CodeForces 687A NP-Hard Problem（二分图判定）

　　这本来一个挺简单的题呢，结果让我给想复杂了，二分图就是把图分成了两部分，然后不同颜色各一边，肯定是满足题目中说的边和点的条件的，真是犯二了。。

代码如下：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<map>
using namespace std;
#define N 100010
vector<int> vt[N];
int clo[N],n;
int out1[N],out2[N];
map<int,int>mp;
queue<int> que;
void Memset(int *a,int k)
{
    for(int i = 1; i <= n; i++) a[i] = k;
}
bool bfs(int x)
{
    clo[x] = 1;
    while(!que.empty()) que.pop();
    que.push(x);
    int now,nxt,len;
    while(!que.empty())
    {
        now = que.front();
        que.pop();
        len = vt[now].size();
        for(int i = 0; i < len; i++)
        {
            nxt = vt[now][i];
            if(clo[nxt] == -1)
            {
                clo[nxt] = clo[now]^1;
                que.push(nxt);
            }
            if(clo[nxt]==clo[now]) return false;
        }
    }
    return true;
}
bool Judge()
{
    Memset(clo,-1);
    for(int i = 1; i <= n; i++)
    {
        if(clo[i]==-1 && bfs(i) == false) return false;
    }
    return true;
}
int main()
{
    int m,a,b,tot;
    cin>>n>>m;
    for(int i = 1; i <= n; i++)
    {
        vt[i].clear();
    }
    mp.clear();
    for(int i = 0; i < m; i++)
    {
        cin>>a>>b;
        mp[a] = 1;
        mp[b] = 1;
        vt[a].push_back(b);
        vt[b].push_back(a);
    }
    if(Judge() == false) printf("-1
");
    else
    {
        int tot1,tot2;
        tot1 = tot2 = 0;
        for(int i = 1; i <= n; i++)
        {
            if(mp[i] == 1)
            {
                if(clo[i] == 1) out1[tot1++] = i;
                if(clo[i] == 0) out2[tot2++] = i;
            }
        }
        cout<<tot1<<endl;
        for(int i = 0;i < tot1;i++){
            if(i==tot1-1) cout<<out1[i]<<endl;
            else cout<<out1[i]<<" ";
        }
        cout<<tot2<<endl;
        for(int i = 0;i < tot2;i++){
            if(i==tot2-1) cout<<out2[i]<<endl;
            else cout<<out2[i]<<" ";
        }
    }
    return 0;
}