算法笔记--圆方树
模板:
struct Circle_Square_Tree {
const static int N = 2e4 + 10;
vector > G[N];
int dp[N], anc[N][18], n;
LL dis[N], cir[N]; //环的大小
bool vis[N];//记录到方点的最短距离是否经过回边
vector g[N];
int dfn[N], low[N], fa[N], cnt, tot;
int a[N];
inline void solve(int u, int v, int d) {
int cnt = 0;
LL sum = d;
for (int i = v; i != u; i = fa[i]) sum += dis[i] - dis[fa[i]], a[++cnt] = i;
a[++cnt] = u;
LL DIS = 0;
++tot;
for (int i = cnt; i >= 1; --i) {
LL D = min(DIS, sum-DIS);
if(D == DIS) vis[a[i]] = true;
else vis[a[i]] = false;
G[a[i]].pb({tot, D});
G[tot].pb({a[i], D});
DIS += dis[a[i-1]]-dis[a[i]];
}
cir[tot] = sum;
}
inline void tarjan(int u, int o) {
fa[u] = o;
dfn[u] = low[u] = ++cnt;
for (int i = 0; i < g[u].size(); ++i) {
int v = g[u][i].fi;
int w = g[u][i].se;
if(v == o) continue;
if(!dfn[v]) {
dis[v] = dis[u] + w;
tarjan(v, u);
low[u] = min(low[u], low[v]);
}
else low[u] = min(low[u], dfn[v]);
if(low[v] > dfn[u]) {
G[u].pb({v, w});
G[v].pb({u, w});
}
}
for (int i = 0; i < g[u].size(); ++i) {
int v = g[u][i].fi;
int w = g[u][i].se;
if(v == o) continue;
if(fa[v] != u && dfn[v] > dfn[u]) {
solve(u, v, w);
}
}
}
inline void dfs(int u, int o) {
anc[u][0] = o;
dp[u] = dp[o] + 1;
for (int i = 1; i < 18; ++i) anc[u][i] = anc[anc[u][i-1]][i-1];
for (int i = 0; i < G[u].size(); ++i) {
int v = G[u][i].fi;
LL w = G[u][i].se;
if(v == o) continue;
dis[v] = dis[u] + w;
dfs(v, u);
}
}
inline void init(int _n) {
tot = n = _n;
cnt = 0;
dis[0] = dp[1] = 0;
tarjan(1, 0);
dfs(1, 1);
}
inline int lca(int u, int v) {
if(dp[u] < dp[v]) swap(u, v);
for (int i = 17; i >= 0; --i) if(dp[anc[u][i]] >= dp[v]) u = anc[u][i];
if(u == v) return u;
for (int i = 17; i >= 0; --i) if(anc[u][i] != anc[v][i]) u = anc[u][i], v = anc[v][i];
return anc[u][0];
}
inline int jump(int u, int lca) {
for (int i = 17; i >= 0; --i) if(dp[anc[u][i]] > dp[lca]) u = anc[u][i];
return u;
}
inline LL query(int u, int v) {
int l = lca(u, v);
if(l <= n) return dis[u]+dis[v]-2*dis[l];
int uu = jump(u, l), vv = jump(v, l);
LL d1 = dis[uu]-dis[l], d2 = dis[vv]-dis[l];
if(!vis[uu]) d1 = cir[l]-d1;
if(!vis[vv]) d2 = cir[l]-d2;
return dis[u]-dis[uu]+dis[v]-dis[vv]+min(abs(d1-d2), cir[l]-abs(d1-d2));
}
}CST;