P4899 [IOI2018] werewolf 狼人

給定一個有$n$個點$m$條邊的無向圖，有$Q$個詢問

每次輸入$S,E,L,R$，表示你在$S$點出發，要到$E$點，且初始時你是人形態，你只能走$[L,n]$的點，

但是我們要以狼的形態走到終點，這個時候只能走$[1,R]$之間的點，

也就是說我們要在$[L,R]$的某個點變身，問能否從$S?>E$。

一個比較容易想到的做法：

顯然，初始狀態我們要滿足$S\in [L,n],E\in [1,R]$，否則就沒有討論的意義了。

假設滿足$S\in [L,n],E\in [1,R]$，

我們對每條邊按照$\{u,v,w=min(u,v)\}$這樣的方式，跑一個降序$kruskal$樹，然后從$S$點出發，向上跳，找到使點權不小于$L$，能到達的點$u$,

同樣的我們對每條邊按照$\{u,v,w=max(u,v\}$的方式，跑一個升序$kruskal$樹，然后從$E$出發，向上跳，找到使點權不大于$R$，能到達的點$v$，

接下來我們只需判斷$u$所代表的點集$S_u$， $v$所代表的點集$S_v$，判斷這兩個點集有沒有交集即可。

設圖一進棧$dfs$序為$L1$，出棧$dfs$序為$R1$，同理圖二為$L2,R2$，

設$x[i]$為在圖一$dfs$序為$i$的點編號，$y[i]$為，點編號為$i$的點在圖二的$dfs$序，接下來我們只要在區間$[L{1}_u,R{1}_u]$的主席樹上查找區間$[L{2}_v,R{2}_v]$是否有值即可。

#include <bits/stdc++.h>using namespace std;const int N = 4e5 + 10;struct Res {int u, v, w;bool operator < (const Res &t) const {return w < t.w;} }edge[N];int n, m, Q, x[N], y[N];int root[N], ls[N << 5], rs[N<< 5], sum[N << 5], num;void update(int &rt, int pre, int l, int r, int x, int v) {rt = ++num;ls[rt] = ls[pre], rs[rt] = rs[pre], sum[rt] = sum[pre] + v;if (l == r) {return ;}int mid = l + r >> 1;if (x <= mid) {update(ls[rt], ls[pre], l, mid, x, v);}else {update(rs[rt], rs[pre], mid + 1, r, x, v);} }int query(int rt1, int rt2, int l, int r, int L, int R) {// cout << l << " " << r << "\n";if (l >= L && r <= R) {return sum[rt2] - sum[rt1];}int ans = 0, mid = l + r >> 1;if (L <= mid) {ans += query(ls[rt1], ls[rt2], l, mid, L, R);}if (R > mid) {ans += query(rs[rt1], rs[rt2], mid + 1, r, L, R);}return ans; }struct Kruskal {int head[N], to[N], nex[N], cnt = 1;int ff[N], value[N], fa[N][21], l[N], r[N], rk[N], tot, nn;void add(int x, int y) {to[cnt] = y;nex[cnt] = head[x];head[x] = cnt++;}int find(int rt) {return ff[rt] == rt ? rt : ff[rt] = find(ff[rt]);}void dfs(int rt, int f) {fa[rt][0] = f, l[rt] = ++tot, rk[tot] = rt;for (int i = 1; i <= 20; i++) {fa[rt][i] = fa[fa[rt][i - 1]][i - 1];}for (int i = head[rt]; i; i = nex[i]) {if (to[i] == f) {continue;}dfs(to[i], rt);}r[rt] = tot;}void kruskal(int rev) {sort(edge + 1, edge + 1 + m);if (rev) {reverse(edge + 1, edge + 1 + m);}for (int i = 1; i < N; i++) {ff[i] = i;}nn = n;for (int i = 1, cur = 1; i <= m && cur < n; i++) {int u = find(edge[i].u), v = find(edge[i].v);if (u != v) {cur++, nn++;ff[u] = ff[v] = nn;value[nn] = edge[i].w;add(nn, u), add(nn, v);if (u <= n) {value[u] = edge[i].w;}if (v <= n) {value[v] = edge[i].w;}}}dfs(nn, 0);}bool judge(int u, int x, int flag) {return flag ? value[u] >= x : value[u] <= x;}int work(int u, int x, int flag) {for (int i = 20; i >= 0; i--) {if (fa[u][i] && judge(fa[u][i], x, flag)) {u = fa[u][i];}}return u;} }a, b;int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);scanf("%d %d %d", &n, &m, &Q);for (int i = 1; i <= m; i++) {scanf("%d %d", &edge[i].u, &edge[i].v);edge[i].u++, edge[i].v++;}for (int i = 1; i <= m; i++) {edge[i].w = min(edge[i].v, edge[i].u);}a.kruskal(1);for (int i = 1; i <= m; i++) {edge[i].w = max(edge[i].v, edge[i].u);}b.kruskal(0);for (int i = 1; i < 2 * n; i++) {x[i] = a.rk[i];y[i] = b.l[i];}for (int i = 1; i < 2 * n; i++) {root[i] = root[i - 1];if (x[i] <= n) {update(root[i], root[i - 1], 1, 2 * n, y[x[i]], 1);}}while (Q--) {int u, v, L, R;scanf("%d %d %d %d", &u, &v, &L, &R);u++, v++, L++, R++;u = a.work(u, L, 1), v = b.work(v, R, 0);int l1 = a.l[u], r1 = a.r[u], l2 = b.l[v], r2 = b.r[v];printf("%d\n", query(root[l1 - 1], root[r1], 1, 2 * n, l2, r2) != 0);}return 0; }

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