树链剖分(模板)
作者:互联网
code
//操作 1: 格式: 1 x y z 表示将树从 x 到 y 结点最短路径上所有节点的值都加上 z。
//操作 2: 格式: 2 x y表示求树从 x 到 y 结点最短路径上所有节点的值之和。
//操作 3: 格式: 3 x z表示将以 x 为根节点的子树内所有节点值都加上 z。
//操作 4: 格式: 4 x表示求以 x 为根节点的子树内所有节点值之和
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ls(x) (x << 1)
#define rs(x) (x << 1 | 1)
int n, m, rt, tot, cnt, id[100005], top[100005], nw[100005], f[100005], dep[100005], sz[100005], mson[100005], a[100005], hd[100005], to[200005], nxt[200005];
ll mod;
struct node
{
int l, r;
ll sum, add;
}t[400005];
int read()
{
int x = 0, fl = 1; char ch = getchar();
while (ch < '0' || ch > '9') { if (ch == '-') fl = -1; ch = getchar();}
while (ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + ch - '0'; ch = getchar();}
return x * fl;
}
void add(int x, int y)
{
tot ++ ;
to[tot] = y;
nxt[tot] = hd[x];
hd[x] = tot;
return;
}
void push_up(int p)
{
t[p].sum = (t[ls(p)].sum + t[rs(p)].sum) % mod;
return;
}
void push_down(int p)
{
if (!t[p].add) return;
t[ls(p)].add = (t[ls(p)].add + t[p].add) % mod;
t[rs(p)].add = (t[rs(p)].add + t[p].add) % mod;
t[ls(p)].sum = (t[ls(p)].sum + 1ll * t[p].add * (t[ls(p)].r - t[ls(p)].l + 1) % mod) % mod;
t[rs(p)].sum = (t[rs(p)].sum + 1ll * t[p].add * (t[rs(p)].r - t[rs(p)].l + 1) % mod) % mod;
t[p].add = 0;
return;
}
void update(int p, int l0, int r0, int d)
{
if (l0 <= t[p].l && t[p].r <= r0)
{
t[p].add = (t[p].add + (ll)d) % mod;
t[p].sum = (t[p].sum + 1ll * (t[p].r - t[p].l + 1) * d % mod) % mod;
return;
}
push_down(p);
int mid = (t[p].l + t[p].r) >> 1;
if (l0 <= mid) update(ls(p), l0, r0, d);
if (r0 > mid) update(rs(p), l0, r0, d);
push_up(p);
return;
}
ll query(int p, int l0, int r0)
{
if (l0 <= t[p].l && t[p].r <= r0) return t[p].sum;
push_down(p);
int mid = (t[p].l + t[p].r) >> 1; ll res = 0ll;
if (l0 <= mid) res = (res + query(ls(p), l0, r0) % mod) % mod;
if (r0 > mid) res = (res + query(rs(p), l0, r0) % mod) % mod;
return res % mod;
}
void build(int p, int l0, int r0)
{
t[p].l = l0; t[p].r = r0;
if (l0 == r0)
{
t[p].sum = (ll)(nw[l0]);
return;
}
int mid = (l0 + r0) >> 1;
build(ls(p), l0, mid);
build(rs(p), mid + 1, r0);
push_up(p);
return;
}
void dfs1(int x, int fa)
{
sz[x] = 1;
int mx = -1;
for (int i = hd[x]; i; i = nxt[i])
{
int y = to[i];
if (y == fa) continue;
dep[y] = dep[x] + 1;
f[y] = x;
dfs1(y, x);
sz[x] += sz[y];
if (sz[y] > mx)
{
mx = sz[y];
mson[x] = y;
}
}
return;
}
void dfs2(int x, int tp)
{
id[x] = ++ cnt;
top[x] = tp;
nw[cnt] = a[x];
if (!mson[x]) return;
dfs2(mson[x], tp);
for (int i = hd[x]; i; i = nxt[i])
{
int y = to[i];
if (y == f[x] || y == mson[x]) continue;
dfs2(y, y);
}
return;
}
void q1()
{
int x = read(), y = read(), z = read();
while (top[x] != top[y])
{
if (dep[top[x]] < dep[top[y]]) swap(x, y);
update(1, id[top[x]], id[x], (ll)z);
x = f[top[x]];
}
if (dep[x] > dep[y]) swap(x, y);
update(1, id[x], id[y], (ll)z);
return;
}
void q2()
{
int x = read(), y = read();
ll res = 0ll;
while (top[x] != top[y])
{
if (dep[top[x]] < dep[top[y]]) swap(x, y);
res = (res + query(1, id[top[x]], id[x])) % mod;
x = f[top[x]];
}
if (dep[x] > dep[y]) swap(x, y);
res = (res + query(1, id[x], id[y])) % mod;
printf("%lld\n", res);
return;
}
void q3()
{
int x = read(), z = read();
update(1, id[x], id[x] + sz[x] - 1, (ll)z);
return;
}
void q4()
{
int x = read();
printf("%lld\n", query(1, id[x], id[x] + sz[x] - 1) % mod);
return;
}
int main()
{
n = read(); m = read(); rt = read(); mod = (ll)read();
for (int i = 1; i <= n; i ++ ) a[i] = read();
for (int i = 1; i <= n - 1; i ++ )
{
int x = read(), y = read();
add(x, y); add(y, x);
}
dfs1(rt, 0); dfs2(rt, rt);
build(1, 1, n);
while (m -- )
{
int opt = read();
if (opt == 1) q1();
else if (opt == 2) q2();
else if (opt == 3) q3();
else q4();
}
return 0;
}
标签:return,剖分,int,top,树链,read,l0,id,模板 来源: https://www.cnblogs.com/andysj/p/13948234.html