树链剖分。

Table of Contents

  1. Description
  2. Solution

Description

有一张无向图, 定义关键路径为删去后图不连通的路径, 现在要依次删去一些边, 随时有一些询问, 求两点间的关键路径数.

Solution

这种题首先要想到倒序处理, 因为删边不好做.

然后来考虑题目的性质, 加入这张图最后被删得只剩一颗树了, 那么每一次加边, 都会让边两端路径上的关键路径变为0.

而在一颗树上两点之间关键路径数都是1.

区间求和, 修改? 树链剖分啊.

首先删边, 然后把剩下的图跑出来一颗生成树, 再用非树边更新, 接着依次执行操作和询问即可.

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#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <cctype>

using namespace std;

const int maxn = 30005;
const int maxm = 1e+5 + 5;

struct option
{
  int typ, a, b;
}q[maxn + 10005];
struct edge
{
  int to, nxt; bool ban;
}e[maxm << 1];
int n, m, ptr, lnk[maxn], opn, cnt;
int dfn[maxn], top[maxn], f[maxn], mxson[maxn], siz[maxn];
int ans[maxn], dep[maxn];

inline void add(int bgn, int end) {
  e[ptr] = (edge) {end, lnk[bgn], 0};
  lnk[bgn] = ptr; ptr++;
  e[ptr] = (edge) {bgn, lnk[end], 0};
  lnk[end] = ptr; ptr++;
}
inline int rd() {
  register int x = 0, f = 0, c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = 1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 + (c ^ 48), c = getchar();
  return f ? -x : x;
}
void dfs(int x, int fa) {
  //printf("%d\n", x);
  siz[x] = 1;
  f[x] = fa;
  dep[x] = dep[fa] + 1;
  for (int p = lnk[x]; ~p; p = e[p].nxt) {
    int y = e[p].to;
    if (y == fa || e[p].ban || dep[y]) continue;
    dfs(y, x);
    siz[x] += siz[y];
    if (siz[y] > siz[mxson[x]]) mxson[x] = y;
  }
}
void dfs2(int x, int init) {
  dfn[x] = ++cnt;
  top[x] = init;
  if (!mxson[x]) return;
  dfs2(mxson[x], init);
  for (int p = lnk[x]; ~p; p = e[p].nxt) {
    int y = e[p].to;
    if (f[y] != x || y == mxson[x] || e[p].ban) continue;
    dfs2(y, y);
  }
}

int s[maxn << 2], tag[maxn << 2];
inline void pushup(int cur) {
  s[cur] = s[cur << 1] + s[cur << 1|1];
}
inline void pushdown(int cur) {
  if (tag[cur]) {
    s[cur << 1] = s[cur << 1|1] = 0;
    tag[cur << 1] = tag[cur << 1|1] = 1;
    tag[cur] = 0;
  }
}
void build(int cur, int l, int r) {
  if (l == r) {
    s[cur] = 1;
    return;
  }
  int mid = (l + r) >> 1;
  build(cur << 1, l, mid);
  build(cur << 1|1, mid + 1, r);
  pushup(cur);
}
void update(int cur, int l, int r, int L, int R) {
  if (L <= l && r <= R) {
    s[cur] = 0;
    tag[cur] = 1;
    return;
  }
  int mid = (l + r) >> 1; pushdown(cur);
  if (L <= mid) update(cur << 1, l, mid, L, R);
  if (R > mid) update(cur << 1|1, mid + 1, r, L, R);
  pushup(cur);
}
int query(int cur, int l, int r, int L, int R) {
  if (L <= l && r <= R) return s[cur];
  int mid = (l + r) >> 1, ret = 0; pushdown(cur);
  if (L <= mid) ret += query(cur << 1, l, mid, L, R);
  if (R > mid) ret += query(cur << 1|1, mid + 1, r, L, R);
  return ret;
}
void updt(int x, int y) {
  while (top[x] != top[y]) {
    if (dep[top[x]] < dep[top[y]]) swap(x, y);
    update(1, 1, n, dfn[top[x]], dfn[x]);
    x = f[top[x]];
  }
  if (dep[x] > dep[y]) swap(x, y);
  if (dep[x] != dep[y]) update(1, 1, n, dfn[x] + 1, dfn[y]);
}
void efs(int x) {
  //printf("e%d\n", x);
  for (int p = lnk[x]; ~p; p = e[p].nxt) {
    int y = e[p].to;
    if (e[p].ban) continue;
    if (f[y] == x) efs(y);
    if (y != f[x] && dep[y] < dep[x])
      updt(x, y);
  }
}
int queryt(int x, int y) {
  int ret = 0;
  //printf("%d %d\n", x, y);
  while (top[x] != top[y]) {
    if (dep[top[x]] < dep[top[y]]) swap(x, y);
    //printf("q %d %d\n", dfn[top[x]], dfn[x]);
    ret += query(1, 1, n, dfn[top[x]], dfn[x]);

    x = f[top[x]];
  }
  if (dep[x] > dep[y]) swap(x, y);
  //printf("q %d %d\n", dfn[x], dfn[y]);
  ret += query(1, 1, n, dfn[x], dfn[y]);
  ret -= query(1, 1, n, dfn[x], dfn[x]);
  return ret;
}

int main() {
  n = rd(); m = rd();
  memset(lnk, -1, sizeof lnk);
  int a, b;
  for (int i = 1; i <= m; ++i) {
    a = rd(); b = rd();
    add(a, b);
  }
  int c;
  while (true) {
    c = rd(); if (c == -1) break;
    q[++opn].typ = c;
    q[opn].a = rd(); q[opn].b = rd();
    if (!c) {
      for (int p = lnk[q[opn].a]; ~p; p = e[p].nxt) {
        if (e[p].to == q[opn].b) {
          //printf("%d %d\n", e[p].to, q[opn].b);
          //puts("try");
          e[p].ban = e[p ^ 1].ban = 1;
          break;
        }
      }
    }
  }
  dfs(1, 0);
  dfs2(1, 1);
  build(1, 1, n);
  efs(1);
  for (int i = opn; i; --i) {
    //puts("try");
    if (q[i].typ) {
      ans[i] = queryt(q[i].a, q[i].b);
    } else {
      updt(q[i].a, q[i].b);
    }
  }
  for (int i = 1; i <= opn; ++i) {
    if (q[i].typ) printf("%d\n", ans[i]);
  }
  return 0;
}