【题解】Bzoj2125最短路

　　处理仙人掌 ---> 首先建立出圆方树。则如果询问的两点 \(lca\) 为圆点，直接计算即可， 若 \(lca\) 为方点，则需要额外判断是走环的哪一侧（此时与两个点在环上的相对位置有关。）

#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 = ; }
    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 = , k = ;
    char c;
    c = getchar();
    while(c < '' || c > '') { if(c == '-') k = -; c = getchar(); }
    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x * k;
}

void Solve(int u, int v, int w)
{
    N ++; int pre = w, ID = ;
    bool flag = ;
    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] = ;
    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][] = ff; dep[u] = dep[ff] + ;
    for(int i = ; i < CNST; i ++) gra[u][i] = gra[gra[u][i - ]][i - ];
    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 - ; ~i; i --)
        if(dep[gra[x][i]] >= dep[y]) x = gra[x][i];
    for(int i = CNST - ; ~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][];
}

signed main()
{
    n = read(), m = read(), Q = read();
    for(int i = ; i <= m; i ++)
    {
        int u = read(), v = read(), w = read();
        E1.add(u, v, w);
    }
    N = n; Tarjan(); dfs(, );
    while(Q --)
    {
        int u = read(), v = read();
        int lca = LCA(u, v);
        if(lca <= n) printf("%lld\n", dis[u] + dis[v] -  * 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 ;
}