https://www.luogu.org/problemnew/show/P1600

亂寫的暴力，這道題暴力寫個60還是比較簡單的

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <cstring>

using namespace std;
const int N = 600001;
const int oo = 99999999
 
;

#define lson jd << 1
#define rson jd << 1 | 1

#define yxy getchar()
#define one_ n <= 993
#define two_ n == 99994
#define three_ n == 99995

int head[N], pre[N], dis[N], tim[N], Answer[N], Askl[N], Askr[N];
bool vis[N];
int n, m, now = 1;
struct Node {
    int u, v, w, nxt;
} G[N];
queue 
 
<int> Q;
vector <int> Vt[N];
int L[N << 2], R[N << 2], W[N << 2], F[N << 2];

inline int read() {
    int x = 0;
    char c = yxy;
    while(c < ‘0‘ || c > ‘9‘) c = yxy;
    while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = yxy;
    return
 
 x;
}

inline void add(int u, int v) {
    G[now].v = v;
    G[now].nxt = head[u];
    head[u] = now ++;
}

inline void spfa(int start,int endd) {
    for(int i = 1; i <= n; i ++) dis[i] = oo, vis[i] = 0;
    dis[start] = 0;
    Q.push(start);
    while(!Q.empty()) {
        int topp = Q.front();
        Q.pop();
        vis[topp] = 0;
        for(int i = head[topp]; ~ i; i = G[i].nxt) {
            if(dis[G[i].v] > dis[topp] + 1) {
                dis[G[i].v] = dis[topp] + 1;
                pre[G[i].v] = topp;
                if(!vis[G[i].v]) {
                    vis[G[i].v] = 1;
                    Q.push(G[i].v);
                }
            }
        }
    }
}

void calc(int start, int endd, int diss) {
    int js = -1;
    pre[start] = 0;
    while(endd) {
        js ++;
        if(tim[endd] == diss - js) Answer[endd] ++;
        endd = pre[endd];
    }
}

void work_1() {
    for(int i = 1; i <= m; i ++) {
        spfa(Askl[i], Askr[i]);
        calc(Askl[i], Askr[i], dis[Askr[i]]);
    }
}

void work_2() {
    for(int i = 1; i <= m; i ++) Vt[Askl[i]].push_back(Askr[i]);
    for(int i = 1; i <= n; i ++) {
        int L_ = i - tim[i], R_ = i + tim[i];
        int siz_ = Vt[L_].size();
        if(L_ > 0)
            for(int j = 0; j < siz_; j ++)
                if(Vt[L_][j] >= i) Answer[i] ++;
        siz_ = Vt[R_].size();
        if(R_ <= n)
            for(int j = 0; j < siz_; j ++)
                if(Vt[R_][j] <= i) Answer[i] ++;
    }
}

int bef[N], top[N], deep[N], size[N], fa[N], son[N], tree[N], spjs;
int cnt[N];

void Dfs_3(int u, int f_, int dep) {
    fa[u] = f_, deep[u] = dep; size[u] = 1;
    for(int i = head[u]; ~ i; i = G[i].nxt) {
        int v = G[i].v;
        if(v != f_) {
            dis[v] = dis[u] + 1;
            Dfs_3(v, u, dep + 1);
            size[u] += size[v];
            cnt[u] += cnt[v];
        }
    }
}

inline void work_3() {
    memset(dis, 0, sizeof dis);
    for(int i = 1; i <= m; i ++) cnt[Askr[i]] ++;
    Dfs_3(1, 0, 0);
    for(int i = 1; i <= n; i ++)
        if(deep[i] == tim[i])
            Answer[i] += cnt[i];
}

int main() {
    n = read();
    m = read();
    for(int i = 1; i <= n; i ++) head[i] = -1;
    for(int i = 1; i <= n - 1; i ++) {
        int u = read();
        int v = read();
        add(u, v);
        add(v, u);
    }
    for(int i = 1; i <= n; i ++) tim[i] = read();
    for(int i = 1; i <= m; i ++) Askl[i] = read(), Askr[i] = read();
    if(one_) work_1();
    else if(two_) work_2();
    else if(three_) work_3(); //起點 == 1 
    for(int i = 1; i <= n; i ++) printf("%d ", Answer[i]);
    return 0;
}

前置知識

Lca + 線段樹 + 差分 + 樹剖

考慮把一條路徑拆成兩段（這是非常常見的解決樹上問題的方法）

分別拆成 S - L 和 L - T （起點 - Lca， Lca - 終點）

這樣就可以得到當滿足

deep[s] - deep[i] = wat[i] => deep[s] = wat[i] + deep[i];

deep[s] + deep[i] - 2 * deep[Lca(s, i)] = wat[i] => deep[s] - 2 * deep[Lca(s, i)] = wat[i] - deep[i];

時玩家才會被 i 觀察員看到

發現 上面兩個式子滿足等式右邊都是定值

因此我們可以 以深度建立線段樹(動態開節點)

查詢時就應該查詢該節點所對應的深度的線段樹的區間lst[] 和 rst[] 之間的總的權值

lst[i] 表示以該節點為子樹的根中樹上編號的下界, 同理rst[]為上界(涉及到DFS序 && 樹剖的知識).

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
const int N = 3e5 + 10;

#define gc getchar()

int n, m, wat[N];
int Askl[N], Askr[N], Lca[N];

int now = 1, head[N];
struct Node {int v, nxt;} G[N << 1];

inline int read() {
    int x = 0; char c = gc;
    while(c < ‘0‘ || c > ‘9‘) c = gc;
    while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = gc;
    return x;
}

inline void Add(int u, int v) {G[now].v = v; G[now].nxt = head[u]; head[u] = now ++;}

int fa[N], deep[N], top[N], size[N], son[N], lst[N], rst[N], tree[N], Spjs;

void Dfs_1(int u, int f_, int dep) {
    fa[u] = f_; deep[u] = dep; size[u] = 1;
    for(int i = head[u]; ~ i; i = G[i].nxt) {
        int v = G[i].v;
        if(v != f_) {
            Dfs_1(v, u, dep + 1);
            size[u] += size[v];
            if(size[son[u]] < size[v]) son[u] = v;
        }
    }
}

void Dfs_2(int u, int tp) {
    top[u] = tp;
    lst[u] = ++ Spjs;
    tree[u] = Spjs;
    if(! son[u]) {rst[u] = Spjs; return ;}
    Dfs_2(son[u], tp);
    for(int i = head[u]; ~ i; i = G[i].nxt) {
        int v = G[i].v;
        if(v != fa[u] && v != son[u]) Dfs_2(v, v);
    }
    rst[u] = Spjs;
}

inline int Ask_Lca(int x, int y) {
    int tp1 = top[x], tp2 = top[y];
    while(tp1 != tp2) {
        if(deep[tp1] < deep[tp2]) swap(x, y), swap(tp1, tp2);
        x = fa[tp1];
        tp1 = top[x];
    }
    return deep[x] < deep[y] ? x : y;
}

int root[N * 3], lson[N * 25], rson[N * 25], W[N * 25], tot, cnt;

void Build_G(int l, int r, int & jd, int x, int yj) {
    if(!x) return ;
    if(!jd) jd = ++ tot;
    W[jd] += yj;
    if(l == r) return ;
    int mid = (l + r) >> 1;
    if(x <= mid) Build_G(l, mid, lson[jd], x, yj);
    else Build_G(mid + 1, r, rson[jd], x, yj);
}

int Sec_A(int jd, int l, int r, int x, int y) {
    if(! jd) return 0;
    if(x <= l && r <= y) return W[jd];
    int mid = (l + r) >> 1;
    if(y <= mid) return Sec_A(lson[jd], l, mid, x, y);
    else if(x > mid) return Sec_A(rson[jd], mid + 1, r, x, y);
    else return Sec_A(lson[jd], l, mid, x, y) + Sec_A(rson[jd], mid + 1, r, x, y);
}

void Clear() {
    tot = 0;
    memset(lson, 0, sizeof lson);
    memset(rson, 0, sizeof rson);
    memset(W, 0, sizeof W);
    memset(root, 0, sizeof root);
}

int Answer[N];

int main() {
    n = read(); m = read();
    for(int i = 1; i <= n; i ++) head[i] = -1;
    for(int i = 1; i <= n - 1; i ++) {
        int u = read(), v = read();
        Add(u, v); Add(v, u);
    }
    for(int i = 1; i <= n; i ++) wat[i] = read();
    for(int i = 1; i <= m; i ++) Askl[i] = read(), Askr[i] = read(); 
    Dfs_1(1, 0, 0);
    Dfs_2(1, 1);
    for(int i = 1; i <= m; i ++) Lca[i] = Ask_Lca(Askl[i], Askr[i]);
    int dep;
    for(int i = 1; i <= m; i ++) {
        dep = deep[Askl[i]];
        Build_G(1, n, root[dep], tree[Askl[i]], 1);
        Build_G(1, n, root[dep], tree[fa[Lca[i]]], -1);
    }
    for(int i = 1; i <= n; i ++) 
        Answer[i] = Sec_A(root[deep[i] + wat[i]], 1, n, lst[i], rst[i]);
    Clear();
    for(int i = 1; i <= m; i ++) {
        dep = deep[Askl[i]] - deep[Lca[i]] * 2 + n * 2;
        Build_G(1, n, root[dep], tree[Askr[i]], 1);
        Build_G(1, n, root[dep], tree[Lca[i]], -1);
    }
    for(int i = 1; i <= n; i ++)
        Answer[i] += Sec_A(root[wat[i] - deep[i] + n * 2], 1, n, lst[i], rst[i]);
    for(int i = 1; i <= n; i ++) 
        cout << Answer[i] << " ";
    return 0;
}

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