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题目链接:
Time Limit: 8000/4000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar());
for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + ‘0‘);
putchar(‘\n‘);
}
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e5+10;
const int maxn=500+10;
const double eps=1e-9;
int n,vis[1010];
LL fx,fy,f[1010];
struct node
{
double ang;
LL x,y;
}po[1010],temp[1010];
int cmp1(node a,node b)
{
return a.ang<b.ang;
}
int cmp(node a,node b)
{
if(a.y==b.y)return a.x<b.x;
return a.y<b.y;
}
int main()
{
int t;
read(t);
f[0]=1;
For(i,1,1008)
{
f[i]=f[i-1]*2%mod;
}
while(t--)
{
read(n);
For(i,1,n)
{
read(po[i].x);read(po[i].y);
}
sort(po+1,po+n+1,cmp);
LL ans=0;
For(i,1,n-1)
{
int cnt=0,s=0;
For(j,i+1,n)
{
if(po[j].x==po[i].x&&po[j].y==po[i].y){s++;continue;}
temp[++cnt].ang=atan2(po[j].y-po[i].y,po[j].x-po[i].x);
temp[cnt].x=po[j].x;
temp[cnt].y=po[j].y;
}
sort(temp+1,temp+cnt+1,cmp1);
fx=po[i].x,fy=po[i].y;
int d=0;
for(int j=1;j<=cnt;)
{
int k,num=s+1;
for(k=j+1;k<=cnt;k++)
{
if((temp[k].y-fy)*(temp[j].x-fx)!=(temp[j].y-fy)*(temp[k].x-fx))break;
num++;
}
j=k;
ans=(ans+f[num]-1+mod)%mod;
d++;
}
ans=(ans-(LL)(d-1)*(f[s]-1+mod)%mod+mod)%mod;
}
cout<<ans<<endl;
}
return 0;
}
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5693133.html