Catching Fish
Time Limit: 10000/5000 MS (java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1930 Accepted Submission(s): 784
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.
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.
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
題解： 依次把圓心枚舉出來，然后取最大值 
【盜圖】原文鏈接
代碼：
#include <iostream>#include <string>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <map>#include <sstream>#define Maxn 100005using namespace std;double pos[305][2];double centre1[2], centre2[2];int n;void GetCentre(double A[], double B[]){ double x0 = (A[0] + B[0]) / 2; double y0 = (A[1] + B[1]) / 2; double a = sqrt( 1 - ( (A[0] - B[0]) * (A[0] - B[0]) + (A[1] - B[1]) * (A[1] - B[1]) ) / 4); if (fabs(A[1] - B[1]) < 1e-6) { centre1[0] = centre2[0] = x0; centre1[1] = y0 + a; centre2[1] = y0 - a; } else { double angel = atan((B[0] - A[0]) / (A[1] - B[1])); centre1[0] = x0 + a * cos(angel); centre1[1] = y0 + a * sin(angel); centre2[0] = x0 - a * cos(angel); centre2[1] = y0 - a * sin(angel); }}int CatchFish(){ int ans1 = 0; int ans2 = 0; for (int i = 0; i < n; i++) { if (sqrt( (pos[i][0] - centre1[0]) * (pos[i][0] - centre1[0]) + (pos[i][1] - centre1[1]) * (pos[i][1] - centre1[1]) ) < 1.0001) ans1++; if (sqrt( (pos[i][0] - centre2[0]) * (pos[i][0] - centre2[0]) + (pos[i][1] - centre2[1]) * (pos[i][1] - centre2[1]) ) < 1.0001) ans2++; } return max(ans1, ans2);}int main(){ int T, i, j; cin >> T; while (T--) { int ans = 1; // 注意n為1時(shí) scanf("%d", &n); for ( i = 0; i < n; i++) { scanf("%lf %lf", &pos[i][0], &pos[i][1]); } for ( i = 0; i < n; i++) for ( j = i + 1; j < n; j++) { // 如果兩點(diǎn)距離超過2，直接略過，不然會(huì)超時(shí) if ((pos[i][0] - pos[j][0]) * (pos[i][0] - pos[j][0]) + (pos[i][1] - pos[j][1]) * (pos[i][1] - pos[j][1]) > 4.0) continue; GetCentre(pos[i], pos[j]); ans = max(ans, CatchFish()); } printf("%d/n", ans ); }}新聞熱點(diǎn)
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