题意
传送门 AcWing 397 逃不掉的路
题解
e − D C C e-DCC e−DCC 中不存在割边,那么 e − D C C e-DCC e−DCC 中各点与相连的割边端点至少存在两条不重叠的路径,而割边是两个相邻连接的 e − D C C e-DCC e−DCC 的必经边。
T a r j a n Tarjan Tarjan 算法求解 e − D C C e-DCC e−DCC,缩点建图,得到一颗树。对每个询问使用倍增 L C A LCA LCA 求解路径端点所在 e − D C C e-DCC e−DCC 间的距离,因为边权为 1 1 1,则这个距离即答案。总时间复杂度 O ( M + ( N + Q ) log N ) O(M+(N+Q)\log N) O(M+(N+Q)logN)。
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100005, maxm = 2 * 200005, maxlg = 18;
int N, M, Q, num, dcc, dfn[maxn], low[maxn], dc[maxn];
int tot, head[maxn], to[maxm], nxt[maxm];
int tot_c, hc[maxn], tc[maxm], nc[maxm];
int lg[maxn], fa[maxn][maxlg], dep[maxn];
bool bg[maxm];
inline void add(int x, int y) {
to[++tot] = y, nxt[tot] = head[x], head[x] = tot; }
inline void add_c(int x, int y) {
tc[++tot_c] = y, nc[tot_c] = hc[x], hc[x] = tot_c; }
void tarjan(int x, int in)
{
low[x] = dfn[x] = ++num;
for (int i = head[x]; i; i = nxt[i])
{
int y = to[i];
if (!dfn[y])
{
tarjan(y, i);
low[x] = min(low[x], low[y]);
if (low[y] > dfn[x])
bg[i] = bg[i ^ 1] = 1;
}
else if (i != (in ^ 1))
low[x] = min(low[x], dfn[y]);
}
}
void dfs(int x)
{
dc[x] = dcc;
for (int i = head[x]; i; i = nxt[i])
{
int y = to[i];
if (!dc[y] && !bg[i])
dfs(y);
}
}
void dfs2(int x, int f, int d)
{
fa[x][0] = f, dep[x] = d;
for (int k = 1; k <= lg[d]; ++k)
fa[x][k] = fa[fa[x][k - 1]][k - 1];
for (int i = hc[x]; i; i = nc[i])
{
int y = tc[i];
if (y != f)
dfs2(y, x, d + 1);
}
}
int lca(int x, int y)
{
if (dep[x] < dep[y])
swap(x, y);
while (dep[x] > dep[y])
x = fa[x][lg[dep[x] - dep[y]]];
if (x == y)
return x;
for (int k = lg[dep[x]]; k >= 0;)
if (fa[x][k] != fa[y][k])
x = fa[x][k], y = fa[y][k], k = lg[dep[x]];
else
--k;
return fa[x][0];
}
int main()
{
scanf("%d%d", &N, &M);
tot = tot_c = 1;
for (int i = 1, x, y; i <= M; ++i)
scanf("%d%d", &x, &y), add(x, y), add(y, x);
tarjan(1, 0);
for (int i = 1; i <= N; ++i)
if (!dc[i])
++dcc, dfs(i);
for (int i = 2; i <= tot; ++i)
{
int x = dc[to[i ^ 1]], y = dc[to[i]];
if (x != y)
add_c(x, y);
}
lg[0] = -1;
for (int i = 1; i <= N; ++i)
lg[i] = lg[i - 1] + (i == (1 << (lg[i - 1] + 1)));
dfs2(1, 0, 0);
scanf("%d", &Q);
for (int i = 1, x, y; i <= Q; ++i)
{
scanf("%d%d", &x, &y);
x = dc[x], y = dc[y];
int z = lca(x, y);
printf("%d\n", dep[x] + dep[y] - (dep[z] << 1));
}
return 0;
}