UVA 11387 (14.2.28) - c++编程基础

II U C ONLINE C ON TEST 2 008
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, print Impossible 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

Problem setter: Manzurur Rahman Khan Original idea: Mohammad Mahmudur Rahman

时隔好几个月再写题解了。。。

题意：就是给出N个点，然后在这N个点间连线，使得每个点的度数为三，并输出每条线连接哪两个点。

思路：由图论知识可得，一条线意味着总度数加2，于是可得无论多少条线，总度数肯定为偶数，若给出奇数个点，3乘以一个奇数，恒为奇数，所以排除奇数个点的情况

然后偶数个点，首先收尾相连，然后每个点都是和对面的那个点相连即可。

AC代码：

#include
  
   

int map[105][105];

int main() {
    int n;
    while(scanf("%d", &n) != EOF) {
        if(n == 0)
            break;
        if(n % 2 == 1 || n == 2) {
            printf("Impossible\n");
        }
        else if(n == 0)
            break;
        else {
            int v = n + n / 2;
            printf("%d\n", v);
            for(int i = 1; i < n; i++)
                printf("%d %d\n", i, i+1);
            printf("%d 1\n", n);
            for(int i = 1; i <= n/2; i++)
                printf("%d %d\n", i, i+n/2);
        }
    }
    return 0;
}