uva10004 Bicoloring（交叉染色法）

题目：

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.

题目翻译：

1976年“四色定理”在计算机的帮助下被证明。 这个定理宣告任何一个地图都可以只用四种颜色来填充， 并且没有相邻区域的颜色是相同的。

现在让你解决一个更加简单的问题。 你必须决定给定的任意相连的图能不能够用两种颜色填充。 就是说，如果给其中一个分配一种颜色， 要让所有直接相连的两个节点不能是相同的颜色。 为了让问题更简单，你可以假设：

1. 没有节点是连接向它自己的。

2. 是无向图。  即如果a连接b, 那么b也是连接a的

3. 图是强连接的。就是说至少有一条路径可走向所有节点。

样例输入：

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

样例输出：

NOT BICOLORABLE.

BICOLORABLE.

#include<stdio.h>
int f,n,m,top,end,queue[200],map[200][200],color[200];
void bfs()
{
int p=queue[top];
if(f==0) return ;
for(int i=0;i<n;i++)
{
if(map[p][i]==1)
{
if(color[i]==2)
{
end++;
queue[end]=i;
color[i]=-color[p];
}
else {if(color[i]==color[p]) f=0;}
}
}
top++;
if(top<=end) bfs();
return ;
}
int main()
{
while(scanf("%d",&n)&&n!=0)
{
for(int i=0;i<n;i++)
{
for(int j=i;j<n;j++)
map[i][j]=map[j][i]=0;
color[i]=2;
}
scanf("%d",&m);
int a,b;
for(int i=0;i<m;i++)
{
scanf("%d%d",&a,&b);
map[a][b]=map[b][a]=1;
color[a]=2;
color[b]=2;
}
top=1;
end=1;
queue[1]=a;
color[a]=1;
f=1;
bfs();
if (f==0) printf("NOT BICOLORABLE.\n");
else printf("BICOLORABLE.\n");
}
return 0;
}