【题解】BZOJ-2125-最短路

题目

https://www.lydsy.com/JudgeOnline/problem.php?id=2125

题意

每次询问仙人掌上两点之间的最短路。

题解

  • 建圆方树。圆方边的长度是圆点到这个方点表示的环的根的最短距离(如果圆点是方点的父亲,这条边长度为 0)。
  • 然后 树剖 / 倍增 求两点间距离。但是如果 LCA 是方点的话,需要找出询问的两个点在 LCA 的哪个儿子的子树中,然后重新计算这两个儿子之间的最短距离。

代码

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#include <bits/stdc++.h>
using namespace std;
using LL = long long;
#define FOR(i, x, y) for (decay<decltype(y)>::type i = (x), _##i = (y); i < _##i; ++i)
#define FORD(i, x, y) for (decay<decltype(x)>::type i = (x), _##i = (y); i > _##i; --i)
#ifdef zerol
#define dbg(args...) do { cout << "\033[32;1m" << #args << " -> "; err(args); } while (0)
#else
#define dbg(...)
#endif
void err() { cout << "\033[39;0m" << endl; }
template<template<typename...> class T, typename t, typename... Args>
void err(T<t> a, Args... args) { for (auto x: a) cout << x << ' '; err(args...); }
template<typename T, typename... Args>
void err(T a, Args... args) { cout << a << ' '; err(args...); }
// -----------------------------------------------------------------------------
const int N = 2E5 + 100;

int n, nn;
namespace R {
struct E { int to, d; };
vector<E> G[N];
int len[N]; // 环的大小
bool lr[N]; // 环的左侧还是右侧
void addedge(int u, int v, int d) {
G[u].push_back({v, d}); G[v].push_back({u, d});
}
int fa[N], dep[N], idx[N], out[N], ridx[N], dd[N], D[N];
namespace hld {
int sz[N], son[N], top[N], clk;
void predfs(int u, int d) {
dep[u] = d; sz[u] = 1;
int& maxs = son[u] = -1;
for (E& e: G[u]) {
int v = e.to;
if (v == fa[u]) continue;
fa[v] = u; dd[v] = e.d; D[v] = D[u] + dd[v];
predfs(v, d + 1);
sz[u] += sz[v];
if (maxs == -1 || sz[v] > sz[maxs]) maxs = v;
}
}
void dfs(int u, int tp) {
top[u] = tp; idx[u] = ++clk; ridx[clk] = u;
if (son[u] != -1) dfs(son[u], tp);
for (E& e: G[u]) {
int v = e.to;
if (v != fa[u] && v != son[u]) dfs(v, v);
}
out[u] = clk;
}
int go(int u, int v) {
int uu = top[u], vv = top[v];
while (uu != vv) {
if (dep[uu] < dep[vv]) { swap(uu, vv); swap(u, v); }
u = fa[uu]; uu = top[u];
}
if (dep[u] < dep[v]) swap(u, v);
return v;
}
int finds(int u, int rt) {
while (top[u] != top[rt]) {
u = top[u];
if (fa[u] == rt) return u;
u = fa[u];
}
return ridx[idx[rt] + 1];
}
}
int psd[N];
void init() {
hld::clk = 0; hld::predfs(1, 1); hld::dfs(1, 1);
FOR (i, 1, nn + 1) psd[i] = dd[ridx[i]] + psd[i - 1];
}
int query(int u, int v) {
int lca = hld::go(u, v);
int ans = D[u] + D[v] - 2 * D[lca];
if (lca > n) {
int uu = hld::finds(u, lca), vv = hld::finds(v, lca);
ans -= dd[uu] + dd[vv];
if (lr[uu] == lr[vv]) ans += abs(dd[uu] - dd[vv]);
else ans += min(dd[uu] + dd[vv], len[lca] - dd[uu] - dd[vv]);
}
return ans;
}
}

namespace X {
struct E { int to, d, nxt; };
E e[N * 2];
int hd[N], ecnt;
void addedge(int u, int v, int d) {
e[ecnt] = {v, d, hd[u]}; hd[u] = ecnt++;
e[ecnt] = {u, d, hd[v]}; hd[v] = ecnt++;
}
int idx[N], clk, fa[N], dep[N];
bool ring[N];
void init() {
ecnt = clk = 0;
memset(hd, -1, sizeof hd);
memset(dep, 0, sizeof dep);
}
void dfs(int u, int feid) {
idx[u] = ++clk;
for (int i = hd[u]; ~i; i = e[i].nxt) {
if ((i ^ feid) == 1) continue;
int v = e[i].to;
if (!idx[v]) {
fa[v] = u; dep[v] = dep[u] + e[i].d; ring[u] = false;
dfs(v, i);
if (!ring[u]) R::addedge(u, v, e[i].d);
} else if (idx[v] < idx[u]) {
++nn; R::addedge(v, nn, 0); // 强行把环的根放在最前面
R::len[nn] = dep[u] - dep[v] + e[i].d;
for (int x = u; x != v; x = fa[x]) {
ring[x] = true;
int L = dep[x] - dep[v], R = dep[u] - dep[x] + e[i].d;
int d = min(L, R); R::lr[x] = d == L;
R::addedge(nn, x, d);
}
ring[v] = true;
}
}
}
}

int main() {
#ifdef zerol
freopen("in", "r", stdin);
#endif
X::init();
int m, Qn; cin >> n >> m >> Qn; nn = n;
FOR (_, 0, m) {
int u, v, d; scanf("%d%d%d", &u, &v, &d);
X::addedge(u, v, d);
}
X::dfs(1, -1);

R::init();
while (Qn--) {
int u, v; scanf("%d%d", &u, &v);
printf("%d\n", R::query(u, v));
}
}
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