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;

}