给定一个有$n$个点$m$条边的无向图，有$Q$个询问

每次输入$S,E,L,R$，表示你在$S$点出发，要到$E$点，且初始时你是人形态，你只能走$[L,n]$的点，

但是我们要以狼的形态走到终点，这个时候只能走$[1,R]$之间的点，

也就是说我们要在$[L,R]$的某个点变身，问能否从$S->E$。

一个比较容易想到的做法：

显然，初始状态我们要满足$S\in [L,n],E\in [1,R]$，否则就没有讨论的意义了。

假设满足$S\in [L,n],E\in [1,R]$，

我们对每条边按照{u,v,w=min(u,v)}\{u, v, w = min(u, v)\}{u,v,w=min(u,v)}这样的方式，跑一个降序$kruskal$树，然后从$S$点出发，向上跳，找到使点权不小于$L$，能到达的点$u$,

同样的我们对每条边按照{u,v,w=max(u,v}\{u, v, w = max(u, v\}{u,v,w=max(u,v}的方式，跑一个升序$kruskal$树，然后从$E$出发，向上跳，找到使点权不大于$R$，能到达的点$v$，

接下来我们只需判断$u$所代表的点集$S_u$， $v$所代表的点集$S_v$，判断这两个点集有没有交集即可。

设图一进栈$dfs$序为$L1$，出栈$dfs$序为$R1$，同理图二为$L2,R2$，

设$x[i]$为在图一$dfs$序为$i$的点编号，$y[i]$为，点编号为$i$的点在图二的$dfs$序，接下来我们只要在区间$[L{1}_u,R{1}_u]$的主席树上查找区间$[L{2}_v,R{2}_v]$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;
}