洛谷P10953 逃不掉的路 题解 边双连通分量（缩点）+ LCA

题目链接：https://www.luogu.com.cn/problem/P10953

解题思路：

缩点之后是棵树，答案是两点对应的的点在树上的距离。

因为缩点之后的树上的每一条边都对应一座桥。

示例程序：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;

int n, m, q, ts, dcc, dfn[maxn], low[maxn], bl[maxn], fa[maxn][17], dep[maxn];
vector<int> g[maxn], G[maxn];
stack<int> stk;

void tarjan(int u, int p) {
    dfn[u] = low[u] = ++ts;
    stk.push(u);
    for (auto v : g[u]) {
        if (v == p)
            continue;
        if (!dfn[v])
            tarjan(v, u), low[u] = min(low[u], low[v]);
        else
            low[u] = min(low[u], dfn[v]);
    }
    if (low[u] == dfn[u]) {
        dcc++;
        int v;
        do {
            v = stk.top();
            stk.pop();
            bl[v] = dcc;
        } while (u != v);
    }
}

void dfs(int u, int p, int d) {
    fa[u][0] = p;
    dep[u] = d;
    for (auto v : G[u])
        if (v != p)
            dfs(v, u, d+1);
}

int lca(int x, int y) {
    if (dep[x] < dep[y])
        swap(x, y);
    for (int i = 16; i >= 0; i--) {
        if (dep[x] - dep[y] >= (1<<i))
            x = fa[x][i];
    }
    if (x == y) return x;
    for (int i = 16; i>= 0; i--) {
        if (fa[x][i] != fa[y][i])
            x = fa[x][i], y = fa[y][i];
    }
    return fa[x][0];
}

int dis(int x, int y) {
    int z = lca(x, y);
    return dep[x] + dep[y] - 2 * dep[z];
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 0, u, v; i < m; i++) {
        scanf("%d%d", &u, &v);
        g[u].push_back(v);
        g[v].push_back(u);
    }
    tarjan(1, -1);
    for (int u = 1; u <= n; u++) {
        for (auto v : g[u]) {
            int x = bl[u], y = bl[v];
            if (x != y) {
                G[x].push_back(y);
            }
        }
    }
    dfs(1, 0, 0);
    for (int j = 1; j <= 16; j++)
        for (int i = 1; i <= dcc; i++)
            fa[i][j] = fa[ fa[i][j-1] ][j-1];
    scanf("%d", &q);
    for (int i = 0, a, b; i < q; i++) {
        scanf("%d%d", &a, &b);
        printf("%d\n", dis(bl[a], bl[b]));
    }
    return 0;
}