Codeforces-687A NP-Hard Problem(二分图染色)

A. NP-Hard Problem

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Recently, Pari and Arya did some research about NP-Hard problems and they found the minimum vertex cover problem very interesting.
Suppose the graph G is
given. Subset A of its vertices is called a vertex
cover of this graph, if for each edge uv there is at least
one endpoint of it in this set, i.e. 

 or 

 (or
both).
Pari and Arya have won a great undirected graph as an award in a team contest. Now they have to split it in two parts, but both of them want their parts of the graph to be a vertex
cover.
They have agreed to give you their graph and you need to find two disjoint subsets of its
vertices A and B,
such that both A and B are
vertex cover or claim it's impossible. Each vertex should be given to no more than one of the friends (or you can even keep it for yourself).

Input
The first line of the input contains two integers n and m (2 ≤ n ≤ 100 000, 1 ≤ m ≤ 100 000) —
the number of vertices and the number of edges in the prize graph, respectively.
Each of the next m lines
contains a pair of integers ui and vi (1  ≤  ui,  vi  ≤  n),
denoting an undirected edge between ui and vi.
It's guaranteed the graph won't contain any self-loops or multiple edges.

Output
If it's impossible to split the graph between Pari and Arya as they expect, print "-1"
(without quotes).
If there are two disjoint sets of vertices, such that both sets are vertex cover, print their descriptions. Each description must contain two lines. The first line contains a single
integer k denoting the number of vertices in that vertex cover, and the second
line contains kintegers — the indices of vertices. Note that because of m ≥ 1,
vertex cover cannot be empty.

Examples

input

4 2
1 2
2 3

output

1
2
2
1 3

input

3 3
1 2
2 3
1 3

output

-1

Note
In the first sample, you can give the vertex number 2 to
Arya and vertices numbered 1 and 3 to
Pari and keep vertex number 4 for yourself (or give it someone, if you wish).
In the second sample, there is no way to satisfy both Pari and Arya.

二分图染色，待……必搞定

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<vector>

using namespace std;

const int N = 100005;

int n,m,x,y,num1,num2;

int vis
,flag = 1;

vector<int> G
;

void dfs(int x,int k)

{

    if(vis[x])

    {

        if(vis[x] != k)

            flag = 0;

        return;

    }

    if(!flag)

        return;

    vis[x] = k;

    if(k == 1)

        num1++;

    else

        num2++;

    for(int i = 0;i < G[x].size() && flag;i++)

        dfs(G[x][i],-k);

    return;

}

int main()

{

    scanf("%d%d",&n,&m);

    for(int i = 1;i <= m;i++)

    {

        scanf("%d%d",&x,&y);

        G[x].push_back(y);

        G[y].push_back(x);

    }

    for(int i = 1;i <= n;i++)

    {

        if(!vis[i])

            dfs(i,1);

    }

    if(!flag)

    {

        printf("-1\n");

        return 0;

    }

    printf("%d\n",num1);

    for(int i = 1,j = 1;i <= n;i++)

    {

        if(vis[i] == 1)

        {

            if(j == num1)

                printf("%d\n",i);

            else

                printf("%d ",i);

            j++;

        }

    }

    printf("%d\n",num2);

    for(int i = 1,j = 1;i <= n;i++)

    {

        if(vis[i] == -1)

        {

            if(j == num2)

                printf("%d\n",i);

            else

                printf("%d ",i);

            j++;

        }

    }

    return 0;

}