處理仙人掌 ---> 首先建立出圓方樹。則如果詢問的兩點 \(lca\) 為圓點，直接計算即可， 若 \(lca\) 為方點，則需要額外判斷是走環的哪一側（此時與兩個點在環上的相對位置有關。）

#include <bits/stdc++.h>
using namespace std;
#define maxn 200000
#define int long long
#define CNST 20
int n, m, Q, gra[maxn][CNST];
int N, dfn[maxn], low[maxn], timer;
int S[maxn], dis[maxn], bk[maxn];
 
int dep[maxn], fa[maxn], id[maxn];
int A, B;

struct edge
{
    int cnp, head[maxn], to[maxn], last[maxn], w[maxn];
    edge() { cnp = 1; }
    void add(int u, int v, int ww)
    {
        to[cnp] = v, last[cnp] = head[u], w[cnp] = ww, head[u] = cnp ++;
        to[cnp] = u, last[cnp] = head[v], w[cnp] = ww, head[v] = cnp ++;
    }
}E1, E2;

 
int read()
{
    int x = 0, k = 1;
    char c;
    c = getchar();
    while(c < ‘0‘ || c > ‘9‘) { if(c == ‘-‘) k = -1; c = getchar(); }
    while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar();
    return x * k;
}

void Solve(int u, int v, int w)
{
    N ++; int
 
 pre = w, ID = 0;
    bool flag = 0;
    for(int i = v; i != fa[u]; i = fa[i])
    {
        S[i] = pre; pre += bk[i];
        id[i] = ++ ID;
    }
    S[N] = S[u]; S[u] = 0;
    for(int i = v; i != fa[u]; i = fa[i]) 
        E2.add(N, i, min(S[i], S[N] - S[i]));
}

void Tarjan(int u)
{
    dfn[u] = low[u] = ++ timer; 
    for(int i = E1.head[u]; i; i = E1.last[i])
    {
        int v = E1.to[i]; if(v == fa[u]) continue;
        if(!dfn[v]) bk[v] = E1.w[i], fa[v] = u, Tarjan(v), low[u] = min(low[u], low[v]);
        else low[u] = min(low[u], dfn[v]);
        if(low[v] > dfn[u]) E2.add(u, v, E1.w[i]);
    }
    for(int i = E1.head[u]; i; i = E1.last[i])
    {
        int v = E1.to[i];
        if(fa[v] != u && dfn[v] > dfn[u]) Solve(u, v, E1.w[i]);
    }
}

void dfs(int u, int ff)
{
    gra[u][0] = ff; dep[u] = dep[ff] + 1;
    for(int i = 1; i < CNST; i ++) gra[u][i] = gra[gra[u][i - 1]][i - 1]; 
    for(int i = E2.head[u]; i; i = E2.last[i])
    {
        int v = E2.to[i];
        if(v != ff) 
            bk[v] = E2.w[i], dis[v] = dis[u] + E2.w[i], dfs(v, u);
    }
}

int LCA(int x, int y)
{
    if(dep[x] < dep[y]) swap(x, y);
    for(int i = CNST - 1; ~i; i --)
        if(dep[gra[x][i]] >= dep[y]) x = gra[x][i];
    for(int i = CNST - 1; ~i; i --)
        if(gra[x][i] != gra[y][i]) x = gra[x][i], y = gra[y][i];
    A = x, B = y;
    return x == y ? x : gra[x][0];
}

signed main()
{
    n = read(), m = read(), Q = read();
    for(int i = 1; i <= m; i ++)
    {
        int u = read(), v = read(), w = read();
        E1.add(u, v, w);
    }
    N = n; Tarjan(1); dfs(1, 0);
    while(Q --)
    {
        int u = read(), v = read();
        int lca = LCA(u, v);
        if(lca <= n) printf("%lld\n", dis[u] + dis[v] - 2 * dis[lca]);
        else
        {
            int ans = dis[u] + dis[v] - dis[A] - dis[B];
            if(id[A] <= id[B]) swap(A, B);
            ans += min(S[A] - S[B], S[lca] - S[A] + S[B]);
            printf("%lld\n", ans);
        }
    }
    return 0;
}

【題解】Bzoj2125最短路