Poj 1328 (radar installation) ry + greedy

[Description ]:

Given n small islands and their locations, and entering the radar radiation area, how many radar stations are required to cover all small islands?

[Train of Thought Analysis ]:

The first thing that comes to mind in this question is the use of greedy algorithms, but how can we get greedy? When I started to solve this problem, I had only one idea: to make every radar cover more points, but how to cover more is a typical mathematical problem, I did not think about it. I finally learned how to do it after reading the question on the Internet. Let's take a look at how to create a diagram:

Through this graph, we first use a mathematical knowledge, that is, to create a circle with a small island as the center and a radar radiation range as the center. The circle has an intersection with the X axis, using the intersection as the radar station laying point, the radar station can certainly cover any small island. At the same time, if the circles built on the center of each small island do not overlap, then there must be multiple radar stations, if the right point of the circle and the X axis of a small island is smaller than the right point of the circle of the Center of another small island, the two islands will certainly be covered by the same radar.

Of course, when processing data, we need to sort the data.

[AC code ]:

1 # include <iostream> 2 # include <math. h> 3 # include <algorithm> 4 using namespace STD; 5 # define Max 1005 6 struct radar 7 {8 double left; 9 double right; 10 double X; 11 Double Y; 12} R [Max]; 13 int CMP (radar A, radar B) 14 {15 return. left <B. left; // sort the left point of the circle and the X axis of a small island in ascending order. 16} 17 int main () 18 {19 int COUNT = 1, n, D; 20 while (CIN> N> D & (N | D) 21 {22 int flag = 0, sum = 1; 23 Double X, Y; 24 For (INT I = 0; I <n; I ++) 25 {2 6 CIN> x> Y; 27 if (Y> d) Flag = 1; 28 R [I]. X = x; 29 R [I]. y = y; 30} 31 32 for (INT I = 0; I <n; I ++) 33 {34 R [I]. left = R [I]. x * 1.0-sqrt (D * d-r [I]. y * R [I]. y); // calculates the Left and Right intersections between the circle and the X axis. 35 R [I]. right = R [I]. x * 1.0 + SQRT (D * d-r [I]. y * R [I]. y); 36} 37 sort (R, R + N, CMP); 38 If (FLAG) {cout <"case" <count ++ <": -1 "<Endl; continue;} 39 double temp = R [0]. right; 40 for (INT I = 1; I <n; I ++) 41 {42 if (R [I]. left> temp) 43 {44 sum ++; temp = R [I]. right; // If two circles do not intersection, Then add 1 to the number of radars. 45} 46 else if (R [I]. right <temp) 47 {48 temp = R [I]. right; // if the right point of a circle is smaller than the right point of a circle, the circle is updated. 49} 50} 51 cout <"case" <count ++ <":" <sum <Endl; 52} 53 return 0; 54}