本节内容过于暴力没什么好说的。借着这个专题改掉写倍增的陋习，虽然写链剖代码长了点不过常数小还是很香。
10130. 「一本通 4.4 例 1」点的距离
#include <bits/stdc++.h>
using namespace std;
const int N = 1000005;
namespace tree {
int n;
vector<int> g[N];
int top[N], wson[N], dep[N], siz[N], vis[N], fa[N], dfn[N], idfn[N], ind;
void read() {
cin >> n;
for (int i = 1; i < n; i++) {
int u, v;
cin >> u >> v;
g[u].push_back(v);
g[v].push_back(u);
}
}
void dfs1(int p) {
vis[p] = 1;
siz[p] = 1;
for (int i = 0; i < g[p].size(); i++) {
int q = g[p][i];
if (vis[q])
continue;
fa[q] = p;
dep[q] = dep[p] + 1;
dfs1(q);
siz[p] += siz[q];
if (siz[q] > siz[wson[p]])
wson[p] = q;
}
}
void dfs2(int p) {
vis[p] = 1;
dfn[p] = ++ind;
idfn[ind] = p;
if (wson[p]) {
top[wson[p]] = top[p];
dfs2(wson[p]);
}
for (int i = 0; i < g[p].size(); i++) {
int q = g[p][i];
if (vis[q])
continue;
top[q] = q;
dfs2(q);
}
}
void presolve() {
memset(vis, 0, sizeof vis);
dfs1(1);
memset(vis, 0, sizeof vis);
top[1] = 1;
dfs2(1);
}
int lca(int p, int q) {
while (top[p] != top[q]) {
if (dep[top[p]] > dep[top[q]])
swap(p, q);
q = fa[top[q]];
}
if (dep[p] > dep[q])
swap(p, q);
return p;
}
} // namespace tree
int main() {
ios::sync_with_stdio(false);
tree::read();
tree::presolve();
int q, t1, t2;
cin >> q;
for (int i = 1; i <= q; i++) {
cin >> t1 >> t2;
cout << tree::dep[t1] + tree::dep[t2] - 2 * tree::dep[tree::lca(t1, t2)] << endl;
}
}
10131. 「一本通 4.4 例 2」暗的连锁
#include <bits/stdc++.h>
using namespace std;
int n, m, u, v, ind;
const int N = 1000005;
int fa[N], dep[N], siz[N], wson[N], top[N], dfn[N], idfn[N], vis[N], sum[N];
vector<int> g[N];
long long ans = 0;
void dfs1(int p) {
vis[p] = 1;
for (int i = 0; i < g[p].size(); i++) {
int q = g[p][i];
if (vis[q])
continue;
fa[q] = p;
dep[q] = dep[p] + 1;
dfs1(q);
if (siz[q] > siz[wson[p]])
wson[p] = q;
siz[p] += siz[q];
}
}
void dfs2(int p) {
vis[p] = 1;
dfn[p] = ++ind;
idfn[ind] = p;
if (wson[p]) {
top[wson[p]] = top[p];
dfs2(wson[p]);
}
for (int i = 0; i < g[p].size(); i++) {
int q = g[p][i];
if (vis[q])
continue;
top[q] = q;
dfs2(q);
}
}
int lca(int p, int q) {
while (top[p] != top[q]) {
if (dep[top[q]] > dep[top[p]])
swap(p, q);
p = fa[top[p]];
}
if (dep[q] > dep[p])
swap(p, q);
return q;
}
void dfs3(int p) {
vis[p] = 1;
for (int i = 0; i < g[p].size(); i++) {
int q = g[p][i];
if (vis[q])
continue;
dfs3(q);
sum[p] += sum[q];
}
ans += (sum[p] == 1 ? 1 : 0) + (sum[p] == 0 ? m * (p > 1) : 0);
}
int main() {
ios::sync_with_stdio(false);
cin >> n >> m;
for (int i = 1; i < n; i++) {
cin >> u >> v;
g[u].push_back(v);
g[v].push_back(u);
}
dfs1(1);
memset(vis, 0, sizeof vis);
top[1] = 1;
dfs2(1);
for (int i = 1; i <= m; i++) {
cin >> u >> v;
sum[u]++;
sum[v]++;
sum[lca(u, v)] -= 2;
}
memset(vis, 0, sizeof vis);
dfs3(1);
cout << ans << endl;
}
10132. 「一本通 4.4 例 3」异象石
这个题值得去思考一下。首先维护集合并对答案进行转化的方法值得借鉴（个人柑橘有种欧拉序的味道），然后是对集合边界的特殊处理需要注意。
#include <bits/stdc++.h>
using namespace std;
#define int long long
int n, m, x, y, z, ind;
char op;
const int N = 1000005;
int fa[N], dep[N], dis[N], top[N], dfn[N], fin[N], idfn[N], vis[N], wson[N], siz[N];
vector<pair<int, int> > g[N];
void dfs1(int p) {
vis[p] = 1;
for (int i = 0; i < g[p].size(); i++) {
int q = g[p][i].first;
if (vis[q])
continue;
fa[q] = p;
dep[q] = dep[p] + 1;
dis[q] = dis[p] + g[p][i].second;
dfs1(q);
siz[p] += siz[q];
if (siz[q] > siz[wson[p]])
wson[p] = q;
}
}
void dfs2(int p) {
vis[p] = 1;
dfn[p] = ++ind;
idfn[ind] = p;
if (wson[p]) {
top[wson[p]] = top[p];
dfs2(wson[p]);
}
for (int i = 0; i < g[p].size(); i++) {
int q = g[p][i].first;
if (vis[q])
continue;
top[q] = q;
dfs2(q);
}
fin[p] = ++ind;
}
int lca(int p, int q) {
while (top[p] != top[q]) {
if (dep[top[p]] < dep[top[q]])
swap(p, q);
p = fa[top[p]];
}
if (dep[p] < dep[q])
swap(p, q);
return q;
}
int dist(int p, int q) {
// cout<<"dis <"<<p<<","<<q<<"> = "<<dis[p]+dis[q]-2*dis[lca(p,q)]<<endl;
return dis[p] + dis[q] - 2 * dis[lca(p, q)];
}
set<int> s;
long long ans = 0;
signed main() {
ios::sync_with_stdio(false);
cin >> n;
for (int i = 1; i < n; i++) {
cin >> x >> y >> z;
g[x].push_back(make_pair(y, z));
g[y].push_back(make_pair(x, z));
}
dfs1(1);
memset(vis, 0, sizeof vis);
top[1] = 1;
dfs2(1);
cin >> m;
for (int i = 1; i <= m; i++) {
cin >> op;
if (op == '+') {
cin >> x;
set<int>::iterator pre, suf;
if (s.size()) {
pre = suf = s.lower_bound(dfn[x]);
if (suf == s.end())
suf = s.begin();
if (pre == s.begin())
pre = --s.end();
else
--pre;
int p = idfn[*pre], q = idfn[*suf];
ans += dist(p, x) + dist(q, x) - dist(p, q);
}
s.insert(dfn[x]);
}
if (op == '-') {
cin >> x;
set<int>::iterator pre, suf;
s.erase(dfn[x]);
if (s.size()) {
pre = suf = s.lower_bound(dfn[x]);
if (suf == s.end())
suf = s.begin();
if (pre == s.begin())
pre = --s.end();
else
--pre;
int p = idfn[*pre], q = idfn[*suf];
ans -= dist(p, x) + dist(q, x) - dist(p, q);
}
}
if (op == '?') {
cout << ans / 2 << endl;
}
// cout<<ans<<endl;
}
}
10133. 「一本通 4.4 例 4」次小生成树
我仍然没找到我91分的原因是什么
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1000005;
struct Edge {
int u, v, w;
bool operator<(const Edge &b) { return w < b.w; }
};
int n, m, t1, t2, t3, t4;
Edge e[N];
bool f[N]; // 1 - on tree
vector<pair<int, int> > g[N];
namespace Kruskal {
int fa[N], ans;
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
void merge(int p, int q) {
p = find(p);
q = find(q);
fa[p] = q;
}
void solve() {
sort(e + 1, e + m + 1);
for (int i = 1; i <= n; i++) fa[i] = i;
for (int i = 1; i <= m; i++) {
if (find(e[i].u) != find(e[i].v)) {
merge(e[i].u, e[i].v);
ans += e[i].w;
f[i] = true;
g[e[i].u].push_back(make_pair(e[i].v, e[i].w));
g[e[i].v].push_back(make_pair(e[i].u, e[i].w));
}
}
}
} // namespace Kruskal
namespace SegmentTree {
int a[N << 2], b[N << 2];
void build(int p, int l, int r, int *src) {
if (l == r)
a[p] = src[l];
else {
build(p * 2, l, (l + r) / 2, src);
build(p * 2 + 1, (l + r) / 2 + 1, r, src);
a[p] = max(a[p * 2], a[p * 2 + 1]);
if (a[p * 2] < a[p])
b[p] = a[p * 2];
else if (a[p * 2 + 1] < a[p])
b[p] = a[p * 2 + 1];
else
b[p] = max(b[p * 2], b[p * 2 + 1]);
}
}
int query(int p, int l, int r, int ql, int qr) {
if (l > qr || r < ql)
return 0;
if (l >= ql && r <= qr)
return a[p];
return max(query(p * 2, l, (l + r) / 2, ql, qr), query(p * 2 + 1, (l + r) / 2 + 1, r, ql, qr));
}
int querys(int p, int l, int r, int ql, int qr) {
if (l > qr || r < ql)
return 0;
if (l >= ql && r <= qr)
return b[p];
return max(querys(p * 2, l, (l + r) / 2, ql, qr), querys(p * 2 + 1, (l + r) / 2 + 1, r, ql, qr));
}
} // namespace SegmentTree
namespace Tree {
int ind, fa[N], siz[N], dep[N], dis[N], vis[N], wson[N], top[N], dfn[N];
void dfs1(int p) {
vis[p] = 1;
siz[p] = 1;
for (int i = 0; i < g[p].size(); i++) {
int q = g[p][i].first;
if (vis[q])
continue;
dep[q] = dep[p] + 1;
dis[q] = g[p][i].second;
fa[q] = p;
dfs1(q);
siz[p] += siz[q];
if (siz[wson[p]] < siz[q])
wson[p] = q;
}
}
void dfs2(int p) {
vis[p] = 1;
dfn[p] = ++ind;
if (wson[p]) {
top[wson[p]] = top[p];
dfs2(wson[p]);
}
for (int i = 0; i < g[p].size(); i++) {
int q = g[p][i].first;
if (vis[q])
continue;
top[q] = q;
dfs2(q);
}
}
void presolve() {
dfs1(1);
memset(vis, 0, sizeof vis);
top[1] = 1;
dfs2(1);
}
int query(int p, int q) {
int ret = 0;
while (top[p] != top[q]) {
if (dep[top[p]] < dep[top[q]])
swap(p, q);
ret = max(ret, SegmentTree::query(1, 1, n, dfn[top[p]], dfn[p]));
p = fa[top[p]];
}
if (dep[p] > dep[q])
swap(p, q);
return max(ret, SegmentTree::query(1, 1, n, dfn[p] + 1, dfn[q]));
}
int querys(int p, int q) {
int ret = 0, lim = query(p, q);
while (top[p] != top[q]) {
if (dep[top[p]] < dep[top[q]])
swap(p, q);
if (SegmentTree::query(1, 1, n, dfn[top[p]], dfn[p]) < lim)
ret = max(ret, SegmentTree::query(1, 1, n, dfn[top[p]], dfn[p]));
else
ret = max(ret, SegmentTree::querys(1, 1, n, dfn[top[p]], dfn[p]));
p = fa[top[p]];
}
if (dep[p] > dep[q])
swap(p, q);
if (SegmentTree::query(1, 1, n, dfn[p] + 1, dfn[q]) < lim)
return max(ret, SegmentTree::query(1, 1, n, dfn[p] + 1, dfn[q]));
else
return max(ret, SegmentTree::querys(1, 1, n, dfn[p] + 1, dfn[q]));
}
} // namespace Tree
int val[N];
signed main() {
ios::sync_with_stdio(false);
cin >> n >> m;
for (int i = 1; i <= m; i++) {
cin >> t1 >> t2 >> t3;
e[i].u = t1;
e[i].v = t2;
e[i].w = t3;
}
Kruskal::solve();
Tree::presolve();
for (int i = 1; i <= n; i++) {
val[Tree::dfn[i]] = Tree::dis[i];
}
SegmentTree::build(1, 1, n, val);
int ans = 1e+15;
for (int i = 1; i <= m; i++) {
// cout<<i<<", "<<f[i]<<", "<<e[i].w - Tree::query(e[i].u,e[i].v)<<endl;
if (f[i] == false) {
if (e[i].w - Tree::query(e[i].u, e[i].v))
ans = min(ans, e[i].w - Tree::query(e[i].u, e[i].v));
else
ans = min(ans, e[i].w - Tree::querys(e[i].u, e[i].v));
}
}
cout << ans + Kruskal::ans << endl;
}
10134. 「一本通 4.4 练习 1」Dis
#include <bits/stdc++.h>
using namespace std;
int n, m, ind, t1, t2, t3;
const int N = 1000005;
int fa[N], vis[N], dis[N], dep[N], top[N], siz[N], wson[N], dfn[N];
vector<pair<int, int> > g[N];
void dfs1(int p) {
vis[p] = 1;
siz[p] = 1;
for (int i = 0; i < g[p].size(); i++) {
int q = g[p][i].first;
if (vis[q])
continue;
fa[q] = p;
dep[q] = dep[p] + 1;
dis[q] = dis[p] + g[p][i].second;
dfs1(q);
siz[p] += siz[q];
if (siz[q] > siz[wson[p]])
wson[p] = q;
}
}
void dfs2(int p) {
vis[p] = 1;
dfn[p] = ++ind;
if (wson[p]) {
top[wson[p]] = top[p];
dfs2(wson[p]);
}
for (int i = 0; i < g[p].size(); i++) {
int q = g[p][i].first;
if (vis[q])
continue;
top[q] = q;
dfs2(q);
}
}
int lca(int p, int q) {
while (top[p] != top[q]) {
if (dep[top[p]] < dep[top[q]])
swap(p, q);
p = fa[top[p]];
}
if (dep[p] < dep[q])
swap(p, q);
return q;
}
int main() {
cin >> n >> m;
for (int i = 1; i < n; i++) {
cin >> t1 >> t2 >> t3;
g[t1].push_back(make_pair(t2, t3));
g[t2].push_back(make_pair(t1, t3));
}
dfs1(1);
memset(vis, 0, sizeof vis);
top[1] = 1;
dfs2(1);
for (int i = 1; i <= m; i++) {
cin >> t1 >> t2;
cout << dis[t1] + dis[t2] - 2 * dis[lca(t1, t2)] << endl;
}
}
10135. 「一本通 4.4 练习 2」祖孙询问
#include <bits/stdc++.h>
using namespace std;
const int N = 1000005;
int n, m, ind, t1, t2, t3, t4;
int vis[N], dep[N], siz[N], top[N], wson[N], dfn[N], fa[N];
vector<int> g[N];
void dfs1(int p) {
vis[p] = 1;
siz[p] = 1;
for (int i = 0; i < g[p].size(); i++) {
int q = g[p][i];
if (vis[q])
continue;
fa[q] = p;
dep[q] = dep[p] + 1;
dfs1(q);
siz[p] += siz[q];
if (siz[wson[p]] < siz[q])
wson[p] = q;
}
}
void dfs2(int p) {
vis[p] = 1;
dfn[p] = ++ind;
if (wson[p]) {
top[wson[p]] = top[p];
dfs2(wson[p]);
}
for (int i = 0; i < g[p].size(); i++) {
int q = g[p][i];
if (vis[q])
continue;
top[q] = q;
dfs2(q);
}
}
int lca(int p, int q) {
while (top[p] != top[q]) {
if (dep[top[p]] < dep[top[q]])
swap(p, q);
p = fa[top[p]];
}
if (dep[p] > dep[q])
swap(p, q);
return p;
}
map<int, int> mp;
int root;
int x[N], y[N], cnt, Index;
int main() {
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> t1 >> t2;
if (t2 == -1) {
root = t1;
} else {
++cnt;
x[cnt] = t1;
y[cnt] = t2;
mp[t1]++;
mp[t2]++;
}
}
for (map<int, int>::iterator it = mp.begin(); it != mp.end(); it++) {
it->second = ++Index;
}
for (int i = 1; i <= cnt; i++) {
x[i] = mp[x[i]];
y[i] = mp[y[i]];
g[x[i]].push_back(y[i]);
g[y[i]].push_back(x[i]);
}
root = mp[root];
dfs1(root);
memset(vis, 0, sizeof vis);
top[root] = root;
dfs2(root);
cin >> m;
for (int i = 1; i <= m; i++) {
cin >> t1 >> t2;
t1 = mp[t1];
t2 = mp[t2];
int l = lca(t1, t2);
if (l == t1 && l != t2)
cout << 1 << endl;
else if (l == t2 && l != t1)
cout << 2 << endl;
else
cout << 0 << endl;
}
}
10136. 「一本通 4.4 练习 3」聚会
分类以后总结一下发现就是最深的那个LCA
#include <bits/stdc++.h>
using namespace std;
int n, m, t1, t2, t3, t4, ind;
const int N = 1000005;
int fa[N], dep[N], siz[N], vis[N], wson[N], top[N], dfn[N];
vector<int> g[N];
void dfs1(int p) {
vis[p] = 1;
siz[p] = 1;
for (int i = 0; i < g[p].size(); i++) {
int q = g[p][i];
if (vis[q])
continue;
fa[q] = p;
dep[q] = dep[p] + 1;
dfs1(q);
siz[p] += siz[q];
if (siz[wson[p]] < siz[q])
wson[p] = q;
}
}
void dfs2(int p) {
vis[p] = 1;
dfn[p] = ++ind;
if (wson[p]) {
top[wson[p]] = top[p];
dfs2(wson[p]);
}
for (int i = 0; i < g[p].size(); i++) {
int q = g[p][i];
if (vis[q])
continue;
top[q] = q;
dfs2(q);
}
}
int lca(int p, int q) {
while (top[p] != top[q]) {
if (dep[top[p]] < dep[top[q]])
swap(p, q);
p = fa[top[p]];
}
if (dep[p] < dep[q])
swap(p, q);
return q;
}
int dist(int p, int q) { return dep[p] + dep[q] - 2 * dep[lca(p, q)]; }
int main() {
ios::sync_with_stdio(false);
scanf("%d%d", &n, &m);
for (int i = 1; i < n; i++) {
scanf("%d%d", &t1, &t2);
g[t1].push_back(t2);
g[t2].push_back(t1);
}
dfs1(1);
memset(vis, 0, sizeof vis);
top[1] = 1;
dfs2(1);
for (int i = 1; i <= m; i++) {
scanf("%d%d%d", &t1, &t2, &t3);
int l1 = lca(t1, t2), l2 = lca(t2, t3), l3 = lca(t3, t1);
if (dep[l2] > dep[l1])
l1 = l2;
if (dep[l3] > dep[l1])
l1 = l3;
printf("%d %d\n", l1, dist(l1, t1) + dist(l1, t2) + dist(l1, t3));
}
}