E. Vasya and a Tree (前缀和思维) Educational Codeforces Round 54 (Rated for Div. 2)

传送门：http://codeforces.com/problemset/problem/1076/E

题目大意：

       一棵树，m次操作。每次操作将距离该点小于等于d的儿子点权加x,输出全部的点权。

题目思路：

       首先离线查询，落实到每个点上，然后dfs的时候先把这些查询，加到差分的前缀和数组上，然后往下走的时候要加上相应的差分数组，同理回溯的时候要减去差分值，差分数组也要讲之前加的减去。复杂度O（n+m）

        同时也可以树状数组维护前缀和，每个点单独查询。

差分数组做法：

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const ll MAXN =  3e5+5;
vector<ll>v[MAXN];
struct node
{
    ll d,x;
    node(){}
    node(ll a,ll b){
        d = a;x = b;
    }
}C[MAXN];
vector<node>g[MAXN];
ll sum[MAXN],ans[MAXN];
ll c[MAXN];
ll Max = 0;
void add(ll x,ll y)
{
    for( ;x <= MAXN ;x += x&-x)
        c[x] += y;
}
ll ask(ll x)
{
    ll ans = 0 ;
    for( ; x;x -= (x&-x))ans += c[x];
    return ans;
}
ll as = 0;
void dfs(ll x,ll fa,ll dep)
{
    ll len = v[x].size();
    ll len2 = g[x].size();
    for(ll i=0;i<len2;i++){
        node to = g[x][i];
        sum[dep] += to.x;
        sum[min(Max+1,dep+to.d+1)] -= to.x;
    }
    as += sum[dep];
    ans[x] = as;
    for(ll i=0;i<len;i++){
        ll to = v[x][i];
        if(to == fa)continue;
        dfs(to,x,dep+1);
    }
    as -= sum[dep];
    for(ll i=0;i<len2;i++){
        node to = g[x][i];
        sum[dep] -= to.x;
        sum[min(Max+1,dep+to.d+1)] += to.x;
    }
}
void dfs_d(ll x,ll fa,ll dep)
{
    Max = max(Max,dep);
    ll len = v[x].size();
    for(ll i=0;i<len;i++){
        ll to = v[x][i];
        if(to == fa)continue;
        dfs_d(to,x,dep+1);
    }
}
int main()
{
    ll n;
    scanf("%lld",&n);
    for(ll i=1;i<n;i++){
        ll a,b;
        scanf("%lld%lld",&a,&b);
        v[a].push_back(b);
        v[b].push_back(a);
    }
    ll m;
    scanf("%lld",&m);
    while(m--){
         ll d,nd,x;
         scanf("%lld%lld%lld",&nd,&d,&x);
         g[nd].push_back(node(d,x));
    }
    dfs_d(1,1,1);
    dfs(1,1,1);
    for(ll i=1;i<=n;i++){
        printf("%lld ",ans[i]);
    }
    printf("\n");
}

树状数组做法：

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const ll MAXN =  3e5+5;
vector<ll>v[MAXN];
struct node
{
    ll d,x;
    node(){}
    node(ll a,ll b){
        d = a;x = b;
    }
}C[MAXN];
vector<node>g[MAXN];
ll sum[MAXN],ans[MAXN];
ll c[MAXN];
void add(ll x,ll y)
{
    for( ;x <= MAXN ;x += x&-x)
        c[x] += y;
}
ll ask(ll x)
{
    ll ans = 0 ;
    for( ; x;x -= (x&-x))ans += c[x];
    return ans;
}
void dfs(ll x,ll fa,ll dep)
{
    ll len = v[x].size();
    ll len2 = g[x].size();
    for(ll i=0;i<len2;i++){
        node to = g[x][i];
        add(dep,to.x);
        add(dep+to.d+1,-to.x);
    }
    ans[x] = ask(dep);
    for(ll i=0;i<len;i++){
        ll to = v[x][i];
        if(to == fa)continue;
        dfs(to,x,dep+1);
    }
    for(ll i=0;i<len2;i++){
        node to = g[x][i];
        add(dep,-to.x);
        add(dep+to.d+1,to.x);
    }
}
int main()
{
    ll n;
    scanf("%lld",&n);
    for(ll i=1;i<n;i++){
        ll a,b;
        scanf("%lld%lld",&a,&b);
        v[a].push_back(b);
        v[b].push_back(a);
    }
    ll m;
    scanf("%lld",&m);
    while(m--){
         ll d,nd,x;
         scanf("%lld%lld%lld",&nd,&d,&x);
         g[nd].push_back(node(d,x));
    }
    dfs(1,1,1);
    for(ll i=1;i<=n;i++){
        printf("%lld ",ans[i]);
    }
    printf("\n");
}