uva10004 Bicoloring（交织染色法）

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;
}

uva10004 Bicoloring（交织染色法）

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