Problem C. Painting Cottages
Time Limit: 2 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/gym/100342/attachments
Description
The new cottage settlement is organized near the capital of Flatland. The construction company that is building the settlement has decided to paint some cottages pink and others — light blue. However, they cannot decide which cottages must be painted which color. The director of the company claims that the painting is nice if there is at least one pink cottage, at least one light blue cottage, and it is possible to draw a straight line in such a way that pink cottages are at one side of the line, and light blue cottages are at the other side of the line (and no cottage is on the line itself). The main architect objects that there are several possible nice paintings.
Help them to find out how many different nice paintings are there
Input
The first line of the input file contains n — the number of the cottages (1 ≤ n ≤ 300). The following n lines contain the coordinates of the cottages — each line contains two integer numbers xi and yi (−104 ≤ xi , yi ≤ 104 ).
Output
Output one integer number — the number of different nice paintings of the cottages.
Sample Input
4
0 0
1 0
1 1
0 1
Sample Output
12
HINT
题意
给你n个坐标即点，求出有多少种划分方法
题解：
q神说题意转化一下，就是这个n个点，能够连多少个不同的线段，
对于覆盖的线段不算在内，ORZ q神
代码：
1 #include <cstdio> 2 #include <cmath> 3 #include <cstring> 4 #include <ctime> 5 #include <iostream> 6 #include <algorithm> 7 #include <set> 8 #include <vector> 9 #include <queue> 10 #include <typeinfo> 11 #include <map> 12 #include <stack> 13 typedef __int64 ll; 14 #define inf 1000000000000 15 using namespace std; 16 inline ll read() 17 { 18 ll x=0,f=1; 19 char ch=getchar(); 20 while(ch<'0'||ch>'9') 21 { 22 if(ch=='-')f=-1; 23 ch=getchar(); 24 } 25 while(ch>='0'&&ch<='9') 26 { 27 x=x*10+ch-'0'; 28 ch=getchar(); 29 } 30 return x*f; 31 } 32 33 //************************************************************************************** 34 35 36 struct node 37 { 38 double x,y; 39 }; 40 int gcd(int a,int b) 41 { 42 if(b==0) return a; 43 else return gcd(b,a%b); 44 } 45 map< pair<int ,int > ,int> H; 46 node a[333]; 47 int main() 48 { 49 freopen("cottages.in","r",stdin); 50 freopen("cottages.out","w",stdout); 51 int n=read(); 52 for(int i=0;i<n;i++) 53 cin>>a[i].x>>a[i].y; 54 int ans=0; 55 for(int i=0;i<n;i++) 56 { 57 H.clear(); 58 for(int j=i+1;j<n;j++) 59 { 60 int aa=a[i].x-a[j].x; 61 int bb=a[i].y-a[j].y; 62 int cc=gcd(aa,bb); 63 if(H[make_pair(aa/cc,bb/cc)]==0) 64 { 65 ans++; 66 H[make_pair(aa/cc,bb/cc)]=1; 67 } 68 } 69 } 70 cout<<ans*2<<endl; 71 }