UVA - 10004 Bicoloring

  Bicoloring 

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, llines 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 number a ( ).

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 

330 11 22 0980 10 20 30 40 50 60 70 80

Sample Output 

NOT BICOLORABLE.BICOLORABLE.题意：二染色
no node will have an edge to itself. 不存在环
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. 图是连通
the graph will be strongly connected. That is, there will be at least one path from any node to any other node. 双向环
如果没有染色，将它染色，而且颜色相反如果已经染色，是否与现在染色的点的颜色相同，相同，则退出
#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#define N 1000using namespace std;int vis[N];int color[N];int a[N][N];int n,m;int dfs(int u) {for (int i = 0; i < n; i++)  {   if (a[u][i]) {if (vis[i] == 0) {vis[i] = 1;color[i] = !color[u];dfs(i);}else if (color[u] == color[i])return 0;}}return 1;}int main() {while (scanf("%d",&n) != EOF) {if (n == 0)break;memset(vis,0,sizeof(vis));memset(a,0,sizeof(a));scanf("%d",&m);for (int i = 0; i < m; i++) {int u, v;scanf ("%d%d",&u,&v);a[u][v] = a[v][u] = 1;}color[0] = 1;vis[0] = 1;if (dfs(0))printf("BICOLORABLE.\n");elseprintf("NOT BICOLORABLE.\n");}return 0;}