ural 1106. Two Teams 二分图染色

链接：http://acm.timus.ru/problem.aspx?space=1&num=1106
描述：有n(n<=100)个人，每个人有一个或多个朋友（朋友关系是相互的）。将其分成两组，使每一组都有朋友在另一个组。

思路：大意就是求一个子图使其是二分图。直接用dfs染色。 实际上不是二分图，因为本题每个子集里边可以有边相连，只要满足题目给的条件就行了。比二分图简单了一些。

//g++ 4.7.2

#include <cstdio>
#include <iostream>
#include <cstring>
#include <vector>
using namespace std;
const int M = 100 + 10;
int color[M], vis[M];		//color[i]表示结点i的颜色，1表示黑色，2白色
vector<int> G[M];
void dfs(int u)
{
	vis[u] = 1;
	for (int i = 0; i < G[u].size(); ++i)
	{
		int v = G[u][i];
		if (!vis[v])
		{
			color[v] = 3 - color[u];
			dfs(v);
		}
	}
}
int main()
{
	int n, t;
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i)
		while (scanf("%d", &t) && t)
		{
			G[i].push_back(t);
		}
	memset(vis, 0, sizeof(vis));
	memset(color, 0, sizeof(color));
	for (int i = 1; i <= n; ++i)
		if (!vis[i])
		{
			color[i]=1;					//每个新连通分量起始点都要设置为1
			dfs(i);
		}
	int sum = 0;
	for (int i = 1; i <= n; ++i)
		if (color[i] == 1)
			++sum;
	printf("%d\n", sum);
	for (int i = 1; i <= n; ++i)
		if (color[i] == 1)
			printf("%d ", i);
	return 0;
}

还有一种方法差不多，看着像dfs实际不是，本题只需要对每个结点的邻接点染色就行,可以不用递归。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <vector>
using namespace std;
const int M = 100 + 10;
int color[M], vis[M];
vector<int> G[M];
void coloring(int u)
{
	vis[u] = 1;
	color[u] = 1;
	for (int i = 0; i < G[u].size(); ++i)
	{
		int v = G[u][i];
		if (!vis[v])
			color[v] = 3 - color[u];
		vis[v] = 1;
	}
}
int main()
{
	int n, t;
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i)
		while (scanf("%d", &t) && t)
		{
			G[i].push_back(t);
		}
	memset(vis, 0, sizeof(vis));
	memset(color, 0, sizeof(color));
	for (int i = 1; i <= n; ++i)
		if (!vis[i])
			coloring(i);
	int sum = 0;
	for (int i = 1; i <= n; ++i)
		if (color[i] == 1)
			++sum;
	printf("%d\n", sum);
	for (int i = 1; i <= n; ++i)
		if (color[i] == 1)
			printf("%d ", i);
	return 0;
}