Problem Portal
Portal1: Luogu
Description
As stated, it is known a containing \ (N \) nodes of the tree (and no communication ring), each node comprising a numerical value, needs to support the following operations:
Operation \ (1 \) : Format: 1 x y zshows a tree from x to y values of all nodes in the shortest path are nodes plus \ (Z \) ;
Operation \ (2 \) : Format: 2 x yindicates finding a tree node from x to y value on the shortest path and all the nodes;
Operation \ (3 \) : Format: 3 x zrepresents all nodes will be x values are within the subtree root node plus \ (Z \) ;
Operation \ (4 \) : Format: 4 xindicates finding all nodes within the x value is a root node of the subtree and.
Input
The first line contains \ (4 \) positive integer \ (N \) , \ (M \) , \ (R & lt \) , \ (P \) , each represent the number of tree nodes, the number of operations, root modulo the number and node number ( i.e., all of this output are modulo ).
The next line contains the \ (N \) non-negative integers, represents an initial value sequentially, on each node.
Next \ (N - 1 \) lines contains two integers \ (X \) , \ (Y \) , represents the point \ (X \) and the point \ (Y \) is connected there is an edge between (guaranteed acyclic and communication)
Next \ (M \) lines each comprising a number of positive integers, each row represents an operation, the following format:
Operation \ (1 \) :1 x y z
Operation \ (2 \) :2 x y
Operation \ (3 \) :3 x z
Operation \ (4 \) :4 x
Output
Comprising a plurality of output lines, respectively, sequentially illustrating each operation \ (2 \) or operation \ (4 \) The results obtained (p \ (P \) modulo).
Sample Input
5 5 2 24
7 3 7 8 0
1 2
1 5
3 1
4 1
3 4 2
3 2 2
4 5
1 5 1 3
2 1 3Sample Output
2
21Solution
Template tree chain split title.
Some concepts:
Son weight: a maximum of in each non-leaf node sons, son node to the root of the subtree of nodes for node weight son son point;
Light son: in a non-leaf, heavy non-son node;
Multiple edges: a father node connecting edge has a weight sons;
Light Edge: non-reentrant edge;
Heavy chain: Heavy adjacent edges connected together, the connection node of one heavy chain son called heavy chain.
dfs1Features:
Obtaining depth of each node;
Obtaining a parent node of each node;
Determined size of each non-leaf sub-tree node;
Find each weighing son of a non-leaf node number.
dfs2Features:
Each processing chain;
The new tag ID of each node;
Obtaining each of the top node where the chain;
The initial value of the node to the new number in the updated.
Code
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int MAXN = 2000005;
struct EDGE {
int u, v, nxt;
} edge[MAXN];
struct node {
int l, r, w, size, f;
} tree[MAXN];
int n, m, root, mod, cnt, num = 1, a[MAXN], b[MAXN], tot[MAXN], son[MAXN], top[MAXN], idx[MAXN], dep[MAXN], head[MAXN], father[MAXN];
inline void addedge(int u, int v) {
edge[num].u = u; edge[num].v = v; edge[num].nxt = head[u]; head[u] = num++;
}
//dep[i]表示i结点的深度
//father[i]表示i结点的父亲结点
//son[]表示重儿子的编号
inline int dfs1(int now, int f, int deep) {
dep[now] = deep;
father[now] = f;
tot[now] = 1;
int Maxson = -1;
for (int i = head[now]; ~i; i = edge[i].nxt) {
if (edge[i].v == f) continue;
tot[now] += dfs1(edge[i].v, now, deep + 1);
if (tot[edge[i].v] > Maxson) {
Maxson = tot[edge[i].v];
son[now] = edge[i].v;
}
}
return tot[now];
}
inline void dfs2(int now, int topf) {
idx[now] = ++cnt;
a[cnt] = b[now];
top[now] = topf;
if (!son[now]) return ;
dfs2(son[now], topf);
for (int i = head[now]; ~i; i = edge[i].nxt)
if (!idx[edge[i].v]) dfs2(edge[i].v, edge[i].v);
}
inline void pushup(int root) {
tree[root].w = (tree[root << 1].w + tree[root << 1 | 1].w + mod) % mod;
}
inline void build(int root, int l, int r) {
tree[root].l = l; tree[root].r = r; tree[root].size = r - l + 1;
if (l == r) {
tree[root].w = a[l];
return ;
}
int mid = l + r >> 1;
build(root << 1, l, mid);
build(root << 1 | 1, mid + 1, r);
pushup(root);
}
inline void pushdown(int root) {
if (!tree[root].f) return ;
tree[root << 1].w = (tree[root << 1].w + tree[root << 1].size * tree[root].f) % mod;
tree[root << 1 | 1].w = (tree[root << 1 | 1].w + tree[root << 1 | 1].size * tree[root].f) % mod;
tree[root << 1].f = (tree[root << 1].f + tree[root].f) % mod;
tree[root << 1 | 1].f = (tree[root << 1 | 1].f + tree[root].f) % mod;
tree[root].f = 0;
}
inline void update_add(int root, int ansl, int ansr, int val) {
if (ansl <= tree[root].l && tree[root].r <= ansr) {
tree[root].w += tree[root].size * val;
tree[root].f += val;
return ;
}
pushdown(root);
int mid = tree[root].l + tree[root].r >> 1;
if (ansl <= mid) update_add(root << 1, ansl, ansr, val);
if (ansr > mid) update_add(root << 1 | 1, ansl, ansr, val);
pushup(root);
}
//线段树操作
inline void tree_add(int x, int y, int val) {
while (top[x] != top[y]) {
if (dep[top[x]] < dep[top[y]]) swap(x, y);
update_add(1, idx[top[x]], idx[x], val);
x = father[top[x]];
}
if (dep[x] > dep[y]) swap(x, y);
update_add(1, idx[x], idx[y], val);
}
inline int query_sum(int root, int ansl, int ansr) {
int ret = 0;
if (ansl <= tree[root].l && tree[root].r <= ansr) return tree[root].w;
pushdown(root);
int mid = tree[root].l + tree[root].r >> 1;
if (ansl <= mid) ret = (ret + query_sum(root << 1, ansl, ansr)) % mod;
if (ansr > mid) ret = (ret + query_sum(root << 1 | 1, ansl, ansr)) % mod;
return ret;
}
inline void tree_sum(int x, int y) {
int ret = 0;
while (top[x] != top[y]) {
if (dep[top[x]] < dep[top[y]]) swap(x, y);
ret = (ret + query_sum(1, idx[top[x]], idx[x])) % mod;
x = father[top[x]];
}
if (dep[x] > dep[y]) swap(x, y);
ret = (ret + query_sum(1, idx[x], idx[y])) % mod;
printf("%d\n", ret);
}
int main() {
memset(head, -1, sizeof(head));
scanf("%d%d%d%d", &n, &m, &root, &mod);
for (int i = 1; i <= n; i++)
scanf("%d", &b[i]);
for (int i = 1; i < n; i++) {
int x, y;
scanf("%d%d", &x, &y);
addedge(x, y);
addedge(y, x);
}
dfs1(root, 0, 1);
dfs2(root, root);
build(1, 1, n);
while (m--) {
int opt, x, y, val;
scanf("%d", &opt);
if (opt == 1) {
scanf("%d%d%d", &x, &y, &val);
val %= mod;
tree_add(x, y, val);
} else
if (opt == 2) {
scanf("%d%d", &x, &y);
tree_sum(x, y);
} else
if (opt == 3) {
scanf("%d%d", &x, &val);
update_add(1, idx[x], idx[x] + tot[x] - 1, val % mod);
} else {
scanf("%d", &x);
printf("%d\n", query_sum(1, idx[x], idx[x] + tot[x] - 1));
}
}
return 0;
}