标签：uva   c++   

                                uva 10004 Bicoloring

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 

An input with n = 0 will mark the end of the input and isnot to be processed.

Output 

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.

题目大意：二染色。

条件：

1、不存在环

2、图是连通图

3、双向图

解题思路（转）：

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<stdio.h>
#include<string.h>
int G[205][205], vis[205], color[205];
int m, n;
int DFS(int a) {
	for (int i = 0; i < m; i++) {
		if (G[a][i]) {
			if (!vis[i]) {
				vis[i] = 1;
				color[i] = !color[a];
				DFS(i);
			}
			else if (color[i] == color[a]) return 0;
		}
	}
	return 1;
}
int main() {
	while (scanf("%d\n%d", &m, &n) == 2) {
		memset(G, 0, sizeof(G));
		memset(vis, 0, sizeof(vis));
		int a, b;
		for (int i = 0; i < n; i++) {
			scanf("%d %d\n", &a, &b);
			G[a][b] = G[b][a] = 1;
		}
		color[0] = 1;
		vis[0] = 1;
		if (DFS(0)) printf("BICOLORABLE.\n");
		else printf("NOT BICOLORABLE.\n");
	}
	return 0;
}

uva 10004 Bicoloring（DFS）

标签：uva   c++   

原文地址：http://blog.csdn.net/llx523113241/article/details/43197551