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题目链接:
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 570 Accepted Submission(s): 183
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┆ ┏┓ ┏┓ ┆
┆┏┛┻━━━┛┻┓ ┆
┆┃ ┃ ┆
┆┃ ━ ┃ ┆
┆┃ ┳┛ ┗┳ ┃ ┆
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┆┃ ┻ ┃ ┆
┆┗━┓ ┏━┛ ┆
┆ ┃ ┃ ┆
┆ ┃ ┗━━━┓ ┆
┆ ┃ AC代马 ┣┓┆
┆ ┃ ┏┛┆
┆ ┗┓┓┏━┳┓┏┛ ┆
┆ ┃┫┫ ┃┫┫ ┆
┆ ┗┻┛ ┗┻┛ ┆
************************************************ */
#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=5e4+10;
const int maxn=2e3+14;
const double eps=1e-12;
double temp[2*maxn];
int n;
struct node
{
double x,y;
}po[maxn];
int main()
{
while(scanf("%d",&n)!=EOF)
{
For(i,1,n)
{
scanf("%lf%lf",&po[i].x,&po[i].y);
}
LL ans1=0,ans2=0;
For(i,1,n)
{
int cnt=0;
For(j,1,n)
{
if(i==j)continue;
temp[++cnt]=atan2(po[j].y-po[i].y,po[j].x-po[i].x);
if(temp[cnt]<0)temp[cnt]+=2*PI;
}
sort(temp+1,temp+cnt+1);
For(j,1,cnt)
{
temp[j+cnt]=temp[j]+2*PI;
}
int l=1,r=1,le=1;
For(j,1,cnt)
{
while(temp[r]-temp[j]<PI&&r<=2*cnt)r++;
while(temp[l]-temp[j]<0.5*PI&&l<=2*cnt)l++;
while(temp[le]-temp[j]<=eps&&le<=2*cnt)le++;
ans1=ans1+r-l;
ans2=ans2+l-le;
}
}
cout<<(ans2-2*ans1)/3<<"\n";
}
return 0;
}
hdu-5784 How Many Triangles(计算几何+极角排序)
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5733811.html