[hdu5215]无向图找奇偶环

题意：如标题

思路：对于奇环，一个二分图判定就ok了，有奇环<=>非二分图。对于偶环，考虑环必定出现在双联通分量里面，可以先求出图的双联通分量，对于一个双联通分量，对于双联通分量里面的每个环，如果是偶环，则偶环已找到，否则假定存在多个奇环，则可以任选两个奇环，把共享边去掉，一定可以得到一个新偶环，这种情况下偶环也是存在的。所以不存在偶环的情况只可能是双联通分量是一个大奇环，特点是：边数=点数，且为奇。于是先dfs一下标记所有桥，用并查集标记所有双联通分量，对每个双联通分量，计算它的点数，对每条边，如果它的两个端点属于同一个双联通分量，则对应双联通分量边数+1。由于是无向边，每条边会被考虑两次。对每个双联通分量，条件改成!((cnt_v*2=cnt_e)&1)，如果上述式子为true，则表示存在偶环。

#pragma comment(linker, "/STACK:102400000,102400000")

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <deque>
#include <cmath>
#include <ctime>
#include <cctype>
#include <set>
#include <bitset>
#include <functional>
#include <numeric>
#include <stdexcept>
#include <utility>
#include <vector>

using namespace std;

#define mem0(a) memset(a, 0, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define define_m int m = (l + r) >> 1
#define rep_up0(a, b) for (int a = 0; a < (b); a++)
#define rep_up1(a, b) for (int a = 1; a <= (b); a++)
#define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
#define rep_down1(a, b) for (int a = b; a > 0; a--)
#define all(a) (a).begin(), (a).end()
#define lowbit(x) ((x) & (-(x)))
#define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
#define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
#define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
#define pchr(a) putchar(a)
#define pstr(a) printf("%s", a)
#define sstr(a) scanf("%s", a)
#define sint(a) scanf("%d", &a)
#define sint2(a, b) scanf("%d%d", &a, &b)
#define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define pint(a) printf("%d\n", a)
#define test_print1(a) cout << "var1 = " << a << endl
#define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
#define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl

typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> vi;

const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
const int maxn = 1e5 + 7;
const int md = 10007;
const int inf = 1e9 + 7;
const LL inf_L = 1e18 + 7;
const double pi = acos(-1.0);
const double eps = 1e-6;

template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
template<class T>T condition(bool f, T a, T b){return f?a:b;}
template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
int make_id(int x, int y, int n) { return x * n + y; }

struct UFS {
    vector<int> F;
    void init(int n) { F.resize(n + 5); for (int i = 0; i <= n; i ++) F[i] = i; }
    int get(int u) { if (F[u] == u) return u; return F[u] = get(F[u]); }
    void add(int u, int v) { F[get(u)] = get(v); }
};

struct Graph {
    vector<vector<int> > G;
    void clear() { G.clear(); }
    void resize(int n) { G.resize(n + 2); }
    void add(int u, int v) { G[u].push_back(v); }
    vector<int> & operator [] (int u) { return G[u]; }
};

Graph G;
int n, m;

int color[maxn];
bool BG_chk(int u, int c) {
    color[u] = c;
    int sz = G[u].size();
    rep_up0(i, sz) {
        int v = G[u][i];
        if (color[v] == c) return false;
        if (color[v]) continue;
        if (!BG_chk(v, 3 - c)) return false;
    }
    return true;
}

UFS us;
int pre[maxn], low[maxn], dfs_clock;
int getBridge(int u, int fa) {
    int lowu = pre[u] = ++ dfs_clock;
    int child = 0, sz = G[u].size();
    rep_up0(i, sz) {
        int v = G[u][i];
        if (!pre[v]) {
            child ++;
            int lowv = getBridge(v, u);
            min_update(lowu, lowv);
            if (lowv <= pre[u]) us.add(u, v);
        }
        else {
            if (pre[v] < pre[u] && v != fa) {
                min_update(lowu, pre[v]);
            }
        }
    }
    return low[u] = lowu;
}

int cnt_v[maxn], cnt_e[maxn], vis[maxn];
bool findEvenRing() {
    dfs_clock = 0;
    mem0(pre);
    rep_up1(i, n) {
        if (!pre[i]) {
            getBridge(i, 0);
        }
    }
    mem0(cnt_e);
    mem0(cnt_v);
    rep_up1(i, n) {
        int u = us.get(i);
        cnt_v[u] ++;
    }
    rep_up1(i, n) {
        int sz = G[i].size();
        rep_up0(j, sz) {
            int u = G[i][j], tmp;
            if ((tmp = us.get(i)) == us.get(u)) {
                cnt_e[tmp] ++;
            }
        }
    }
    mem0(vis);
    rep_up1(i, n) {
        int u = us.get(i);
        if (vis[u]) continue;
        vis[u] = true;
        if ((cnt_v[u] * 2 != cnt_e[u] || !(cnt_v[u] & 1)) && cnt_v[u] >= 4) return true;
    }
    return false;
}

int main() {
    //freopen("in.txt", "r", stdin);
    int T;
    cin >> T;
    while (T --) {
        cin >> n >> m;
        G.clear();
        G.resize(n);
        us.init(n);
        rep_up0(i, m) {
            int u, v;
            sint2(u, v);
            G.add(u, v);
            G.add(v, u);
        }
        bool odd = false, even = findEvenRing();
        mem0(color);
        rep_up1(i, n) {
            if (!color[i]) {
                odd = odd || !BG_chk(i, 1);
            }
        }
        puts(odd? "YES" : "NO");
        puts(even? "YES" : "NO");
    }
    return 0;
}