10004--Bicoloring

交叉染色法。。。。

In 1976 the ``Four Color Map Theorem" was proven with the assistance of a computer. This theorem states that every map can be colored using only four colors, in such a way that no region is colored using the same color as a neighbor region.

Here you are asked to solve a simpler similar problem. You have to decide whether a given arbitrary connected graph can be bicolored. That is, if one can assign colors (from a palette of two) to the nodes in such a way that no two adjacent nodes have the same color. To simplify the problem you can assume:

• no node will have an edge to itself.
• the graph is nondirected. That is, if a node a is said to be connected to a node b, then you must assume that b is connected to a.
• the graph will be strongly connected. That is, there will be at least one path from any node to any other node.

Input

  The input consists of several test cases. Each test case starts with a line containing the number n (1 < n < 200) of different nodes. The second line contains the number of edges l. After this, l lines will follow, each containing two numbers that specify an edge between the two nodes that they represent. A node in the graph will be labeled using a numbera ().

An input with n = 0 will mark the end of the input and is not to be processed.

Output

  You have to decide whether the input graph can be bicolored or not, and print it as shown below.

Sample Input

3
3
0 1
1 2
2 0
9
8
0 1
0 2
0 3
0 4
0 5
0 6
0 7
0 8
0

Sample Output

NOT BICOLORABLE.
BICOLORABLE.

/****************************************
* author:crazy_石头
* algorithm:交叉染色法
* date:2013/10/16
* problem:UVA 10004
条件：
1、不存在环
2、图是连通图
3、双向图思路：
任意取一个点进行染色，如果发现要涂某一块时这个块已经被涂了色，并且与我们要使用的颜色不同的话，就说明这个图不能被染成BICOLORABLE的。
（1）如果没有染色，将它染色，并将它周围的点变成相反色。
（2）如果已经染色，判断是否与现在染色的点的颜色相同，相同，则退出，否则继续。
*****************************************/
#include
#include
#includeusing namespace std;#define N 90005
#define M 4000005
#define INF INT_MAXconst int maxn=1000+5;int map[maxn][maxn];
int vis[maxn];
int color[maxn];int n,m;int dfs(int u)
{for(int i=0;i

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