[Luogu] 天天爱跑步
https://www.luogu.org/problemnew/show/P1600
https://www.zybuluo.com/wsndy-xx/note/1135243
乱写的暴力，这道题暴力写个60还是比较简单的
#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(); //起点 == 1 for(int i = 1; i <= n; i ++) printf("%d ", Answer[i]); return 0; }
前置知识
Lca + 线段树 + 差分 + 树剖
考虑把一条路径拆成两段（这是非常常见的解决树上问题的方法）
分别拆成 S - L 和 L - T （起点 - Lca， Lca - 终点）
这样就可以得到当满足
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];
时玩家才会被 i 观察员看到
发现 上面两个式子满足等式右边都是定值
因此我们可以 以深度建立线段树(动态开节点)
查询时就应该查询该节点所对应的深度的线段树的区间lst[] 和 rst[] 之间的总的权值
lst[i] 表示以该节点为子树的根中树上编号的下界, 同理rst[]为上界(涉及到DFS序 && 树剖的知识).
#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; }
