[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。

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

  2 

  3 #include <iostream>

  4 #include <cstdio>

  5 #include <algorithm>

  6 #include <cstdlib>

  7 #include <cstring>

  8 #include <map>

  9 #include <queue>

 10 #include <deque>

 11 #include <cmath>

 12 #include <ctime>

 13 #include <cctype>

 14 #include <set>

 15 #include <bitset>

 16 #include <functional>

 17 #include <numeric>

 18 #include <stdexcept>

 19 #include <utility>

 20 #include <vector>

 21 

 22 using namespace std;

 23 

 24 #define mem0(a) memset(a, 0, sizeof(a))

 25 #define mem_1(a) memset(a, -1, sizeof(a))

 26 #define lson l, m, rt << 1

 27 #define rson m + 1, r, rt << 1 | 1

 28 #define define_m int m = (l + r) >> 1

 29 #define rep_up0(a, b) for (int a = 0; a < (b); a++)

 30 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)

 31 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)

 32 #define rep_down1(a, b) for (int a = b; a > 0; a--)

 33 #define all(a) (a).begin(), (a).end()

 34 #define lowbit(x) ((x) & (-(x)))

 35 #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) {}

 36 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}

 37 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}

 38 #define pchr(a) putchar(a)

 39 #define pstr(a) printf("%s", a)

 40 #define sstr(a) scanf("%s", a)

 41 #define sint(a) scanf("%d", &a)

 42 #define sint2(a, b) scanf("%d%d", &a, &b)

 43 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)

 44 #define pint(a) printf("%d\n", a)

 45 #define test_print1(a) cout << "var1 = " << a << endl

 46 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl

 47 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl

 48 

 49 typedef long long LL;

 50 typedef pair<int, int> pii;

 51 typedef vector<int> vi;

 52 

 53 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};

 54 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };

 55 const int maxn = 1e5 + 7;

 56 const int md = 10007;

 57 const int inf = 1e9 + 7;

 58 const LL inf_L = 1e18 + 7;

 59 const double pi = acos(-1.0);

 60 const double eps = 1e-6;

 61 

 62 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}

 63 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}

 64 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}

 65 template<class T>T condition(bool f, T a, T b){return f?a:b;}

 66 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}

 67 int make_id(int x, int y, int n) { return x * n + y; }

 68 

 69 struct UFS {

 70     vector<int> F;

 71     void init(int n) { F.resize(n + 5); for (int i = 0; i <= n; i ++) F[i] = i; }

 72     int get(int u) { if (F[u] == u) return u; return F[u] = get(F[u]); }

 73     void add(int u, int v) { F[get(u)] = get(v); }

 74 };

 75 

 76 struct Graph {

 77     vector<vector<int> > G;

 78     void clear() { G.clear(); }

 79     void resize(int n) { G.resize(n + 2); }

 80     void add(int u, int v) { G[u].push_back(v); }

 81     vector<int> & operator [] (int u) { return G[u]; }

 82 };

 83 

 84 Graph G;

 85 int n, m;

 86 

 87 int color[maxn];

 88 bool BG_chk(int u, int c) {

 89     color[u] = c;

 90     int sz = G[u].size();

 91     rep_up0(i, sz) {

 92         int v = G[u][i];

 93         if (color[v] == c) return false;

 94         if (color[v]) continue;

 95         if (!BG_chk(v, 3 - c)) return false;

 96     }

 97     return true;

 98 }

 99 

100 UFS us;

101 int pre[maxn], low[maxn], dfs_clock;

102 int getBridge(int u, int fa) {

103     int lowu = pre[u] = ++ dfs_clock;

104     int child = 0, sz = G[u].size();

105     rep_up0(i, sz) {

106         int v = G[u][i];

107         if (!pre[v]) {

108             child ++;

109             int lowv = getBridge(v, u);

110             min_update(lowu, lowv);

111             if (lowv <= pre[u]) us.add(u, v);

112         }

113         else {

114             if (pre[v] < pre[u] && v != fa) {

115                 min_update(lowu, pre[v]);

116             }

117         }

118     }

119     return low[u] = lowu;

120 }

121 

122 int cnt_v[maxn], cnt_e[maxn], vis[maxn];

123 bool findEvenRing() {

124     dfs_clock = 0;

125     mem0(pre);

126     rep_up1(i, n) {

127         if (!pre[i]) {

128             getBridge(i, 0);

129         }

130     }

131     mem0(cnt_e);

132     mem0(cnt_v);

133     rep_up1(i, n) {

134         int u = us.get(i);

135         cnt_v[u] ++;

136     }

137     rep_up1(i, n) {

138         int sz = G[i].size();

139         rep_up0(j, sz) {

140             int u = G[i][j], tmp;

141             if ((tmp = us.get(i)) == us.get(u)) {

142                 cnt_e[tmp] ++;

143             }

144         }

145     }

146     mem0(vis);

147     rep_up1(i, n) {

148         int u = us.get(i);

149         if (vis[u]) continue;

150         vis[u] = true;

151         if ((cnt_v[u] * 2 != cnt_e[u] || !(cnt_v[u] & 1)) && cnt_v[u] >= 4) return true;

152     }

153     return false;

154 }

155 

156 int main() {

157     //freopen("in.txt", "r", stdin);

158     int T;

159     cin >> T;

160     while (T --) {

161         cin >> n >> m;

162         G.clear();

163         G.resize(n);

164         us.init(n);

165         rep_up0(i, m) {

166             int u, v;

167             sint2(u, v);

168             G.add(u, v);

169             G.add(v, u);

170         }

171         bool odd = false, even = findEvenRing();

172         mem0(color);

173         rep_up1(i, n) {

174             if (!color[i]) {

175                 odd = odd || !BG_chk(i, 1);

176             }

177         }

178         puts(odd? "YES" : "NO");

179         puts(even? "YES" : "NO");

180     }

181     return 0;

182 }

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