洛谷 P3178 [HAOI2015]树上操作
作者:互联网
P3178 [HAOI2015]树上操作
Solution
树链剖分
树链剖分板子题,比板子还板子
关于树链剖分我就不多说了,如果又不会的话可以看我的博客 浅谈树链剖分
回归正题,我们发现题目只要求单点加,子树加,以及查询一点到根节点路径和。
单点加不就是区间加把左右端点改成那个点吗?
子树加不就是板子吗??
查询一点到根节点路径和不就是路径查询简化版吗???
综上所述,这道题是板子的板子
那么我们这道题就已经做完了,只需要把板子里多余的函数删掉就好啦。
别忘了开 \(long \ long\)
完整代码
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define ll long long
#define N 500010
#define ls rt << 1
#define rs rt << 1 | 1
#define INF 0x3f3f3f3f
using namespace std;
struct node{
ll v, nxt;
}edge[N << 1];
ll head[N], tot;
ll n, m;
ll w[N], tw[N];
ll dep[N], siz[N], fa[N], son[N];
ll top[N], id[N], cnt;
ll a[N << 2], lazy[N << 2];
inline ll read(){
ll x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
while(ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return f * x;
}
inline void add(ll x, ll y){
edge[++tot] = (node){y, head[x]};
head[x] = tot;
}
void dfs1(ll x, ll f){
fa[x] = f;
dep[x] = dep[f] + 1;
siz[x] = 1;
for(ll i = head[x]; i; i = edge[i].nxt){
ll y = edge[i].v;
if(y == f) continue;
dfs1(y, x);
siz[x] += siz[y];
if(!son[x] || siz[y] > siz[son[x]])
son[x] = y;
}
}
void dfs2(ll x, ll topfa){
top[x] = topfa;
id[x] = ++cnt;
tw[cnt] = w[x];
if(!son[x]) return;
dfs2(son[x], topfa);
for(ll i = head[x]; i; i = edge[i].nxt){
ll y = edge[i].v;
if(y == fa[x] || y == son[x]) continue;
dfs2(y, y);
}
}
inline void pushup(ll rt){
a[rt] = a[ls] + a[rs];
}
inline void pushdown(ll l, ll r, ll rt){
if(lazy[rt]){
ll mid = (l + r) >> 1;
a[ls] += lazy[rt] * (mid - l + 1);
a[rs] += lazy[rt] * (r - mid);
lazy[ls] += lazy[rt];
lazy[rs] += lazy[rt];
lazy[rt] = 0;
}
}
void build(ll l, ll r, ll rt){
if(l == r){
a[rt] = tw[l];
return;
}
ll mid = (l + r) >> 1;
build(l, mid, ls);
build(mid + 1, r, rs);
pushup(rt);
}
void update(ll L, ll R, ll k, ll l, ll r, ll rt){
if(L <= l && r <= R){
a[rt] += k * (r - l + 1);
lazy[rt] += k;
return;
}
pushdown(l, r, rt);
ll mid = (l + r) >> 1;
if(L <= mid) update(L, R, k, l, mid, ls);
if(R > mid) update(L, R, k, mid + 1, r, rs);
pushup(rt);
}
ll query(ll L, ll R, ll l, ll r, ll rt){
if(L <= l && r <= R)
return a[rt];
pushdown(l, r, rt);
ll mid = (l + r) >> 1;
ll res = 0;
if(L <= mid) res += query(L, R, l, mid, ls);
if(R > mid) res += query(L, R, mid + 1, r, rs);
return res;
}
void update_Son(ll x, ll k){
update(id[x], id[x] + siz[x] - 1, k ,1, n, 1);
}
ll query_Range(ll x, ll y){
ll res = 0;
while(top[x] != top[y]){
if(dep[top[x]] < dep[top[y]]) swap(x, y);
res += query(id[top[x]], id[x], 1, n, 1);
x = fa[top[x]];
}
if(dep[x] > dep[y]) swap(x, y);
res += query(id[x], id[y], 1, n, 1);
return res;
}
signed main(){
n = read(), m =read();
for(ll i = 1; i <= n; i++)
w[i] = read();
for(ll i = 1; i < n; i++){
ll u, v;
u = read(), v = read();
add(u, v), add(v, u);
}
dfs1(1, 0);
dfs2(1, 1);
build(1, n, 1);
while(m--){
ll op, x, k;
op = read();
if(op == 1){
x = read(), k = read();
update(id[x], id[x], k, 1, n, 1);
}else if(op == 2){
x = read(), k = read();
update_Son(x, k);
}else{
x = read();
printf("%lld\n", query_Range(1, x));
}
}
return 0;
}
标签:P3178,rt,lazy,洛谷,res,ll,mid,HAOI2015,top 来源: https://www.cnblogs.com/xixike/p/15108371.html