[题解]AT_abc298_h [ABC298Ex] Sum of Min of Length

模拟赛怒码 7KB 错解，赛后 10min AC。

思路

首先观察 $l,r$ 不同的关系对于结果的构成有什么影响，记 $gf = LCA(l,r)$。

1. $l = r$。很显然，答案就是以 $l$ 为根的节点的深度和。

2. $gf \neq l \wedge gf \neq r$。在 $l$ 子树中的节点在式子中一定会取 $dist(i,l)$，在 $r$ 子树中的节点在式子中一定会取 $dist(i,r)$。如果 $d_l \leq d_r$，则在 $gf$ 之上的节点（包括 $gf$ 除开 $l,r$ 所在的子树）；否则是同理的。其次对于 $l \rightsquigarrow r$ 这段路径，长度为 $len$，则前 $\frac{len}{2}$ 个节点及其不在链上的子树归 $l$，后 $\frac{len}{2}$ 个节点及其不在链上的子树归 $r$。

3. $gf = l \vee gf = r$。与情况 2 同理，除了在处理 $gf$ 子树的时候需要除去包含链的子树。

发现这些贡献可以离线下来换根。发现需要动态求子树深度和，所以用线段树换根即可。注意判断中点与需要处理的点的位置关系。

Code

#include <bits/stdc++.h>
#define re register
#define int long long

using namespace std;

const int N = 2e5 + 10,M = 4e5 + 10,K = 24,inf = 1e9 + 10;
int n,q,ans[N];
int num,id[N],d[N],sz[N],val[N];
int lg[N],f[N][K];
int idx,h[N],ne[M],e[M];

struct Query{
    int u,id;
    int gf,midx,midy;
};
vector<Query> Q[N];

inline int read(){
    int r = 0,w = 1;
    char c = getchar();
    while (c < '0' || c > '9'){
        if (c == '-') w = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9'){
        r = (r << 3) + (r << 1) + (c ^ 48);
        c = getchar();
    }
    return r * w;
}

inline void add(int a,int b){
    ne[idx] = h[a];
    e[idx] = b;
    h[a] = idx++;
}

struct seg{
    #define ls(u) (u << 1)
    #define rs(u) (u << 1 | 1)

    struct node{
        int l,r;
        int sum,tag;
    }tr[N << 2];

    inline void calc(int u,int k){
        tr[u].sum += k * (tr[u].r - tr[u].l + 1); tr[u].tag += k;
    }

    inline void pushup(int u){
        tr[u].sum = tr[ls(u)].sum + tr[rs(u)].sum;
    }

    inline void pushdown(int u){
        if (tr[u].tag){
            calc(ls(u),tr[u].tag); calc(rs(u),tr[u].tag);
            tr[u].tag = 0;
        }
    }

    inline void build(int u,int l,int r){
        tr[u] = {l,r};
        if (l == r) return tr[u].sum = val[l] - 1,void();
        int mid = l + r >> 1;
        build(ls(u),l,mid); build(rs(u),mid + 1,r);
        pushup(u);
    }

    inline void modify(int u,int l,int r,int k){
        if (l <= tr[u].l && tr[u].r <= r) return calc(u,k);
        pushdown(u);
        int mid = tr[u].l + tr[u].r >> 1;
        if (l <= mid) modify(ls(u),l,r,k);
        if (r > mid) modify(rs(u),l,r,k);
        pushup(u);
    }

    inline int query(int u,int l,int r){
        if (l <= tr[u].l && tr[u].r <= r) return tr[u].sum;
        pushdown(u);
        int res = 0;
        int mid = tr[u].l + tr[u].r >> 1;
        if (l <= mid) res += query(ls(u),l,r);
        if (r > mid) res += query(rs(u),l,r);
        return res;
    }

    #undef ls
    #undef rs
}T;

inline void get(int u,int fa){
    sz[u] = 1; f[u][0] = fa;
    val[id[u] = ++num] = d[u] = d[fa] + 1;
    for (re int i = 1;i <= lg[d[u]];i++) f[u][i] = f[f[u][i - 1]][i - 1];
    for (re int i = h[u];~i;i = ne[i]){
        int j = e[i];
        if (j == fa) continue;
        get(j,u); sz[u] += sz[j];
    }
}

inline int lca(int x,int y){
    while (d[x] != d[y]){
        if (d[x] < d[y]) swap(x,y);
        x = f[x][lg[d[x] - d[y]]];
    }
    if (x == y) return x;
    for (re int i = lg[d[x]];~i;i--){
        if (f[x][i] != f[y][i]) x = f[x][i],y = f[y][i];
    }
    return f[x][0];
}

inline bool check(int u,int v){
    if (d[v] > d[u]) return false;
    while (d[u] != d[v]) u = f[u][lg[d[u] - d[v]]];
    return (u == v);
}

inline void dfs(int u,int fa){
    for (auto p:Q[u]){
        int res = 0;
        int v = p.u,gf = p.gf;
        int midx = p.midx,midy = p.midy;
        if (u == v) ans[p.id] = T.tr[1].sum;
        else if (check(u,midx) && !check(v,midx)) res = T.query(1,id[midx],id[midx] + sz[midx] - 1);
        else res = T.tr[1].sum - T.query(1,id[midy],id[midy] + sz[midy] - 1);
        ans[p.id] += res;
    }
    for (re int i = h[u];~i;i = ne[i]){
        int j = e[i];
        if (j == fa) continue;
        T.modify(1,1,n,1); T.modify(1,id[j],id[j] + sz[j] - 1,-2);
        dfs(j,u);
        T.modify(1,1,n,-1); T.modify(1,id[j],id[j] + sz[j] - 1,2);
    }
}

signed main(){
    memset(h,-1,sizeof(h));
    n = read();
    for (re int i = 2;i <= n;i++) lg[i] = lg[i >> 1] + 1;
    for (re int i = 1;i < n;i++){
        int a,b; a = read(),b = read();
        add(a,b); add(b,a);
    }
    get(1,0); T.build(1,1,n);
    q = read();
    for (re int i = 1;i <= q;i++){
        int x,y;
        int gf = lca(x = read(),y = read());
        int len = d[x] + d[y] - 2 * d[gf] - 1;
        int mid = len / 2,midx,midy;
        int u = x,v = y;
        if (d[u] <= d[v]){
            for (re int j = 20;~j;j--){
                if ((1ll << j) <= mid){
                    v = f[v][j]; mid -= (1ll << j);
                }
            }
            midx = f[v][0],midy = v;
        }
        else{
            for (re int j = 20;~j;j--){
                if ((1ll << j) <= mid){
                    u = f[u][j]; mid -= (1ll << j);
                }
            }
            midx = u,midy = f[u][0];
        }
        Q[x].push_back({y,i,gf,midx,midy});
        Q[y].push_back({x,i,gf,midy,midx});
    }
    dfs(1,0);
    for (re int i = 1;i <= q;i++) printf("%lld\n",ans[i]);
    return 0;
}