UVA 10004 Bicoloring

题目如下：

Bicoloring

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

Here you are asked to solve a simpler similar problem. Youhave 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 sucha way that no two adjacent nodes have the same color. To simplify the problemyou can assume:

• no node will have an edge to itself.
• the graph is nondirected. That is, if a node a is said to be connectedto 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 leastone path from any node to any other node.

Input 

The input consists of several test cases. Each test casestarts with a line containing the number n (1 < n < 200) of differentnodes. The second line contains the number of edges l. After this, l lineswill follow, each containing two numbers that specify an edge between the twonodes 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 isnot to be processed.

Output 

You have to decide whether the input graph can bebicolored 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.

问是否可以用两种颜色将一个无向强联通图染色，其实可以转化为这个图是否是二分图。二分图的充要条件是图中的每个圈都是偶数，所以用DFS找出图中每个圈的长度，再判断是否是偶数，值得注意的是vis标记数组不应该标记节点，因为节点可能访问多次，而应该为二维数组标记边，在找圈过程中边是不能重复遍历的，1A。

AC的代码如下：

#include 

using namespace std;

int graph[200][200];
int vis[200][200];
vector ans;
int n,l;
void dfs(int u,int v,int length)
{
    if(u==v&&length!=0)
    {
        ans.push_back(length);
        return;
    }
    for(int i=0; i>n)
    {
        if(n==0)
            break;
        cin>>l;
        memset(graph,0,sizeof graph);

        int a,b;
        for(int i=0; i>a>>b;
            graph[a][b]=graph[b][a]=1;
        }
        int flag=1;
        for(int i=0; i