link ：http://acm.timus.ru/problem.aspx?space=1&num=1106

describe ： Yes n(n<=100) personal , Everyone has one or more friends （ Friendship is mutual ）. Divide them into two groups , Make each group have friends in another group .

Ideas ： The main idea is to find a subgraph so that it is a bipartite graph . Direct use dfs dyeing . It's not actually a bipartite graph , Because each subset of this problem can be connected by edges , As long as the conditions given by the title are met . It's a little simpler than bipartite graphs .

//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] Represents a node i The color of the ,1 According to black ,2 white

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; // The starting point of each new connected component should be set to 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;

}

There's another way that's more or less , Look like dfs The actual is not , This problem only needs to dye the adjacent points of each node , You don't need recursion .

#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;

}

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