Portal
Title description
There are n snack machines in Baidu Science Park, and the snack machines are connected to each other through n−1 roads. Each snack machine has a value v, which represents the value of providing snacks for Xiaodu Bear.
As snacks are frequently consumed and replenished, the value v of the snack machine will change from time to time. Xiaodu Xiong can only start from the snack machine numbered 0, and each snack machine passes through at most once. In addition, Xiaodu Xiong will have a preference for the snacks of a certain snack machine and requires that the snack machine must be on the route.
Plan a route for Xiaodu Bear to maximize the total value of the route.
analysis
For this question, we can convert the tree structure into a chain structure to maintain the distance from the following node to each node, and then use line segments to maintain the maximum value of the interval, and modify the value of a point. In fact, this point and Modification of the subtree at this point is also converted into interval modification, so we can use dfs to convert this tree into a chain, and then use the line segment tree to perform interval query and modification
Code
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <queue>
#include <cstring>
#define debug(x) cout<<#x<<":"<<x<<endl;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
const ll INF = 0x3f3f3f3f3f3f;
const int N = 100010;
int h[N],ne[N * 2],e[N * 2],idx;
ll d[N],s[N];
int in[N],out[N];
int n,m;
int cnt;
struct Node{
int l,r;
ll sum,add;
}tr[N << 2];
void add(int x,int y){
ne[idx] = h[x],e[idx] = y,h[x] = idx++;
}
void dfs(int u,int fa,ll pp){
in[u] = ++cnt;
d[cnt] = pp;
for(int i = h[u];~i;i = ne[i]){
int j = e[i];
if(j == fa) continue;
dfs(j,u,pp + s[j]);
}
out[u] = cnt;
}
void pushup(int u){
tr[u].sum = max(tr[u << 1].sum,tr[u << 1 | 1].sum);
}
void down(int u){
auto &f = tr[u],&l = tr[u << 1],&r = tr[u << 1 | 1];
if(f.add){
l.add += f.add,l.sum += f.add;
r.add += f.add,r.sum += f.add;
f.add = 0;
}
}
void build(int u,int l,int r){
tr[u].l = l,tr[u].r = r,tr[u].add = 0;
if(l == r){
tr[u].sum = d[l];
return;
}
int mid = l + r >> 1;
build(u << 1,l,mid);
build(u << 1 | 1,mid + 1,r);
pushup(u);
}
void modify(int u,int l,int r,ll x){
if(tr[u].l >= l && tr[u].r <= r){
tr[u].sum += x;
tr[u].add += x;
return;
}
down(u);
int mid = tr[u].l + tr[u].r >> 1;
if(l <= mid) modify(u << 1,l,r,x);
if(r > mid) modify(u << 1 | 1,l,r,x);
pushup(u);
}
ll query(int u,int l,int r){
if(tr[u].l >= l && tr[u].r <= r){
return tr[u].sum;
}
down(u);
int mid = tr[u].l + tr[u].r >> 1;
ll ans = -INF;
if(l <= mid) ans = query(u << 1,l,r);
if(r > mid) ans = max(query(u << 1 | 1,l,r),ans);
return ans;
}
int main(){
int t;
scanf("%d",&t);
for(int ppp = 1;ppp <= t;ppp++){
memset(h,-1,sizeof h);
idx = cnt = 0;
scanf("%d%d",&n,&m);
printf("Case #%d:\n",ppp);
for(int i = 1;i < n;i++){
int x,y;
scanf("%d%d",&x,&y);
add(x,y),add(y,x);
}
for(int i = 0;i < n;i++) scanf("%lld",&s[i]);
dfs(0,-1,s[0]);
build(1,1,n);
while(m--){
int op,x,y;
scanf("%d%d",&op,&x);
if(!op){
scanf("%d",&y);
ll add = 1ll * y - s[x];
modify(1,in[x],out[x],add);
s[x] = y;
}
else printf("%lld\n",query(1,in[x],out[x]));
}
}
}