UOJ 407(IOI2018 D1T3)
作者：互联网
给定一张$n$个点，$m$条边的无向连通图以及$q$次询问，每次询问给出$S,E,L,R$，问你是否能从$S$出发，不经过编号小于$L$的点到达某个编号大于等于$L$且小于等于$R$的点，此时切换状态，不经过编号大于$R$的点到达$E$。
$$n\le200000,m\le400000,q\le200000$$
这种有限制的连通性问题一般考虑Kruskal重构树，对两个限制建出两棵树，倍增找出每个询问能到达的点的区间，然后就是一个经典的二维数点问题了，排序后树状数组解决。
1 constexpr int MAXN = 200000 + 5, MAXM = 400000 + 5, LOG = 20;
2
3 int lg[MAXN * 2];
4
5 struct Tree {
6 static constexpr int MAX_N = MAXM;
7
8 int hed[MAX_N], nxt[MAX_N * 2], to[MAX_N * 2], val[MAX_N], fa[LOG][MAX_N], dep[MAX_N], L[MAX_N], R[MAX_N], cnt, num;
9
10 void addEdge(int u, int v) {
11 ++ cnt;
12 to[cnt] = v;
13 nxt[cnt] = hed[u];
14 hed[u] = cnt;
15 }
16
17 void DFS(int u, int total) {
18 for (int i = 1; 1 << i <= dep[u]; ++ i) {
19 fa[i][u] = fa[i - 1][fa[i - 1][u]];
20 }
21 if (u <= total) {
22 L[u] = R[u] = ++ num;
23 } else {
24 L[u] = INF, R[u] = 0;
25 }
26 for (int e = hed[u]; e; e = nxt[e]) {
27 int v = to[e];
28 if (v != fa[0][u]) {
29 fa[0][v] = u;
30 dep[v] = dep[u] + 1;
31 DFS(v, total);
32 chkmin(L[u], L[v]);
33 chkmax(R[u], R[v]);
34 }
35 }
36 }
37
38 int find(int u, int w, const std::function<bool(int, int)> &comp) {
39 Rep(i, lg[dep[u]], 0) {
40 if (fa[i][u] && comp(val[fa[i][u]], w)) {
41 u = fa[i][u];
42 }
43 }
44 return u;
45 }
46 } T1, T2;
47
48 struct DSU {
49 int fa[MAXN], bel[MAXN];
50
51 void init(int n) {
52 For(i, 1, n) {
53 fa[i] = bel[i] = i;
54 }
55 }
56
57 int find(int u) {
58 return fa[u] == u ? u : fa[u] = find(fa[u]);
59 }
60
61 bool same(int u, int v) {
62 return find(u) == find(v);
63 }
64
65 void merge(int u, int v, int x) {
66 if (!same(u, v)) {
67 fa[fa[v]] = fa[u], bel[fa[u]] = x;
68 }
69 }
70 } U;
71
72 struct BinaryIndexedTree {
73 int a[MAXN], length;
74
75 #define lowbit(x) ((x) & -(x))
76 void init(int n) {
77 length = n;
78 }
79
80 void modify(int i, int x) {
81 for (; i <= length; i += lowbit(i)) {
82 a[i] += x;
83 }
84 }
85
86 int query(int i) {
87 int result = 0;
88 for (; i; i -= lowbit(i)) {
89 result += a[i];
90 }
91 return result;
92 }
93 #undef lowbit
94 } BIT;
95
96 IL void adjust(std::vector<int> &v) {
97 for (auto &x : v) {
98 ++ x;
99 }
100 }
101
102 std::vector<int> check_validity(int N, std::vector<int> X, std::vector<int> Y,
103 std::vector<int> S, std::vector<int> E,
104 std::vector<int> L, std::vector<int> R) {
105 adjust(X), adjust(Y), adjust(S), adjust(E), adjust(L), adjust(R);
106 std::vector<std::pair<int, int>> edge(X.size());
107 FOR(i, 0, (int)edge.size()) {
108 edge[i] = std::make_pair(X[i], Y[i]);
109 }
110 lg[0] = -1;
111 FOR(i, 1, N << 1) {
112 lg[i] = lg[i >> 1] + 1;
113 }
114 int Q = S.size();
115 std::vector<int> answer(Q);
116 static bool ok[MAXN];
117 memset(ok, 1, sizeof ok);
118 FOR(i, 0, Q) {
119 if (S[i] < L[i] || E[i] > R[i]) {
120 ok[i] = 0;
121 }
122 }
123 static std::vector<int> left[MAXN], right[MAXN];
124 std::sort(edge.begin(), edge.end(), [](const std::pair<int, int> &a, const std::pair<int, int> &b) -> bool {
125 return mymin(a.first, a.second) > mymin(b.first, b.second);
126 });
127 U.init(N);
128 For(i, 1, N) {
129 T1.val[i] = i;
130 }
131 int cnt = 0;
132 for (auto &x : edge) {
133 if (!U.same(x.first, x.second)) {
134 ++ cnt;
135 int p = U.fa[x.first], q = U.fa[x.second];
136 T1.val[N + cnt] = mymin(T1.val[U.bel[p]], T1.val[U.bel[q]]);
137 T1.addEdge(U.bel[p], N + cnt);
138 T1.addEdge(N + cnt, U.bel[p]);
139 T1.addEdge(U.bel[q], N + cnt);
140 T1.addEdge(N + cnt, U.bel[q]);
141 U.merge(x.first, x.second, N + cnt);
142 }
143 if (cnt >= N - 1) {
144 break;
145 }
146 }
147 T1.DFS((N << 1) - 1, N);
148 FOR(i, 0, Q) {
149 if (!ok[i]) {
150 continue;
151 }
152 int u = T1.find(S[i], L[i], [](int a, int b) -> bool {
153 return a >= b;
154 });
155 left[T1.L[u]].push_back(i);
156 right[T1.R[u]].push_back(i);
157 }
158 std::sort(edge.begin(), edge.end(), [](const std::pair<int, int> &a, const std::pair<int, int> &b) -> bool {
159 return mymax(a.first, a.second) < mymax(b.first, b.second);
160 });
161 U.init(N);
162 For(i, 1, N) {
163 T2.val[i] = i;
164 }
165 cnt = 0;
166 for (auto &x : edge) {
167 if (!U.same(x.first, x.second)) {
168 ++ cnt;
169 int p = U.fa[x.first], q = U.fa[x.second];
170 T2.val[N + cnt] = mymax(T2.val[U.bel[p]], T2.val[U.bel[q]]);
171 T2.addEdge(U.bel[p], N + cnt);
172 T2.addEdge(N + cnt, U.bel[p]);
173 T2.addEdge(U.bel[q], N + cnt);
174 T2.addEdge(N + cnt, U.bel[q]);
175 U.merge(x.first, x.second, N + cnt);
176 }
177 if (cnt >= N - 1) {
178 break;
179 }
180 }
181 T2.DFS((N << 1) - 1, N);
182 std::vector<std::pair<int, int>> interval(Q);
183 FOR(i, 0, Q) {
184 if (!ok[i]) {
185 continue;
186 }
187 int u = T2.find(E[i], R[i], [](int a, int b) -> bool {
188 return a <= b;
189 });
190 interval[i] = std::make_pair(T2.L[u], T2.R[u]);
191 }
192 static std::vector<int> point[MAXN];
193 For(i, 1, N) {
194 point[T1.L[i]].push_back(T2.L[i]);
195 }
196 BIT.init(N);
197 For(i, 1, N) {
198 for (auto &x : left[i]) {
199 answer[x] = BIT.query(interval[x].second) - BIT.query(interval[x].first - 1);
200 }
201 for (auto &x : point[i]) {
202 BIT.modify(x, 1);
203 }
204 for (auto &x : right[i]) {
205 answer[x] = !!(BIT.query(interval[x].second) - BIT.query(interval[x].first - 1) - answer[x]);
206 }
207 }
208 return answer;
209 }
标签：std,cnt,IOI2018,bel,int,T1,407,fa,UOJ 来源： https://www.cnblogs.com/sjkmost/p/10372007.html