Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Ad hoc networks are wireless networks with no fixed infrastructure. Each device in the network functions as a router that discovers and maintains routes for other nodes. Suppose that the communication range (radius) of each device is R unit length. Device A can communicate with device B directly, when the distance between A and B is less than or equal to R. Specifically, device A can communicate with device B directly if (x₁ - x₂)² + (y₁ - y₂)² R², where (x₁, y₁) and (x₂, y₂) are, respectively, the locations of A and B.

The hidden terminal problem in ad hoc networks is caused by concurrent transmissions of two devices that cannot communicate with each other directly, but transmit to the same destination. For example, device A cannot communicate with device B directly (i.e. the distance between A and B is greater than R), but device A can communicate with device C directly (i.e. the distance between A and C is less than or equal to R) and device B can communicate with device C directly. Such three devices are referred to here a hidden-terminal set. For example, there are four devices A, B, C, D, deployed in the plane at (0, 0), (0, 1), (1, 0), (1, 1), respectively. Given that the radius of each device is R = 1 unit length, A can communicate with B (C) directly, and similarly D can communicate with B (C) directly. However, A cannot communicate with D directly and B cannot communicate with C directly. The number of different hidden-terminal sets in the plane is four (i.e., {A, B, C}, {A, B, D}, {A, C, D}, and {B, C, D}).

Hidden-terminal sets in an ad hoc network seriously results in garbled messages and increases communication delay, thus degrading system performance. Given N devices deployed in a plane, please compute the number of different hidden-terminal sets in the network.

Input 

There are at most 10 test cases. The first line of each instance consists of an integer N(3 N 100), where N is the number of devices in the network. The second line of each instance consists of an integer R(1 R 100), where R is the communication range (radius) of each device. Each of the following N lines consists of two integers x and y (separated by a space) (- 99 x 99, -99 y 99), which indicate the location (x-coordinate and y-coordinate) of a device in the plane. The last test case will be followed by a line containing N = 0.

Output 

The output for each instance should contain an integer denoting the number of different hidden-terminal sets in the network.

Sample Input  

4
1
0 0
0 1
1 0
1 1
5
2
1 1
2 2
3 3
4 4
5 5
0

Sample Output 

4
3

---

窮舉 A(X)B, C-A, C-B

#include <iostream>
#include <cstdio>
using namespace std;

int main() {
    int N, R;
    int i, j, k, x[101], y[101];
    while(scanf("%d", &N) == 1 && N) {
        scanf("%d", &R);
        for(i = 0; i < N; i++)
            scanf("%d %d", &x[i], &y[i]);
        int Ans = 0;
        for(i = 0; i < N; i++) {
            for(j = i+1; j < N; j++) {
                if((x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]) > R*R)
                    for(k = 0; k < N; k++) {
                        if(k == i || k == j)    continue;
                        if((x[i]-x[k])*(x[i]-x[k]) + (y[i]-y[k])*(y[i]-y[k]) <= R*R &&
                                (x[k]-x[j])*(x[k]-x[j]) + (y[k]-y[j])*(y[k]-y[j]) <= R*R)
                            Ans++;
                    }
            }
        }
        printf("%d\n", Ans);
    }
    return 0;
}