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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 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #include 11 #include 12 #include 13 #include 14 #include 15 #include 16 #include 17 #include 18 #include 19 #include 20 #include 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 pii; 51 typedef vector 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 T gcd(T a, T b){ return b==0?a:gcd(b,a%b);} 63 template bool max_update(T &a,const T &b){ if(b>a){a = b; return true;}return false;} 64 template bool min_update(T &a,const T &b){ if(b T condition(bool f, T a, T b){ return f?a:b;} 66 template 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 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 > 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 & 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 }
转载于:https://www.cnblogs.com/jklongint/p/4486595.html
