| I I U C O N L I N E C O N T E S T 2 0 0 8 |
|
| Problem C: The 3-Regular Graph |
|
| Input: standard input Output: standard output |
|
|
|
|
| The degree of a vertex in a graph is the number of edges adjacent to the vertex. A graph is 3-regular if all of its vertices have degree 3. Given an integer n, you are to build a simple undirected 3-regular graph with n vertices. If there are multiple solutions, any one will do.
|
|
| Input |
|
| For each test case, the input will be a single integer n as described above. End of input will be denoted by a case where n = 0. This case should not be processed.
|
|
| Output |
|
| If it is possible to build a simple undirected 3-regular graph with n vertices, print a line with an integer e which is the number of edges in your graph. Each of the following e lines describes an edge of the graph. An edge description contains two integers a & b, the two endpoints of the edge. Note that the vertices are indexed from 1 to n. If it is not possible to build a simple undirected 3-regular graph with n vertices, printImpossible in a single line.
|
|
| Constraints |
|
| - 1 ≤ n ≤ 100
|
|
| Sample Input |
Output for Sample Input |
| 4 3 0 |
6 1 2 1 3 1 4 2 3 2 4 3 4 Impossible |
题意：给定n个点，要求用n个点构造出一个图，每个顶点度数为3，问构造方法。
思路：构造问题，n为奇数是不行的，因为加一条边度数+2，n*3为奇数，肯定不满足，所以只有n为偶数可以，构造方式为，首尾连成圈，这样度数为2了，然后在对角线全相连，度数为3，注意特判2的情况。
代码：
#include#include int n; void solve() { if (n < 3 || n % 2) printf("Impossible\n"); else { int i; printf("%d\n", n + n / 2); for (i = 1; i < n; i++) printf("%d %d\n", i, i + 1); printf("%d 1\n", n); for (i = 1; i <= n / 2; i++) printf("%d %d\n", i, i + n / 2); } } int main() { while (~scanf("%d", &n) && n) { solve(); } return 0; }