给出一棵nn个点有边权的树和树上mm条路径,清零一条边的边权使得mm条路径的最大值最小

n,m300000n,m\le 300000

Luogu P2680

Solution

显然答案具有单调性,考虑二分答案

设当前二分的答案为ansans,考虑如何checkcheck

可以发现我们需要判断是否存在一条长度至少为lengthmaxanslength_{max} - ans的边,使得它被所有length>anslength > ans的边经过

这个东西直接树上差分即可

Code

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#include <bits/stdc++.h>

#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long LL;
typedef pair<int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }
inline int read ()
{
int sum = 0, fl = 1; char ch = getchar();
for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
return sum * fl;
}

const int Maxn = 1000000 + 100;

int N, M, e, Begin[Maxn], To[Maxn << 1], Next[Maxn << 1], W[Maxn << 1];
int dep[Maxn], Dep[Maxn], anc[22][Maxn], Dis[Maxn];
pii Q[Maxn];

inline void add_edge (int x, int y, int z)
{
To[++e] = y;
Next[e] = Begin[x];
Begin[x] = e;
W[e] = z;
}

inline void dfs_pre (int x)
{
dep[x] = dep[anc[0][x]] + 1;
for (int i = 1; i <= 20; ++i) anc[i][x] = anc[i - 1][anc[i - 1][x]];
for (int i = Begin[x]; i; i = Next[i])
{
int y = To[i];
if (y == anc[0][x]) continue;
anc[0][y] = x;
Dep[y] = Dep[x] + W[i];
dfs_pre(y);
}
}

inline int get_lca (int x, int y)
{
if (dep[x] < dep[y]) swap(x, y);
for (int i = 20; i >= 0; --i) if (dep[anc[i][x]] >= dep[y]) x = anc[i][x];
if (x == y) return x;
for (int i = 20; i >= 0; --i) if (anc[i][x] != anc[i][y]) x = anc[i][x], y = anc[i][y];
return anc[0][x];
}

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

int delta[Maxn], now_sum, flag, cnt;

inline void dfs (int x, int f)
{
if (flag) return ;
for (int i = Begin[x]; i; i = Next[i])
{
int y = To[i];
if (y == f) continue;
dfs(y, x);
if (delta[y] == cnt && W[i] >= now_sum)
{
flag = 1;
return ;
}
delta[x] += delta[y];
}
}

inline void Mark (int x, int y)
{
++delta[x];
++delta[y];
delta[get_lca(x, y)] -= 2;
}

inline int Check (int now_ans)
{
flag = 0;
memset(delta, 0, sizeof delta);
int Max = 0;
cnt = 0;
for (int i = 1; i <= M; ++i) if (Dis[i] > now_ans) ++cnt,Mark(Q[i].x, Q[i].y), Chkmax(Max, Dis[i]);
now_sum = Max - now_ans;
dfs(1, 0);
return flag;
}

inline void Solve ()
{
dfs_pre(1);
for (int i = 1; i <= M; ++i) Q[i].x = read(), Q[i].y = read(), Dis[i] = get_dis(Q[i].x, Q[i].y);
int l = 0, r = 3e8, ans = 0;
while (l <= r)
{
int mid = l + r >> 1;
if (Check(mid)) r = mid - 1, ans = mid;
else l = mid + 1;
}
cout<<ans<<endl;
}


inline void Input ()
{
N = read(), M = read();
for (int i = 1; i < N; ++i)
{
int x = read(), y = read(), z = read();
add_edge (x, y, z);
add_edge (y, x, z);
}
}

int main()
{
#ifndef ONLINE_JUDGE
freopen("transport.in", "r", stdin);
freopen("transport.out", "w", stdout);
#endif
Input();
Solve();
return 0;
}