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

 template<class T>T gcd(T a, T b){return b==?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 + ); for (int i = ; 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 + ); }
     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,  - 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 = , 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 = ;
     mem0(pre);
     rep_up1(i, n) {
         if (!pre[i]) {
             getBridge(i, );
         }
     }
     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] *  != cnt_e[u] || !(cnt_v[u] & )) && cnt_v[u] >= ) 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, );
             }
         }
         puts(odd? "YES" : "NO");
         puts(even? "YES" : "NO");
     }
     return ;
 }