Degree Sequence of Graph G
The cargo vessel, Wang Haiyang worked on, sails among 6 ports between which exist 9 routes. At the first sight of his navigation chart, the 6 ports and 9 routes on it reminded him of Graph Theory that he studied in class at university. In the way that Leonhard Euler solved The Seven Bridges of Knoigsberg, Wang Haiyang regarded the navigation chart as a graph of Graph Theory. He considered the 6 ports as 6 nodes and 9 routes as 9 edges of the graph. The graph is illustrated as below.
According to Graph Theory, the number of edges related to a node is defined as Degree number of this node.
Wang Haiyang looked at the graph and thought, If arranged, the Degree numbers of all nodes of graph G can form such a sequence: 4, 4, 3,3,2,2, which is called the degree sequence of the graph. Of course, the degree sequence of any simple graph (according to Graph Theory, a graph without any parallel edge or ring is a simple graph) is a non-negative integer sequence?
Wang Haiyang is a thoughtful person and tends to think deeply over any scientific problem that grabs his interest. So as usual, he also gave this problem further thought, As we know, any a simple graph always corresponds with a non-negative integer sequence. But whether a non-negative integer sequence always corresponds with the degree sequence of a simple graph? That is, if given a non-negative integer sequence, are we sure that we can draw a simple graph according to it.?
Let’s put forward such a definition: provided that a non-negative integer sequence is the degree sequence of a graph without any parallel edge or ring, that is, a simple graph, the sequence is draw-possible, otherwise, non-draw-possible. Now the problem faced with Wang Haiyang is how to test whether a non-negative integer sequence is draw-possible or not. Since Wang Haiyang hasn’t studied Algorithm Design course, it is difficult for him to solve such a problem. Can you help him?
2 6 4 4 3 3 2 2 4 2 1 1 1
2 定理1：总的度mod 2=0。
using namespace std;
bool cmp(int x , int y)
< n ; i++)
< n ; i++)
n , cmp);
= 0 ; j < a ; j++)