Sub-Bipartite Graph FZU

水题 水题 水题

染色

但有个限制条件 就是选出来的边要大于等于 m / 2

所以 再每次加入一个点的时候

判断与这个点相连的是黑色的多 还是白色的多

黑色的多就放到白色那边

反之亦然

保证每次加入点后保留的边的个数尽量多

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <list>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define rb(a) scanf("%lf", &a)
#define rf(a) scanf("%f", &a)
#define pd(a) printf("%d
", a)
#define plld(a) printf("%lld
", a)
#define pc(a) printf("%c
", a)
#define ps(a) printf("%s
", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 105, INF = 0x7fffffff;
vector<int> G[maxn];
int vis[maxn];
int n, m;
int istwo(int u)
{
    queue<int> E;
    E.push(u);
    vis[u] = 1;
    while(!E.empty())
    {
        u = E.front(); E.pop();
        for(int i=0; i<G[u].size(); i++)
        {
            int v = G[u][i];
            if(vis[v] == -1)
            {
                int h1 = 0, h2 = 0;

                for(int j = 0; j < G[v].size(); j++)
                {
                    if(vis[G[v][j]] == 0) h1++;
                    else if(vis[G[v][j]] == 1) h2++;

                }
                if(h1 > h2) vis[v] = 1;
                else vis[v] = 0;
                E.push(v);
            }
        }
    }
    return 1;
}

vector<int> v1, v2;

int main()
{
    int T;
    rd(T);
    while(T--)
    {

        v1.clear();
        v2.clear();
        int u, v;
        rd(n), rd(m);
        for(int i = 0; i <= n; i++) vis[i]= -1, G[i].clear();
        for(int i = 0; i < m; i++)
        {
            rd(u), rd(v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        for(int i = 1; i <= n; i++)
            if(vis[i] == -1)
                istwo(i);
        for(int i = 1; i <= n; i++)
        {
            if(vis[i] == 1) v1.push_back(i);
            else if(vis[i] == 0) v2.push_back(i);
        }
        int len1 = v1.size();
        printf("%d", len1);
        for(int i = 0; i < v1.size(); i++)
        {
            printf(" %d", v1[i]);

        }
        printf("
");
        int len2 = v2.size();
        printf("%d", len2);
        for(int i = 0; i < v2.size(); i++)
        {
            printf(" %d", v2[i]);

        }
        printf("
");

    }

    return 0;
}