timus 1106 Two Teams(二部图)

Two Teams

Time limit: 1.0 second
Memory limit: 64 MB

The group of people consists of N
members. Every member has one or more friends in the group. You are to
write program that divides this group into two teams. Every member of
each team must have friends in another team.

Input

The first line of input contains the only number N (N ≤ 100). Members are numbered from 1 to N. The second, the third,…and the (N+1)th line contain list of friends of the first, the second, …and the Nth member respectively. This list is finished by zero. Remember that friendship is always mutual in this group.

Output

The
first line of output should contain the number of people in the first
team or zero if it is impossible to divide people into two teams. If the
solution exists you should write the list of the first group into the
second
line of output. Numbers should be divided by single space. If there are
more than one solution you may find any of them.

Sample

input	output
7 2 3 0 3 1 0 1 2 4 5 0 3 0 3 0 7 0 6 0	4 2 4 5 6

Problem Author: Dmitry Filimonenkov

【分析】一个简单的二部图染色问题。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <time.h>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define inf 10000000
#define mod 10000
typedef long long ll;
using namespace std;
const int N=;
const int M=;
int power(int a,int b,int c){int ans=;while(b){if(b%==){ans=(ans*a)%c;b--;}b/=;a=a*a%c;}return ans;}
int color[N], vis[N];
vector<int> G[N];
void dfs(int u)
{
    vis[u] = ;
    for (int i = ; i < G[u].size(); ++i)
    {
        int v = G[u][i];
        if (!vis[v])
        {
            color[v] =  - color[u];
            dfs(v);
        }
    }
}
int main()
{
    int n, t;
    scanf("%d", &n);
    for (int i = ; i <= n; ++i)
        while (scanf("%d", &t) && t)
        {
            G[i].push_back(t);
        }
    memset(vis, , sizeof(vis));
    memset(color, , sizeof(color));
    for (int i = ; i <= n; ++i)
        if (!vis[i])
        {
            color[i]=;
            dfs(i);
        }
    int sum = ;
    for (int i = ; i <= n; ++i)
        if (color[i] == )
            ++sum;
    printf("%d\n", sum);
    for (int i = ; i <= n; ++i)
        if (color[i] == )
            printf("%d ", i);
    return ;
}