ACM-ICPC 2018 沈阳赛区网络预赛 J Ka Chang

Ka Chang

思路：

dfs序+树状数组+分块

先dfs处理好每个节点的时间戳

对于每一层，如果这一层的节点数小于sqrt(n)，那么直接按照时间戳在树状数组上更新

如果这一层节点个数大于sqrt(n)，那么直接存一下这一层每个节点的大小（都是一样的），这样的层数不会超过sqrt(n)层

然后查询的时候先在树状数组查询答案，然后再遍历第二种层数，加到答案中

复杂度：n*sqrt(n)*log(n)

代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 1e5 + ;
vector<int>g[N];
int in[N], out[N], deep[N], n, tot = ;
LL val[N], bit[N];
vector<int>dfn[N];
vector<int>big;
void add(int x, int v) {
    while(x <= n) bit[x] += v, x += x&-x;
}
LL query(int x) {
    LL ans = ;
    while(x) ans += bit[x], x -= x&-x;
    return ans;
}
void dfs(int o, int u, int d) {
    deep[u] = d;
    in[u] = ++tot;
    dfn[d].pb(tot);
    for (int v:g[u]) {
        if(v != o) dfs(u, v, d+);
    }
    out[u] = tot;
}
int main() {
    int q, u, v, blo, ty, l, x;
    scanf("%d %d", &n, &q);
    blo = sqrt(n);
    for (int i = ; i < n; i++) {
        scanf("%d %d", &u, &v);
        g[u].pb(v);
        g[v].pb(u);
    }
    dfs(, , );
    for(int i = ; i < n; i++) {
        if(dfn[i].size() > blo) big.pb(i);
    }
    while(q--) {
        scanf("%d", &ty);
        if(ty == ) {
            scanf("%d %d", &l, &x);
            if(dfn[l].size() <= blo) {
                for (int i = ; i < dfn[l].size(); i++) {
                    add(dfn[l][i], x);
                }
            }
            else val[l] += x;
        }
        else {
            scanf("%d", &x);
            LL ans = query(out[x]) - query(in[x]-);
            for (int i = ; i < big.size(); i++) {
                int d = big[i];
                int t = upper_bound(dfn[d].begin(), dfn[d].end(), out[x]) - lower_bound(dfn[d].begin(), dfn[d].end(), in[x]);
                ans += 1LL * t * val[d];
            }
            printf("%lld\n", ans);
        }
    }
    return ;
}