题面
题解
点分治大火题。。。
设白边数量为$a$，黑边为$b$，则$2min(a,b)\geq max(a,b)$
即$2a\geq b\;\&\&2b\geq a$
考虑点分治时如何统计答案：
$2(a_1 +a_2) \geq b_1 + b_2$
$\therefore 2a_1-b_1\geq b_2-2a_2$
另外一边同理
于是我们可以愉快地用$sort+BIT$统计答案了
但是路径有可能重复计算，可以套一个$CDQ$分治什么的来搞一下
代码
#include<cstdio>
#include<cstring>
#include<algorithm>
#define RG register
#define clear(x, y) memset(x, y, sizeof(x));
namespace IO
{
const int BUFSIZE = 1 << 20;
char ibuf[BUFSIZE], *is = ibuf, *it = ibuf;
inline char getchar() { if (is == it) it = (is = ibuf) + fread(ibuf, 1, BUFSIZE, stdin); return *is++; }
}
inline int read()
{
int data = 0, w = 1;
char ch = IO::getchar();
while(ch != '-' && (ch < '0' || ch > '9')) ch = IO::getchar();
if(ch == '-') w = -1, ch = IO::getchar();
while(ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = IO::getchar();
return data*w;
}
const int maxn(1e5 + 10), Mod(1e9 + 7);
inline int fastpow(int x, int y)
{
int ans = 1;
while(y)
{
if(y & 1) ans = 1ll * ans * x % Mod;
x = 1ll * x * x % Mod, y >>= 1;
}
return ans;
}
struct edge { int next, to, val, col; } e[maxn << 1];
int n, head[maxn], e_num, size[maxn], Size, min, root;
struct qry { int r, b, v, o; } stk[maxn], q[maxn << 2];
int top, cntq, cntup, csum[maxn << 2], cmul[maxn << 2], vis[maxn];
inline bool cmpr(const qry &a, const qry &b)
{ return a.r == b.r ? a.b < b.b : a.r < b.r; }
void clean(int x)
{
while(x <= (n << 2 | 1)) csum[x] = 0, cmul[x] = 1, x += x & -x;
}
void add(int x, int v)
{
while(x <= (n << 2 | 1))
++csum[x], cmul[x] = 1ll * cmul[x] * v % Mod, x += x & -x;
}
int query_sum(int x)
{
int ans = 0;
while(x) ans += csum[x], x -= x & -x;
return ans;
}
int query_mul(int x)
{
int ans = 1;
while(x) ans = 1ll * ans * cmul[x] % Mod, x -= x & -x;
return ans;
}
inline void add_edge(int from, int to, int val, int col)
{
e[++e_num] = (edge) {head[from], to, val, col};
head[from] = e_num;
}
void getRoot(int x, int fa)
{
size[x] = 1; int tot = 0;
for(RG int i = head[x]; i; i = e[i].next)
{
int to = e[i].to; if(vis[to] || to == fa) continue;
getRoot(to, x); tot = std::max(tot, size[to]);
size[x] += size[to];
}
tot = std::max(tot, Size - size[x]);
if(tot < min) min = tot, root = x;
}
void getVal(int x, int fa, int r, int b, int v)
{
stk[++top] = (qry) {r, b, v, 0};
for(RG int i = head[x]; i; i = e[i].next)
{
int to = e[i].to; if(vis[to] || to == fa) continue;
getVal(to, x, r + (!e[i].col), b + e[i].col, 1ll * v * e[i].val % Mod);
}
}
int ans = 1;
void CDQ(int l, int r)
{
if(l >= r) return;
int mid = (l + r) >> 1;
CDQ(l, mid); CDQ(mid + 1, r);
int j = l;
for(RG int i = mid + 1; i <= r; i++)
{
if(!q[i].o) continue;
while(q[j].r <= q[i].r && j <= mid)
{
if(!q[j].o) add(q[j].b, q[j].v);
++j;
}
ans = 1ll * ans * query_mul(q[i].b) % Mod *
fastpow(q[i].v, query_sum(q[i].b)) % Mod;
}
for(RG int i = l; i < j; i++) if(!q[i].o) clean(q[i].b);
std::inplace_merge(q + l, q + mid + 1, q + r + 1, cmpr);
}
void calc(int x)
{
int PLUS = n << 1 | 1;
cntq = cntup = 0;
for(RG int i = head[x]; i; i = e[i].next)
{
int to = e[i].to; if(vis[to]) continue;
top = 0; getVal(to, x, !e[i].col, e[i].col, e[i].val);
for(RG int j = 1; j <= top; j++)
{
int a = stk[j].r, b = stk[j].b, c = 2 * a - b, d = 2 * b - a;
q[++cntq] = (qry) {c + PLUS, d + PLUS, stk[j].v, 1};
}
for(RG int j = 1; j <= top; j++)
{
int a = stk[j].r, b = stk[j].b, c = 2 * a - b, d = 2 * b - a;
q[++cntq] = (qry) {-c + PLUS, -d + PLUS, stk[j].v, 0};
if(2 * std::min(a, b) >= std::max(a, b))
ans = 1ll * ans * stk[j].v % Mod;
}
}
CDQ(1, cntq);
}
void dfs(int x)
{
vis[x] = 1; calc(x);
for(RG int i = head[x]; i; i = e[i].next)
{
int to = e[i].to; if(vis[to]) continue;
Size = min = size[to]; getRoot(to, x); dfs(root);
}
}
int main()
{
Size = min = n = read();
for(int i = 1, a, b, c, d; i < n; i++)
a = read(), b = read(), c = read(), d = read(),
add_edge(a, b, c, d), add_edge(b, a, c, d);
for(RG int i = 1; i <= (n << 2 | 1); i++) csum[i] = 0, cmul[i] = 1;
getRoot(1, 0); dfs(root);
printf("%d\n", ans);
return 0;
}