2019icpc上海现场赛 F题- A Simple Problem On A Tree

题意：一棵 (n) 个节点树，有三种对 ((u,v)) 路径上的点的权的操作和一个询问。
操作1：将 ((u,v)) 路径上的点的权置为 (w)。
操作2：将 ((u,v)) 路径上的点的权加上 (w)。
操作3：将 ((u,v)) 路径上的点的权乘上 (w)。
询问：求 ((u,v)) 路径上的点的权的立方和。

题解：裸树剖+线段树。
维护立方和：设区间和、平方和、立方和分别为 (a,a^2,a^3)。当加上一个数 (w)，和为 (a + len_{区间长} imes w)，平方和为 ((a+w)^2)，立方和为 ((a+w)^3)。
将平方和展开即 (a^2+2aw+w^2)，若区间和、平方和、立方和分别用 (sum_1[rt],sum_2[rt],sum_3[rt]) （这些值暂时还没有加上(w)）维护，
那么平方和 ((a+w)^2=sum_2[rt]+2 imes sum_1[rt] imes w+w imes w imes len_{区间长})。
同理，立方和即 (a^3+3(a^2w+aw^2)+w^3)，那么立方和 ((a+w)^3=sum_3[rt]+3 imes(sum_2[rt] imes w+sum_1[rt] imes w^2)+w^3 imes len_{区间长})。
所以这样就可以在线段树pushdown的时候维护立方和了，注意要先更新立方和再更新平方和再更新区间和。以及注意标记之间的先后顺序即可。

#include <bits/stdc++.h>
using namespace std;
  
#define debug(x) cerr << #x << " is " << x << '
';
typedef long long LL;
 
#define int long long 

const int N = 1e5 + 5;
const int P = 1e9 + 7;
 
int n, q, opt, x, y, z, cas;
LL sum1[N << 2], sum2[N << 2], sum3[N << 2], tag[N << 2], mul[N << 2], add[N << 2];
int cnt, tot, w[N], nw[N], id[N], top[N], dep[N], pre[N], son[N], siz[N], head[N];
 
struct Graph {
  int v, next;
} edge[N << 1];
 
void addedge(int u, int v) {
  edge[++cnt].v = v;
  edge[cnt].next = head[u];
  head[u] = cnt;
}
 
void init() {
  tot = cnt = 0;
  memset(head, 0, sizeof head);
  memset(siz, 0, sizeof siz);
  memset(nw, 0, sizeof nw);
  memset(top, 0, sizeof top);
  memset(pre, 0, sizeof pre);
  memset(son, 0, sizeof son);
  memset(dep, 0, sizeof dep);
  memset(id, 0, sizeof id);
}
 
#define ls rt << 1
#define rs rt << 1 | 1
 
void pushup(int rt) {
  sum1[rt] = (sum1[ls] + sum1[rs]) % P;
  sum2[rt] = (sum2[ls] + sum2[rs]) % P;
  sum3[rt] = (sum3[ls] + sum3[rs]) % P;
}
 
void downMul(int rt, int z) {
  sum3[rt] = sum3[rt] * z % P * z % P * z % P;
  sum2[rt] = sum2[rt] * z % P * z % P;
  sum1[rt] = sum1[rt] * z % P;
  mul[rt] = mul[rt] * z % P;
  add[rt] = add[rt] * z % P;
}
 
void downAdd(int rt, int len, int z) {
  sum3[rt] = (1LL * sum3[rt] % P + 3LL * (sum2[rt] * z % P + 1LL * sum1[rt] * z % P * z % P) % P + 1LL * len * z % P * z % P * z % P) % P;
  sum2[rt] = (1LL * sum2[rt] % P + 2LL * sum1[rt] % P * z % P + 1LL * len * z % P * z % P) % P;
  sum1[rt] = (sum1[rt] + 1LL * len * z % P) % P;
  add[rt] = (add[rt] + z) % P;
}
 
void pushdown(int rt, int l, int r, int mid) {
  if (tag[rt] == 1) {
    sum1[ls] = sum2[ls] = sum3[ls] = 0;
    sum1[rs] = sum2[rs] = sum3[rs] = 0;
    mul[ls] = mul[rs] = 1;
    add[ls] = add[rs] = 0;
    tag[ls] = tag[rs] = 1;
    tag[rt] = 0;
  }
  if (mul[rt] != 1) {
    downMul(ls, mul[rt]);
    downMul(rs, mul[rt]);
    mul[rt] = 1;
  }
  if (add[rt] != 0) {
    downAdd(ls, mid - l + 1, add[rt]);
    downAdd(rs, r - mid, add[rt]);
    add[rt] = 0;
  }
}
 
void build(int rt, int l, int r) {
  add[rt] = tag[rt] = 0;
  mul[rt] = 1;
  if (l == r) {
    sum1[rt] = nw[l] % P;
    sum2[rt] = nw[l] * nw[l] % P;
    sum3[rt] = sum2[rt] * nw[l] % P;
    return ;
  }
  int mid = l + r >> 1;
  build(ls, l, mid);
  build(rs, mid + 1, r);
  pushup(rt);
}
 
void updateAdd(int rt, int l, int r, int x, int y, int z) {
  if (x <= l && r <= y) {
    sum3[rt] = (1LL * sum3[rt] % P + 3LL * (sum2[rt] * z % P + 1LL * sum1[rt] * z % P * z % P) % P + 1LL * (r - l + 1) * z % P * z % P * z % P) % P;
    sum2[rt] = (1LL * sum2[rt] % P + 2LL * sum1[rt] % P * z % P + 1LL * (r - l + 1) * z % P * z % P) % P;
    sum1[rt] = (sum1[rt] + 1LL * (r - l + 1) * z % P) % P;
    add[rt] = (add[rt] + z) % P;
    return ;
  }
  int mid = l + r >> 1;
  pushdown(rt, l, r, mid);
  if (x <= mid) updateAdd(ls, l, mid, x, y, z);
  if (y > mid) updateAdd(rs, mid + 1, r, x, y, z);
  pushup(rt);
}
 
void updateMul(int rt, int l, int r, int x, int y, int z) {
  if (x <= l && r <= y) {
    sum3[rt] = sum3[rt] * z % P * z % P * z % P;
    sum2[rt] = sum2[rt] * z % P * z % P;
    sum1[rt] = sum1[rt] * z % P;
    mul[rt] = mul[rt] * z % P;
    add[rt] = add[rt] * z % P;
    return ;
  }
  int mid = l + r >> 1;
  pushdown(rt, l, r, mid);
  if (x <= mid) updateMul(ls, l, mid, x, y, z);
  if (y > mid) updateMul(rs, mid + 1, r, x, y, z);
  pushup(rt);
}
 
void updateTag(int rt, int l, int r, int x, int y, int z) {
  if (x <= l && r <= y) {
    sum1[rt] = 1LL * (r - l + 1) * z % P;
    sum2[rt] = 1LL * (r - l + 1) * z % P * z % P;
    sum3[rt] = 1LL * (r - l + 1) * z % P * z % P * z % P;
    tag[rt] = 1, mul[rt] = 1, add[rt] = z;
    return ;
  }
  int mid = l + r >> 1;
  pushdown(rt, l, r, mid);
  if (x <= mid) updateTag(ls, l, mid, x, y, z);
  if (y > mid) updateTag(rs, mid + 1, r, x, y, z);
  pushup(rt);
}
 
LL query(int rt, int l, int r, int x, int y) {
  if (x <= l && r <= y) {
    return sum3[rt];
  }
  int mid = l + r >> 1;
  LL ans = 0;
  pushdown(rt, l, r, mid);
  if (x <= mid) ans = (ans + query(ls, l, mid, x, y)) % P;
  if (y > mid) ans = (ans + query(rs, mid + 1, r, x, y)) % P;
  return ans;
}
 
void dfs1(int u, int f, int d) {
  pre[u] = f, dep[u] = d, siz[u] = 1;
  int mx = -1;
  for (int i = head[u]; i; i = edge[i].next) {
    int v = edge[i].v;
    if (v == f) continue;
    dfs1(v, u, d + 1);
    siz[u] += siz[v];
    if (siz[v] > mx) mx = siz[v], son[u] = v;
   }
}
 
void dfs2(int u, int topf) {
  id[u] = ++tot, nw[tot] = w[u], top[u] = topf;
  if (!son[u]) return ;
  dfs2(son[u], topf);
  for (int i = head[u]; i; i = edge[i].next) {
    int v = edge[i].v;
    if (v == pre[u] || v == son[u]) continue;
    dfs2(v, v);
  }
}
 
void treeAdd(int u, int v, int z) {
  while (top[u] ^ top[v]) {
    if (dep[top[u]] < dep[top[v]]) swap(u, v);
    updateAdd(1, 1, n, id[top[u]], id[u], z);
    u = pre[top[u]];
  }
  if (dep[u] > dep[v]) swap(u, v);
  updateAdd(1, 1, n, id[u], id[v], z);
}
 
void treeMul(int u, int v, int z) {
  while (top[u] ^ top[v]) {
    if (dep[top[u]] < dep[top[v]]) swap(u, v);
    updateMul(1, 1, n, id[top[u]], id[u], z);
    u = pre[top[u]];
  }
  if (dep[u] > dep[v]) swap(u, v);
  updateMul(1, 1, n, id[u], id[v], z);
}
 
void treeTag(int u, int v, int z) {
  while (top[u] ^ top[v]) {
    if (dep[top[u]] < dep[top[v]]) swap(u, v);
    updateTag(1, 1, n, id[top[u]], id[u], z);
    u = pre[top[u]];
  }
  if (dep[u] > dep[v]) swap(u, v);
  updateTag(1, 1, n, id[u], id[v], z);
}
 
LL TreeQuery(int u, int v) {
  LL ans = 0;
  while (top[u] ^ top[v]) {
    if (dep[top[u]] < dep[top[v]]) swap(u, v);
    ans = (ans + query(1, 1, n, id[top[u]], id[u])) % P;
    u = pre[top[u]];
  }
  if (dep[u] > dep[v]) swap(u, v);
  ans = (ans + query(1, 1, n, id[u], id[v])) % P;
  return ans;
}
 
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> n;
    init();
    for (int i = 1, u, v; i <= n - 1; i++) {
      cin >> u >> v;
      addedge(u, v), addedge(v, u);
    }
    for (int i = 1; i <= n; i++) cin >> w[i];
    dfs1(1, -1, 1);
    dfs2(1, 1);
    build(1, 1, n);
    cin >> q;
    cout << "Case #" << ++cas << ":" << '
';
    while (q--) {
      cin >> opt;
      if (opt == 1) {
        cin >> x >> y >> z;
        treeTag(x, y, z);
      } else if (opt == 2) {
        cin >> x >> y >> z;
        treeAdd(x, y, z);
      } else if (opt == 3) {
        cin >> x >> y >> z;
        treeMul(x, y, z);
      } else if (opt == 4) {
        cin >> x >> y;
        cout << TreeQuery(x, y) << '
';
      }
    }
  }
  return 0;
}