maxn];
struct point{
int x,y;
}now;
int N , two[maxn];
point p[maxn];
vector<pair<int ,int> > v;
int main(){
two[0] = 1;
for(int i = 1 ; i < maxn ; ++ i) two[i] = two[i - 1] * 2LL % mod;
int T;
read(T);
while(T--){
read(N);
for(int i = 0 ; i < N ; ++ i){
read(p[i].x),read(p[i].y);
}
int ans = 0;
for(int i=0;i<N;i++) vis[i]=0;
for(int i = 0 ; i < N ; ++ i)
{
if(vis[i]) continue;
int tot=0;
now=p[i];
v.clear();
for(int j=0;j<N;j++)
{
if(now.x==p[j].x&&now.y==p[j].y)
{
tot++;
vis[j]=1;
continue;
}
int x=p[j].x-now.x,y=p[j].y-now.y;
int gc=__gcd(x,y);
if(gc<0) gc=-gc;
x/=gc,y/=gc;
v.push_back(make_pair(x,y));
}
int t1=two[tot]-tot-1,t2;
if(t1<0) t1+=mod;
ans+=t1;
if(ans>=mod) ans-=mod;
sort(v.begin(),v.end());
// cout<<v.size()<<endl;
// for(int i=0;i<v.size();i++) cout<<v[i].first<<" "<<v[i].second<<endl;
int start=0,ed=1;
t1+=tot;
if(t1>=mod) t1-=mod;
while(start<v.size())
{
if(v[start].first>0||(v[start].first==0&&v[start].second>0)) break;
for(ed=start+1;ed<v.size();ed++)
if(v[start]!=v[ed]) break;
t2=two[ed-start]-1;
if(t2<0) t2+=mod;
ans+=(1LL*t1*t2%mod);
if(ans>=mod) ans%=mod;
start=ed;
}
}
println1(ans);
}
return 0;
}
HDU 5738 Eureka 统计共线的子集个数的更多相关文章
- HDU 5738 Eureka
传送门 题目大意: 给出平面上的$n$个点,每个点有唯一的标号($\text{label}$),这$n$个标号的集合记作$S$,点可能重合.求满足下列条件的$S$的子集$T$的数目: 1. $|T|\ ...
- HDU 5738 Eureka(极角排序)
[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=5738 [题目大意] 给出平面中一些点,在同一直线的点可以划分为一个集合,问可以组成多少包含元素不少 ...
- hdu 5738 Eureka 极角排序+组合数学
Eureka Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Subm ...
- HDU 2008 数值统计
题目链接:HDU 2008 Description 统计给定的n个数中,负数.零和正数的个数. Input 输入数据有多组,每组占一行,每行的第一个数是整数n(n<100),表示需要统计的数值的 ...
- 【HDU 5363】Key Set(和为偶数的子集个数)
题 Description soda has a set $S$ with $n$ integers $\{1, 2, \dots, n\}$. A set is called key set if ...
- 给定任意一个字符串,使用 for in 语句来统计字符出现的个数
//找出字符串中的数字 var str = 'haj123sdk54hask33dkhalsd879'; /*function findNum(str){ var arr = []; var tmp ...
- HDU 2008 数字统计
号码值统计 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Subm ...
- HDU 5726 GCD 区间GCD=k的个数
GCD Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submis ...
- vi怎么统计查找字符串的个数
vi怎么统计查找字符串的个数 用vi打开一个比较大的文本,用vi查找指定字符串,现在怎么统计该字符串的个数呢?比如我查找ORA字符串,直接输入 /ORA的时候vi会高亮显示.现在怎么统计ORA的个数呢 ...
随机推荐
- WebViewJavascriptBridge测试示例
android或ios:app与html5通信解决方案 下面只是前端示例代码,后端代码请参考: git https://github.com/marcuswestin/WebViewJavascrip ...
- 20155206 2016-2017-2 《Java程序设计》第8周学习总结
20155206 2016-2017-2 <Java程序设计>第8周学习总结 教材学习内容总结 第十五章 通用API 15.1 日志 日志API简介 java.util.logging包提 ...
- Oracle PLSql配置
1.安装Oracle客户端或者服务端 2.配置环境变量 <1>.一般如果安装了Oracle客户端或者服务端的话,在环境变种的Path中有Oracle的安装路径(计算机-属性-高级系统设置- ...
- CentOS安装SVN客户端(rpm)
http://mirrors.163.com/centos/6/os/x86_64/Packages/ 1.检查是已经安装了svn: rpm -qa subversion subversion-1.7 ...
- linux系统时间不同步解决办法(同步本地时间)
改变/etc/目录下的localtime文件,既可以改变当前的时区 1.方法是到/usr/share/zoneinfo目录下找到你要相对应的时区文件,例如上海在/usr/share/zoneinfo/ ...
- ispoweroftwo 判断2的次幂【转】
转自:https://www.cnblogs.com/troublelost/p/5236391.html 首先结果是: public bool IsPowerOfTwo(int n) { if(n& ...
- 使用java如何操作elasticsearch?简单示例。
在线API:https://www.elastic.co/guide/en/elasticsearch/client/java-api/2.4/transport-client.html教程:http ...
- nginx安装报错:configure: error: the HTTP rewrite module requires the PCRE library
参考:http://blog.51cto.com/williamx/958398 需要安装pcre-devel与openssl-devel yum -y install pcre-devel open ...
- IntelliJ IDEA 中安装junit插件
1.在Intellij IDEA 中安装了 Junit,TestNG插件,首先检查一下intellij IDEA 是否在安装时默认安装了这两个插件,检查方法 打开 settings -->Plu ...
- Windows 10安装pip方法
pip是一款非常方便的python包管理工具,本文主要介绍在windows 10下安装pip方法. 1. 下载pip 地址:https://pypi.python.org/pypi/pip#downl ...