Bzoj 3730 震波 动态点分治

Bzoj 3730 震波

题解：

和在线的边分治差不多。 就是将每层都信息都存下来。

然后对于每一层记录上一层的重心是哪个。

 

对于求和的话， 从自己的那层出发，然后暴力往上爬， 然后计算答案。

对于修改来说，也暴力的往上爬，对于每层所对应的信息来修改

 

用树状数组来统计同一层、不同深度的前缀和。

本来想用线段树，然后TLE了，非常卡。

然后用另一颗树状数组来容斥前缀和。

 

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 5e5;
int head[N], to[N<<1], nt[N<<1], tot, dtot;
int in[N], out[N], dfn[N], deep[N];
int Log[N];
struct ST {
    int dp[N][20], a[N];
    void init(int n) {
        for(int i = -(Log[0]=-1); i <= n; i++)
            Log[i] = Log[i - 1] + ((i & (i - 1)) == 0);
        for(int i = 1; i <= n; ++i) dp[i][0] = a[i];
        for(int j = 1; j <= Log[n]; j++)
            for(int i = 1; i+(1<<j)-1 <= n; i++){
                int x = dp[i][j-1], y = dp[i+(1<<(j-1))][j-1];
                if(deep[x] < deep[y]) dp[i][j] = x;
                else dp[i][j] = y;
            }
    }
    inline int lca(int l, int r) {
        l = in[l], r = in[r];
        if(l > r) swap(l, r);
        int k = Log[r-l + 1];
        if(deep[dp[l][k]] < deep[dp[r-(1<<k)+1][k]]) return deep[dp[l][k]];
        return dp[r-(1<<k)+1][k];
    }
}st;
inline void add(int u, int v){
    to[tot] = v; nt[tot] = head[u]; head[u] = tot++;
}
void dfs(int o, int u){
    in[u] = ++dtot;
    st.a[dtot] = u;
    deep[u] = deep[o] + 1;
    for(int i = head[u]; ~i; i = nt[i]){
        int v = to[i];
        if(o ^ v){
            dfs(u, v);
            st.a[++dtot] = u;
        }
    }
}
inline int dis(int u, int v){
    return deep[u]+deep[v]-2*deep[st.lca(u, v)];
}
int w[N];
int a[N];
int atot;
int Prt[N][2], Pr[N][2];
vector<int> vc[N][2];
int vis[N], tsz;
int sz[N];
int rt, rtnum;
int fa[N];
void Get_root(int o, int u){
    sz[u] = 1;
    int Max = 0;
    for(int i = head[u]; ~i; i = nt[i]){
        int v = to[i];
        if(vis[v] || v == o) continue;
        Get_root(u, v);
        Max = max(Max, sz[v]);
        sz[u] += sz[v];
    }
    Max = max(Max, tsz - sz[u]);
    if(Max < rtnum){
        rt = u;
        rtnum = Max;
    }
}
int Maxdep;
void dfs(int g, int o, int u){
    sz[u] = 1;
    Maxdep = max(Maxdep, dis(g, u));
    for(int i = head[u]; ~i; i = nt[i]){
        int v = to[i];
        if(vis[v] || v == o) continue;
        dfs(g, u, v);
        sz[u] += sz[v];
    }
}
void Add(int g, int o, int u){
    a[dis(g, u)] += w[u];
    for(int i = head[u]; ~i ; i = nt[i]){
        int v = to[i];
        if(vis[v] || v == o) continue;
        Add(g, u, v);
    }
}
void Run(vector<int> & v){
    for(int i = 0; i <= Maxdep; ++i)
        v.pb(a[i]);
    for(int i = Maxdep; i >= 1; --i){
        int j = i;
        j += i & (-i);
        while(j <= Maxdep){
            v[j] += a[i];
            j += j & (-j);
        }
    }
}
void solve(int o, int u, int num){
    tsz = num;
    rtnum = num + 1;
    Get_root(0, u);
    fa[rt] = o;
    vis[rt] = 1;
    Maxdep = 0;
    dfs(rt, 0, rt);/// Find_Max_Deep
    for(int i = 0; i <= Maxdep; ++i)
        a[i] = 0;
    Add(rt, 0, rt);
    Run(vc[rt][0]);
    if(o){
        Maxdep = 0;
        dfs(o, 0, rt);
        for(int i = 0; i <= Maxdep; ++i)
        a[i] = 0;
        Add(o, 0, rt);
        Run(vc[rt][1]);
    }
    int nrt = rt;
    for(int i = head[nrt]; ~i; i = nt[i]){
        int v = to[i];
        if(vis[v]) continue;
        solve(nrt, v, sz[v]);
    }
}
void Updata(vector<int> &v, int x, int val){
    int lim = v.size();
    if(x < 0) return ;
    if(x == 0){
        v[0] += val;
        return ;
    }
    while(x < lim){
        v[x] += val;
        x += x & (-x);
    }
}
int Query(vector<int> &v, int k){
    int lim = v.size();
    k = min(k, lim-1);
    if(k < 0) return 0;
    int ret = v[0];
    while(k > 0){
        ret += v[k];
        k -= k & (-k);
    }
    return ret;
}
void Updata(int x, int val){
    for(int i = x; i; i = fa[i]){
        Updata(vc[i][0], dis(x,i), val);
        if(fa[i]) Updata(vc[i][1], dis(fa[i], x), val);
    }
}
int Query(int x, int k){
    int ret = 0;
    for(int i = x; i; i = fa[i]){
        ret += Query(vc[i][0], k-dis(x,i));
        if(fa[i]) ret -= Query(vc[i][1], k-dis(fa[i], x));
    }
    return ret;
}

struct FastIO {
    static const int S = 1310720;
    int wpos;
    char wbuf[S];
    FastIO() : wpos(0) { }
    inline int xchar() {
        static char buf[S];
        static int len = 0, pos = 0;
        if (pos == len) pos = 0, len = fread(buf, 1, S, stdin);
        if (pos == len) return -1;
        return buf[pos++];
    }
    inline int xint() {
        int c = xchar(), x = 0, s = 1;
        while (c <= 32) c = xchar();
        if (c == '-') s = -1, c = xchar();
        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
        return x * s;
    }
    ~FastIO() {
        if (wpos) fwrite(wbuf, 1, wpos, stdout), wpos = 0;
    }
} io;
int main(){
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; ++i){
        w[i] = io.xint();
    }
    memset(head, -1, sizeof head);
    for(int i = 1, u, v; i < n; ++i){
        u = io.xint(), v = io.xint();
        add(u, v);  add(v, u);
    }
    dfs(0, 1);
    st.init(dtot);
    solve(0, 1, n);
    int lastans = 0;
    int op, x, y;
    for(int i = 1; i <= m; ++i){
        op = io.xint(), x = io.xint(), y = io.xint();
        x ^= lastans, y ^= lastans;
        if(op){
            Updata(x, y-w[x]);
            w[x] = y;
        }
        else{
            lastans = Query(x,y);
            printf("%d\n", lastans);
        }
    }
    return 0;
}