[題目鏈接]

https://www.lydsy.com/JudgeOnline/problem.php?id=4326

[算法]

首先，此題的答案是具有單調性的，因此可以二分答案mid

檢驗答案時，我們判斷每條路徑的長度是否大於mid，若大於mid，則說明至少要將這條路徑上的一條邊變為“蟲洞”

因此，我們可以對所有長度大於mid的路徑做樹上差分，若一條邊差分後的值 = 大於mid的路徑總數，那麽判斷最長路徑 - 這條邊的長度 <= mid

時間復雜度 : O(N log LEN)( LEN為每條邊的長度之和 )

[代碼]

此份代碼在BZOJ上通過了所有測試點，但在UOJ上由於常數原因只能拿到95分 , 此題需要一些常數優化

#include<bits/stdc++.h>
using namespace std;
#define MAXN 300010
#define MAXLOG 18

struct edge
{
        int to,w,nxt;
} e[MAXN << 1];

int i,n,m,tot,l,r,mid,ans,cnt,maxlen,num;
int u[MAXN],v[MAXN],a[MAXN],b[MAXN],t[MAXN],head[MAXN],depth[MAXN],sum[MAXN],s[MAXN],len[MAXN],dis[MAXN];
 
int anc[MAXN][MAXLOG];

namespace IO
{
    template <typename T> inline void read(T &x)
    {
        int f = 1; x = 0;
        char c = getchar();
        for (; !isdigit(c); c = getchar())
        {
            if (c == ‘-‘) f = -f;
        }
        for (; isdigit(c); c = getchar()) x = (x << 3
 
) + (x << 1) + c - ‘0‘;
        x *= f;
    }
    template <typename T> inline void write(T x)
    {
        if (x < 0)
        {
            putchar(‘-‘);
            x = -x;
        }
        if (x > 9) write(x / 10);
        putchar(x % 10 + ‘0‘);
    }
    template <typename T> inline void writeln(T x)
    {
        write(x);
        puts("");
    }
} ;
inline void addedge(int u,int v,int w)
{
        tot++;
        e[tot] = (edge){v,w,head[u]};
        head[u] = tot;
}
inline void dfs(int u)
{
        int i,v,w;
        for (i = 1; i < MAXLOG; i++)
        {
                if (depth[u] < (1 << i)) break;
                anc[u][i] = anc[anc[u][i - 1]][i - 1];
        }
        for (i = head[u]; i; i = e[i].nxt)
        {
                v = e[i].to;
                w = e[i].w;
                if (v != anc[u][0])
                {
                        len[v] = w;
                        anc[v][0] = u;
                        depth[v] = depth[u] + 1;
                        sum[v] = sum[u] + w;
                        dfs(v);
                }
        }
}
inline void calc(int u)
{
        int i,v;
        for (i = head[u]; i; i = e[i].nxt)
        {
                v = e[i].to;
                if (v != anc[u][0]) 
                {
                        calc(v);
                        s[u] += s[v];
                } 
        }
        if (s[u] == num && len[u] > maxlen) maxlen = len[u];
}
inline int lca(int x,int y)
{
        int i,t;
        if (depth[x] > depth[y]) swap(x,y);
        t = depth[y] - depth[x];
        for (i = 0; i < MAXLOG; i++)
        {
                if (t & (1 << i))
                        y = anc[y][i];        
        }        
        if (x == y) return x;
        for (i = MAXLOG - 1; i >= 0; i--)
        {
                if (anc[x][i] != anc[y][i])
                {
                        x = anc[x][i];
                        y = anc[y][i];
                }
        }
        return anc[x][0];
}
inline int dist(int x,int y)
{
        return sum[x] + sum[y] - 2 * sum[lca(x,y)];
}
inline bool check(int mid)
{
        int i,j,p,q,w,mx = 0;
        num = 0;
        for (i = 1; i <= n; i++) s[i] = 0;
        for (i = 1; i <= m; i++)
        {
                if (dis[i] > mid)
                {
                        mx = max(mx,dis[i]);
                        num++;
                        s[u[i]]++; s[v[i]]++;
                        s[lca(u[i],v[i])] -= 2;
                }
        }
        if (cnt == 0) return true;
        maxlen = 0;
        calc(1);
        return mx - maxlen <= mid;
}
    
int main() 
{
        
        IO :: read(n); IO :: read(m);
        for (i = 1; i < n; i++)
        {
                IO :: read(a[i]); IO :: read(b[i]); IO :: read(t[i]);
                addedge(a[i],b[i],t[i]);
                addedge(b[i],a[i],t[i]);    
                cnt += t[i];    
        }
        dfs(1); 
        for (i = 1; i <= m; i++) 
        {
                IO :: read(u[i]); 
                IO :: read(v[i]);
                dis[i] = dist(u[i],v[i]);
        }
        l = 0; r = cnt;
        while (l <= r)
        {
                mid = (l + r) >> 1;
                if (check(mid))
                {
                        r = mid - 1;
                        ans = mid;
                } else l = mid + 1;
        }
        IO :: writeln(ans);
        
        return 0;
    
}

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