HDU 1077Catching Fish(简单计算几何)
Catching Fish
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1113 Accepted Submission(s): 411
Problem Description
Ignatius likes catching fish very much. He has a fishnet whose shape is a circle of radius one. Now he is about to use his fishnet to catch fish. All the fish are in the lake, and we assume all the fish will not move when Ignatius catching them. Now Ignatius wants to know how many fish he can catch by using his fishnet once. We assume that the fish can be regard as a point. So now the problem is how many points can be enclosed by a circle of radius one.
Note: If a fish is just on the border of the fishnet, it is also caught by Ignatius.
Note: If a fish is just on the border of the fishnet, it is also caught by Ignatius.
Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case starts with a positive integer N(1<=N<=300) which indicate the number of fish in the lake. Then N lines follow. Each line contains two floating-point number X and Y (0.0<=X,Y<=10.0). You may assume no two fish will at the same point, and no two fish are closer than 0.0001, no two fish in a test case are approximately at a distance of 2.0. In other words, if the distance between the fish and the centre of the fishnet is smaller 1.0001, we say the fish is also caught.
Each test case starts with a positive integer N(1<=N<=300) which indicate the number of fish in the lake. Then N lines follow. Each line contains two floating-point number X and Y (0.0<=X,Y<=10.0). You may assume no two fish will at the same point, and no two fish are closer than 0.0001, no two fish in a test case are approximately at a distance of 2.0. In other words, if the distance between the fish and the centre of the fishnet is smaller 1.0001, we say the fish is also caught.
Output
For each test case, you should output the maximum number of fish Ignatius can catch by using his fishnet once.
Sample Input
4 3 6.47634 7.69628 5.16828 4.79915 6.69533 6.20378 6 7.15296 4.08328 6.50827 2.69466 5.91219 3.86661 5.29853 4.16097 6.10838 3.46039 6.34060 2.41599 8 7.90650 4.01746 4.10998 4.18354 4.67289 4.01887 6.33885 4.28388 4.98106 3.82728 5.12379 5.16473 7.84664 4.67693 4.02776 3.87990 20 6.65128 5.47490 6.42743 6.26189 6.35864 4.61611 6.59020 4.54228 4.43967 5.70059 4.38226 5.70536 5.50755 6.18163 7.41971 6.13668 6.71936 3.04496 5.61832 4.23857 5.99424 4.29328 5.60961 4.32998 6.82242 5.79683 5.44693 3.82724 6.70906 3.65736 7.89087 5.68000 6.23300 4.59530 5.92401 4.92329 6.24168 3.81389 6.22671 3.62210
Sample Output
2 5 5 11
Author
Ignatius.L
题目大意：给你n个点的横纵坐标，问你用一个单位圆，最多能使得多少点在圆内，包括圆上的点。
解题思路：开始比较迷茫，不知道用什么方法来解，后来觉得可以枚举，但又想不清楚怎么枚举。这样，我们每次找两个点，看能否根据这两点确定一个单位圆，然后看这个圆能包含其它多少点在这个圆内！
题目地址：Catching Fish
开始是用数组写的，时间直接2s开外了！
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdio>
using namespace std;
int n;
struct node
{
double x;
double y;
};
node a[305];
double dis(node p1,node p2)
{
return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
int cal(int p1,int p2)
{
node t1,t2,t3,t4;
t1=a[p1],t2=a[p2];
double s,tmp,xx,yy;
tmp=dis(t1,t2);
s=tmp/2.0;
s=sqrt(1.0-s*s); //s为圆心到t1,t2弦长的距离
int ans1=0,ans2=0,i;
xx=(t1.y-t2.y)/tmp;
yy=(t2.x-t1.x)/tmp; //(xx,yy)相当于与弦长垂直的单位法向量
t3.x=(t1.x+t2.x)/2.0,t3.y=(t1.y+t2.y)/2.0;
t4.x=t3.x+s*xx,t4.y=t3.y+s*yy; //t4为圆心
for(i=0;i<n;i++)
{
if(dis(t4,a[i])<1.0001)
ans1++;
}
t4.x=t3.x-s*xx,t4.y=t3.y-s*yy; //t4为圆心
for(i=0;i<n;i++)
{
if(dis(t4,a[i])<1.0001)
ans2++;
}
return ans1>ans2?ans1:ans2;
}
int main()
{
int i,j;
int tes;
scanf("%d",&tes);
while(tes--)
{
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%lf%lf",&a[i].x,&a[i].y);
int num;
int res=1;
for(i=0;i<n;i++)
for(j=i+1;j<n;j++)
{
if(dis(a[i],a[j])<2.0001)
{
num=cal(i,j);
if(num>res) res=num;
}
}
printf("%d\n",res);
}
return 0;
}
//2406MS
后来改用数组写了，时间终于降到了1s内，Best solutions里面还是有很多两三百ms的，Orz！！
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdio>
using namespace std;
int n;
double a[305][2];
double dis(double *b1,double *b2)
{
return sqrt((b1[0]-b2[0])*(b1[0]-b2[0])+(b1[1]-b2[1])*(b1[1]-b2[1]));
}
int cal(int p1,int p2)
{
double t1[2],t2[2],t3[2],t4[2];
t1[0]=a[p1][0],t1[1]=a[p1][1],t2[0]=a[p2][0],t2[1]=a[p2][1];
double s,tmp,xx,yy;
tmp=dis(t1,t2);
s=tmp/2.0;
s=sqrt(1.0-s*s); //s为圆心到t1,t2弦长的距离
int ans1=0,ans2=0,i;
xx=(t1[1]-t2[1])/tmp;
yy=(t2[0]-t1[0])/tmp; //(xx,yy)相当于与弦长垂直的单位法向量
t3[0]=(t1[0]+t2[0])/2.0,t3[1]=(t1[1]+t2[1])/2.0;
t4[0]=t3[0]+s*xx,t4[1]=t3[1]+s*yy; //t4为圆心
for(i=0;i<n;i++)
{
if(dis(t4,a[i])<1.0001)
ans1++;
}
t4[0]=t3[0]-s*xx,t4[1]=t3[1]-s*yy; //t4为圆心
for(i=0;i<n;i++)
{
if(dis(t4,a[i])<1.0001)
ans2++;
}
return ans1>ans2?ans1:ans2;
}
int main()
{
int i,j;
int tes;
scanf("%d",&tes);
while(tes--)
{
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%lf%lf",&a[i][0],&a[i][1]);
int num;
int res=1;
for(i=0;i<n;i++)
for(j=i+1;j<n;j++)
{
if(dis(a[i],a[j])<2.0001)
{
num=cal(i,j);
if(num>res) res=num;
}
}
printf("%d\n",res);
}
return 0;
}
//984MS G++