uva 10167 - Birthday Cake

题解：由于解太多,随机抓 A、B, 只要有符合就行了； （首先，Ax+By=0必须表示直线，即A、B不能同时为0；另外，要注意到直线不能过输入中的2N个点；检测点在直线的哪一侧，只需要简单的线性规划的知识）

 1 #include <cstdio>
 2 #include <cstdlib>
 3 
 4 int x[100], y[100];
 5 
 6 int test(int A, int B, int N)
 7 {
 8     static int i, pos, neg, tmp;
 9     pos = 0, neg = 0;
10     for (i = 2*N-1; i >= 0; i--)
11     {
12         tmp = A*x[i] + B*y[i];
13         if (tmp > 0) neg ++;
14         else if(tmp < 0) pos ++;
15         else return 0;
16     }
17     return pos == neg;
18 }
19 
20 void find(int N)
21 {
22     int A, B;
23     while(1)
24     {
25         A = rand()%1001 - 500;
26         B = rand()%1001 - 500;
27         if(test(A, B, N))
28         {
29             printf("%d %d\n", A, B);
30             break;
31         }
32     }
33 }
34 
35 int main()
36 {
37     int N, i;
38     while(scanf("%d", &N) == 1 && N)
39     {
40         for (i = 2*N-1; i >= 0; i--)
41             scanf("%d %d", &x[i], &y[i]);
42         find(N);
43     }
44 }

 

枚举：

 1 #include<iostream>
 2 #include<vector>
 3 using namespace std;
 4 void bruteforce(int &A, int &B, vector<int> &x, vector<int> &y, int n){
 5     for (A = -500; A <= 500; A++){
 6         for (B = -500; B <= 500; B++){
 7             int d = 0, u = 0;
 8             for (int i = 0; i < 2 * n; i++){
 9                 if (A*x[i] + B*y[i] > 0) u++;
10                 if (A*x[i] + B*y[i] < 0) d++;
11             }
12             if (u == n&&d == n) return;
13         }
14     }
15 }
16 int main()
17 {
18     int n;
19     while (cin >> n&&n != 0){
20         vector<int> x(2*n, 0);
21         vector<int> y(2*n, 0);
22         for (int i = 0; i < 2*n; i++){
23             cin >> x[i] >> y[i];
24         }
25         int A, B;
26         bruteforce(A, B, x, y, n);
27         cout << A << ' ' << B << endl;
28     }
29     return 0;
30 }

 

posted @ 2016-01-19 01:36  小小泽  阅读(199)  评论(0)  编辑  收藏  举报