UVA - 10004 Bicoloring （dfs和bfs两种解法）

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 number a ( ).

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.

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。

解析：

取第一个点进行染色，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。

DFS解法：

#include <iostream>
#include <cstdio>
#include <string.h>
using namespace std;
const int N = 210;

int edge[N][N];
int vis[N];
int color[N];
int m,n;

bool dfs(int u) {
	for(int i = 0; i < n; i++) {
		if(edge[u][i]) {
			if(!vis[i]) {
				vis[i] = 1;
				color[i] = !color[u];
				dfs(i);
			}
			else if(color[i] == color[u]){
				return false;
			}
		}
	}
	return true;
}
int main() {
	int u,v;
	while(scanf("%d",&n) != EOF && n) {
		scanf("%d",&m);
		memset(vis,0,sizeof(vis));
		memset(edge,0,sizeof(edge));
		for(int i = 0; i < m; i++) {
			scanf("%d%d",&u,&v);
			edge[u][v] = 1;
			edge[v][u] = 1;
		}
		color[0] = 1;
		vis[0] = 1;
		if(dfs(0)) {
			printf("BICOLORABLE.\n");
		}else {
			printf("NOT BICOLORABLE.\n");
		}
	}
	return 0;
}

BFS解法：

#include <iostream>
#include <stdio.h>
#include <queue>
#include <string.h>
using namespace std;
const int N = 210;
int edge[N][N];
int vis[N];
int color[N];
int m,n;
bool bfs(int s) {
	queue<int> q;
	q.push(s);
	color[s] = 1;
	vis[s] = 1;
	while( !q.empty()) {
		int front = q.front();
		q.pop();
		for(int i = 0; i < n; i++) {
			if(edge[front][i]) {
				if(!vis[i]) {
					color[i] = !color[front];
					vis[i] = 1;
					q.push(i);
				}
				else if(color[i] == color[front]) {
					return false;
				}
			}
		}
	}
	return true;
}

int main() {
	int u,v;
	while(scanf("%d",&n) != EOF && n) {
		scanf("%d",&m);
		memset(edge,0,sizeof(edge));
		memset(vis,0,sizeof(vis));
		memset(color,0,sizeof(color));
		for(int i = 0; i < m; i++) {
			scanf("%d%d",&u,&v);
			edge[u][v] = 1;
			edge[v][u] = 1;
		}
		if( bfs(0) ) {
			printf("BICOLORABLE.\n");
		}else {
			printf("NOT BICOLORABLE.\n");
		}
	}
	return 0;
}