C. NP-Hard Problem

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 k integers — the indices of vertices. Note that because of m ≥ 1, vertex cover cannot be empty.

题意：给你一副可能是不连通的图，问存不存在偶图。
偶图就是对任一一条边的两个节点必须在两个不同的集合中。
孤立节点的话就不用管了。

用染色的办法，对每一个节点的儿子，若没染色，则染色，若染色了，则比较颜色是否相同。用dfs实现~

代码：

#include <cstdio>#include <algorithm>#include <cstring>#include <iostream>#include <vector>const int MAXN=100005;int color[MAXN];int co1[MAXN];int co2[MAXN];int tol;using namespace std;struct Edge{    int u;    int v;}ee[MAXN];vector <int> edge[MAXN];using namespace std;void add(int u,int v){    //ee[tol].u=u;    //ee[tol].v=v;    edge[u].push_back(v);//加边}bool dfs(int now){    for(int i=0;i<edge[now].size();i++)    {        if(color[edge[now][i]]==-1)//若没被染色        {            color[edge[now][i]]=1-color[now];//染和当前颜色相反色            if(!dfs(edge[now][i]))//以染色节点为父亲节点dfs            {                return false;            }        }        else//若染色了        {            if(color[edge[now][i]]==color[now])//若当前颜色和相连点的颜色一样，就不行            {                return false;            }        }    }    return true;}int main (void){    tol=0;    int m,n;    cin>>m>>n;    memset(color,-1,sizeof(color));    for(int i=0;i<n;i++)    {        int u,v;        scanf("%d %d",&u,&v);        add(u,v);        add(v,u);    }    bool sign=true;    for(int i=1;i<=m&&sign;i++)    {        if(color[i]==-1)        {            color[i]=1;            sign&=dfs(i);        }    }    if(!sign)    {        printf("-1\n");    }    else    {        int ans1=0;        int ans2=0;        for(int i=1;i<=m;i++)        {            if(color[i]==1)            {                co1[ans1++]=i;            }            if(color[i]==0)            {                co2[ans2++]=i;            }        }        printf("%d\n",ans1);        for(int i=0;i<ans1;i++)        {            printf("%d%c",co1[i],i==ans1-1?'\n':' ');        }         printf("%d\n",ans2);        for(int i=0;i<ans2;i++)        {        printf("%d%c",co2[i],i==ans2-1?'\n':' ');        }    }    return 0;}