[BZOJ4372][烁烁的游戏][动态树分治+线段树+LCA]

[BZOJ4372][烁烁的游戏][动态树分治+线段树+LCA]

[b]题目大意：[/b]

给定一颗n个节点的树，边权均为1，初始每个点权值为0 。

其中操作Q x询问x点的点权，操作 M x d w把x点周围距离不超过w的点权值加上w。

[b]思路：[/b]

应该是一棵蛮裸的动态树分治，和BZOJ3730 : http://blog.csdn.net/g1n0st/article/details/56674271类似吧。

也是在每个节点上开一棵动态权值线段树。只不过这道题是初始值为0，而且要求的是区间修改，单点询问。

坑：

WTF求重心的时候存每个节点最大的儿子的数组没有清空，到时求到的并不是重心。。。效率瞬间hhhh，还以为是常数问题优化了好久。

代码：

#include <bits/stdc++.h>
using namespace std;
const int Maxn = 200005;
inline int Max(const int &a, const int &b) {
return a > b ? a : b;
}
inline int Min(const int &a, const int &b) {
return a < b ? a : b;
}
inline void swap(int &a, int &b) {
static int c;
c = a;
a = b;
b = c;
}

int n, k;
int head[Maxn], sub;
struct Edge {
int to, nxt;
Edge(void) {}
Edge(const int &to, const int &nxt) : to(to), nxt(nxt) {}
} edge[Maxn << 1];
inline void add(int a, int b) {
edge[++sub] = Edge(b, head[a]), head[a] = sub;
}
int siz[Maxn], son[Maxn], root, S, rt[Maxn][2], p[Maxn];
bool vis[Maxn];

namespace IO {
inline char get(void) {
static char buf[1000000], *p1 = buf, *p2 = buf;
if (p1 == p2) {
p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin);
if (p1 == p2) return EOF;
}
return *p1++;
}
inline void read(int &x) {
x = 0; static char c; bool f = 0;
for (; !(c >= '0' && c <= '9'); c = get()) if (c == '-') f = 1;
for (; c >= '0' && c <= '9'; x = x * 10 + c - '0', c = get()); if (f) x = -x;
}
inline void read(char &x) {
x = get();
while (!(x >= 'A' && x <= 'Z')) x = get();
}
inline void write(int x) {
if (!x) return (void)puts("0");
if (x < 0) putchar('-'), x = -x;
static short s[12], t;
while (x) s[++t] = x % 10, x /= 10;
while (t) putchar('0' + s[t--]);
putchar('\n');
}
};
namespace Q {
#define Maxw 20000500
int s[Maxw], ls[Maxw], rs[Maxw], sz;
inline void insert(int &o, int l, int r, int L, int R, int v) {
if (!o) o = ++sz;
if (l >= L && r <= R) {
s[o] += v;
return;
}
int mid = (l + r) >> 1;
if (mid >= L) insert(ls[o], l, mid, L, R, v);
if (mid < R) insert(rs[o], mid + 1, r, L, R, v);
}
inline int query(int o, int l, int r, int v) {
if (!o || l == r) return s[o];
int mid = (l + r) >> 1;
if (mid >= v) return s[o] + query(ls[o], l, mid, v);
else return s[o] + query(rs[o], mid + 1, r, v);
}
#undef Maxw
};
namespace LCA {
int ff[Maxn][25], dep[Maxn];
inline void dfs1(int u, int fa) {
ff[u][0] = fa;
for (int i = 1; i <= 20; i++) {
ff[u][i] = ff[ff[u][i - 1]][i - 1];
}
for (int i = head[u], v; i; i = edge[i].nxt) {
v = edge[i].to;
if (v == fa) continue;
dep[v] = dep[u] + 1;
dfs1(v, u);
}
}
inline int lca(int x, int y) {
if (dep[x] < dep[y]) swap(x, y);
int tmp = dep[x] - dep[y];
for (int k = 0, j = 1; j <= tmp; j <<= 1, k++)
if (tmp & j) x = ff[x][k];
while (x != y) {
int j = 0;
while (ff[x][j] != ff[y][j]) j++;
if (j) j--;
x = ff[x][j], y = ff[y][j];
}
return x;
}
inline int dist(int x, int y) {
return dep[x] + dep[y] - (dep[lca(x, y)] << 1);
}
};

inline void getroot(int u, int fa) {
siz[u] = 1; son[u] = 0;
for (int i = head[u], v; i; i = edge[i].nxt) {
v = edge[i].to;
if (v == fa || vis[v]) continue;
getroot(v, u);
siz[u] += siz[v];
son[u] = Max(son[u], siz[v]);
}
son[u] = Max(son[u], S - siz[u]);
if (son[u] < son[root]) root = u;
}
inline void solve(int x) {
vis[x] = 1;
for (int i = head[x], v; i; i = edge[i].nxt) {
v = edge[i].to;
if (vis[v]) continue;
S = siz[v]; root = 0;
getroot(v, 0);
p[root] = x;
solve(root);
}
}
inline void Modify(int x, int d, int w) {
Q::insert(rt[x][0], 0, n, 0, d, w);
int dist = 0;
for (int i = x; p[i]; i = p[i]) {
dist = LCA::dist(x, p[i]);
if (dist > d) continue;
Q::insert(rt[i][1], 0, n, 0, d - dist, w);
Q::insert(rt[p[i]][0], 0, n, 0, d - dist, w);
}
}
inline int Ask(int x) {
int dist = 0, ret = 0;
ret += Q::query(rt[x][0], 0, n, 0);
for (int i = x; p[i]; i = p[i]) {
dist = LCA::dist(x, p[i]);
ret += Q::query(rt[p[i]][0], 0, n, dist);
ret -= Q::query(rt[i][1], 0, n, dist);
}
return ret;
}
int main(void) {
//freopen("4372.in", "r", stdin);
//freopen("4372.out", "w", stdout);
IO::read(n), IO::read(k);
for (int i = 1; i < n; i++) {
int a, b;
IO::read(a), IO::read(b);
add(a, b), add(b, a);
}
LCA::dfs1(1, 0);
S = son[0] = n;
getroot(1, 0);
solve(root);
char op; int x, d, w;
while (k--) {
if (IO::read(op), op == 'Q') {
IO::read(x);
IO::write(Ask(x));
}
else {
IO::read(x), IO::read(d), IO::read(w);
Modify(x, d, w);
}
}
return 0;
}

完。

By g1n0st