两种形态都对经过的路程有限制，我们可以联想到克鲁斯卡尔重构树。
我们考虑将点权转化为边权，因为我们走这条边的话两个端点都要符合条件，所以人形态是边权为边的两个端点的较小值，狼形态相反。
人形态时要建一个最大生成树，狼形态相反
然后我们就可以知道人形态时起点可以到达哪些点，狼形态时哪些点可以到达终点。
怎么判又没有交集呢？因为树上一个子树的 \(dfs\) 序是连续的，我们可以以第一棵树为基础建立主席树，查询的时候只用看相应的区间是否为空就行了。
#include<algorithm>
#include<iostream>
#include<cstdio>
using namespace std;
int n, m, q, x, y, num, l, r, cnt, s, t;
const int N = 400010;
int fa[N], root[N], lson[N * 21], rson[N * 21], siz[N * 21];
struct bian {int x, y, z;} e[N];
int my1(bian a, bian b) {return a.z > b.z;}
int my2(bian a, bian b) {return a.z < b.z;}
int find(int x) {return x == fa[x] ? fa[x] : fa[x] = find(fa[x]);}
struct Tu
{
int tot, cnt;
int head[N], to[N], nt[N], val[N], fa[N][21], siz[N], dfn[N], nfd[N]; //19
void add(int f, int t)
{
to[++tot] = t; nt[tot] = head[f]; head[f] = tot;
}
void dfs(int x)
{
siz[x] = 1; dfn[x] = ++cnt; nfd[cnt] = x;
for (int i = 1; i <= 19; ++i)
fa[x][i] = fa[fa[x][i - 1]][i - 1];
for (int i = head[x]; i; i = nt[i])
if (to[i] != fa[x][0])dfs(to[i]), siz[x] += siz[to[i]];
}
int find(int x, int v, int opt)
{
for (int i = 19; i >= 0; --i)
{
if (opt == 1) {if (fa[x][i] && val[fa[x][i]] >= v)x = fa[x][i];}
else {if (fa[x][i] && val[fa[x][i]] <= v)x = fa[x][i];}
}
return x;
}
} X1, X2;
void klske()
{
for (int i = 1; i <= m; ++i)e[i].z = min(e[i].x, e[i].y); //人
for (int i = 1; i <= (n << 1); ++i)fa[i] = i; num = n;
sort(e + 1, e + 1 + m, my1);
for (int i = 1; i <= m; ++i)
{
x = find(e[i].x); y = find(e[i].y);
if (x == y)continue;
X1.val[++num] = e[i].z;
X1.add(num, x); X1.add(num, y);
X1.fa[x][0] = num; X1.fa[y][0] = num;
fa[x] = num; fa[y] = num;
if (num - n == n - 1)break;
}
X1.dfs(find(1));
for (int i = 1; i <= m; ++i)e[i].z = max(e[i].x, e[i].y); //狼
for (int i = 1; i <= (n << 1); ++i)fa[i] = i; num = n;
sort(e + 1, e + 1 + m, my2);
for (int i = 1; i <= m; ++i)
{
x = find(e[i].x); y = find(e[i].y);
if (x == y)continue;
X2.val[++num] = e[i].z;
X2.add(num, x); X2.add(num, y);
X2.fa[x][0] = num; X2.fa[y][0] = num;
fa[x] = num; fa[y] = num;
if (num - n == n - 1)break;
}
X2.dfs(find(1));
}
void Insert(int pre, int &k, int l, int r, int pos)
{
k = ++cnt;
siz[k] = siz[pre] + 1;
if (l == r)return;
int mid = (l + r) >> 1;
if (pos <= mid)rson[k] = rson[pre], Insert(lson[pre], lson[k], l, mid, pos);
else lson[k] = lson[pre], Insert(rson[pre], rson[k], mid + 1, r, pos);
}
int ask(int pre, int k, int l, int r, int x, int y)
{
if (x <= l && r <= y)return siz[k] - siz[pre];
int mid = (l + r) >> 1, res = 0;
if (x <= mid)res += ask(lson[pre], lson[k], l, mid, x, y);
if (mid + 1 <= y)res += ask(rson[pre], rson[k], mid + 1, r, x, y);
return res;
}
int main()
{
cin >> n >> m >> q;
for (int i = 1; i <= m; ++i)
{
scanf("%d%d", &x, &y); ++x; ++y;
e[i].x = x; e[i].y = y;
}
klske();
for (int i = 1; i <= (n << 1); ++i)
if (X1.nfd[i] <= n)Insert(root[i - 1], root[i], 1, n << 1, X2.dfn[X1.nfd[i]]);
else root[i] = root[i - 1];
while (q--)
{
scanf("%d%d%d%d", &s, &t, &l, &r); ++s; ++t; ++l; ++r;
s = X1.find(s, l, 1); t = X2.find(t, r, 2);
puts(ask(root[X1.dfn[s] - 1], root[X1.dfn[s] + X1.siz[s] - 1], 1, n << 1, X2.dfn[t], X2.dfn[t] + X2.siz[t] - 1) ? "1" : "0");
}
return 0;
}