WC2014紫荆花之恋

发表于 更新于 分类于 阅读次数：

[WC2014]紫荆花之恋 题目要求$r_i + r_j \geq dis_{i,j}$的点对有多少，那么考虑枚举路径上一个点把路径分开，就是求$r_i -dis_i \geq r_j + dis_j$的点对数量。那么在这个点上维护平衡树，就可以二分查找了。去重，考虑在儿子节点上挂一个平衡树，维护的信息和父亲的一样，但是只包含儿子的子树中的点。枚举$i$，在分割点的平衡树和分割点包含$i$的子树平衡树中分别查找 考虑点分治优化这一过程。新增一个点后，只需计算它在点分树上每个祖先处形成的贡献之和。就可以快速求出答案了。将原来挂在儿子上有关父亲的信息挂在点分树的儿子上即可。注意到加入点可能会使点分树退化为不优秀的性质，考虑替罪羊树的思想，定期重构即可 一些实现上的细节：需要写三个东西：平衡树，倍增求距，点分树。要采用高效平衡树。注意平衡树回收删掉的节点节省内存。建议模块化代码方便一块一块的调试。。。点分树部分很多迷之细节，想清楚再写，对着rqy的代码调了半天才发现问题。。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
#include <bits/stdc++.h>

struct IO {
  const static int S = 3e7;
  char buc_in[S], buc_out[S], *p_in = buc_in, *p_out = buc_out;
  IO() { fread(buc_in, 1, S, stdin); }
  ~IO() { fwrite(buc_out, 1, p_out - buc_out, stdout); }
  inline int rd() {
    int ret = 0;
    while (!isdigit(*p_in)) ++p_in;
    while (isdigit(*p_in)) ret = ret * 10 + *p_in++ - '0';
    return ret;
  }
  inline void out(long long x) {
    static char buc[20], *p = buc;
    do *p++ = x % 10 + '0';
    while (x /= 10);
    while (p != buc)
      *p_out++ = *--p;
    *p_out++ = '\n';
  }
} io;

const int N = 2e5 + 233, P = 1e9;

namespace BTree {
  int pool[N * 100], *top = pool;
  inline void init_pool() {
    for (int i = 1; i < N * 100; ++i)
      *top++ = i;
  }
  inline int get() { return *--top; }
  inline void del(int x) { *top++ = x; }

  struct Node {
    int ch[2], val, size;
    inline void init() {
      ch[0] = ch[1] = val = size = 0;
    }
  } t[N * 100];

  inline void pushup(int x) {
    t[x].size = t[t[x].ch[0]].size + t[t[x].ch[1]].size + 1;
  }

  int stk[N], tmp[N], ver;

  void dfs(int x) {
    if (t[x].ch[0]) dfs(t[x].ch[0]);
    stk[++ver] = x, tmp[ver] = t[x].val;
    if (t[x].ch[1]) dfs(t[x].ch[1]);
  }

  void build(int &x, int l, int r) {
    int mid = (l + r) >> 1;
    x = stk[ver--]; t[x].init();
    t[x].val = tmp[mid];
    if (l < mid) build(t[x].ch[0], l, mid - 1);
    if (r > mid) build(t[x].ch[1], mid + 1, r);
    pushup(x);
  }

  int re, rfa, dir;

  void _insert(int &x, int val) {
    if (!x) {
      t[x = get()].val = val;
      return pushup(x);
    }
    _insert(t[x].ch[val > t[x].val], val);
    pushup(x);
    if (t[t[x].ch[0]].size > 0.75 * t[x].size  t[t[x].ch[1]].size > 0.75 * t[x].size)
      re = x, rfa = dir = 0;
    else if (t[x].ch[val > t[x].val] == re)
      rfa = x, dir = val > t[x].val;
  }

  inline void insert(int &x, int val) {
    re = rfa = dir = 0;
    _insert(x, val);
    if (re) {
      ver = 0; dfs(re); build(re, 1, ver);
      if (rfa)
        t[rfa].ch[dir] = re;
      else
        x = re;
    }
  }

  int query(int x, int val) {
    int ret = 0;
    while (x) {
      if (val < t[x].val)
        x = t[x].ch[0];
      else {
        ret += t[t[x].ch[0]].size + 1;
        x = t[x].ch[1];
      }
    }
    return ret;
  }

  void clear(int x) {
    if (t[x].ch[0]) clear(t[x].ch[0]);
    if (t[x].ch[1]) clear(t[x].ch[1]);
    t[x].init(); del(x);
  }
}

int R[N];

struct Graph {
  int to, nxt, cost;
} G[N * 2];
int head[N], etot;

inline void addedge(int x, int y, int c) {
  G[++etot] = { y, head[x], c }, head[x] = etot;
  G[++etot] = { x, head[y], c }, head[y] = etot;
}

namespace FTree {
  const int up = 17;
  int dad[N][up + 1], dep[N], dis[N];

  inline void link(int x, int f, int c) {
    addedge(x, f, c);
    dep[x] = dep[f] + 1;
    dis[x] = dis[f] + c;
    dad[x][0] = f;
    for (int i = 1; i <= up; ++i)
      dad[x][i] = dad[dad[x][i - 1]][i - 1];
  }

  inline int get_lca(int x, int y) {
    if (dep[x] < dep[y]) std::swap(x, y);
    for (int i = up; ~i; --i)
      if (dep[dad[x][i]] >= dep[y])
        x = dad[x][i];
    if (x == y) return x;
    for (int i = up; ~i; --i)
      if (dad[x][i] != dad[y][i])
        x = dad[x][i], y = dad[y][i];
    return dad[x][0];
  }

  inline int get_dis(int x, int y) {
    int lca = get_lca(x, y);
    return dis[x] + dis[y] - dis[lca] * 2;
  }
}

namespace DTree {

  int root1[N], root2[N], fa[N], vis1[N], vis2[N], tag;
  std::set<int> son[N];

  void clear(int x) {
    vis1[x] = tag;
    for (int y : son[x]) {
      BTree::clear(root2[y]);
      root2[y] = 0;
      clear(y);
    }
    BTree::clear(root1[x]);
    root1[x] = 0;
    son[x].clear();
  }

  int pos, min_tree, all;

  int get_size(int x, int dad) {
    int ret = 1;
    for (int i = head[x]; i; i = G[i].nxt) {
      int y = G[i].to;
      if (vis1[y] == tag && vis2[y] != tag && y != dad) {
        ret += get_size(y, x);
      }
    }
    return ret;
  }

  int get_pos(int x, int dad) {
    int ret = 1, max = 0;
    for (int i = head[x]; i; i = G[i].nxt) {
      int y = G[i].to;
      if (vis1[y] == tag && vis2[y] != tag && y != dad) {
        int tmp = get_pos(y, x);
        ret += tmp;
        max = std::max(max, tmp);
      }
    }
    max = std::max(max, all - ret);
    if (max < min_tree) {
      min_tree = max;
      pos = x;
    }
    return ret;
  }

  void insert(int x, int dad, int dep, int &r) {
    BTree::insert(r, dep - R[x]);
    for (int i = head[x]; i; i = G[i].nxt) {
      int y = G[i].to;
      if (vis1[y] == tag && vis2[y] != tag && y != dad) {
        insert(y, x, dep + G[i].cost, r);
      }
    }
  }

  int find_center(int x) {
    all = get_size(x, 0);
    min_tree = P;
    get_pos(x, 0);
    x = pos;
    vis2[x] = tag;

    insert(x, 0, 0, root1[x]);
    for (int i = head[x]; i; i = G[i].nxt) {
      int y = G[i].to;
      if (vis1[y] == tag && vis2[y] != tag) {
        int r = 0; insert(y, x, G[i].cost, r);
        int z = find_center(y);
        fa[z] = x; son[x].insert(z);
        root2[z] = r;
      }
    }
    return x;
  }

  inline void rebuild(int x) {
    ++tag;
    clear(x);
    int dad = fa[x], r = root2[x];
    root2[x] = 0;
    if (dad)
      son[dad].erase(x);
    x = find_center(x);
    fa[x] = dad;
    root2[x] = r;
    if (dad)
      son[dad].insert(x);
  }

  inline long long insert(int x, int f) {
    long long ret = 0;
    fa[x] = f;
    son[f].insert(x);
    for (int i = x; i; i = fa[i]) {
      if (fa[i]) {
        int tmp = FTree::get_dis(x, fa[i]);
        ret += BTree::query(root1[fa[i]], R[x] - tmp);
        ret -= BTree::query(root2[i], R[x] - tmp);
        BTree::insert(root2[i], tmp - R[x]);
      }
      int tmp = FTree::get_dis(x, i);
      BTree::insert(root1[i], tmp - R[x]);
    }
    int y = 0;
    for (int i = x; fa[i]; i = fa[i])
      if (BTree::t[root1[i]].size > 0.75 * BTree::t[root1[fa[i]]].size)
        y = fa[i];
    if (y) rebuild(y);
    return ret;
  }
}

int main() {
  BTree::init_pool();
  io.rd(); int n = io.rd();
  long long last_ans = 0;
  for (int i = 1; i <= n; ++i) {
    int f = io.rd() ^ (last_ans % P), c = io.rd();
    R[i] = io.rd();
    FTree::link(i, f, c);
    last_ans += DTree::insert(i, f);
    io.out(last_ans);
  }
  return 0;
}