*【学习笔记】(19) 启发式合并
启发式合并
启发式合并核心思想就一句话:把小集合的合并到大的里。
启发式合并思想可以放到很多数据结构里,链表、线段树、甚至平衡树都可以。
考虑时间复杂度,设总共有 \(n\) 个元素,由于每次集合的大小至少翻倍,所以至多会合并 \(logn\) 次,总的复杂度就是 \(O(nlogn)\) 的(结合线段树合并就是 \(O(nlog^2n)\) 的)
dsu on tree 就是树上启发式合并。
dsu on tree 优雅的思想
对于以 u 为根的子树
①. 先统计它轻子树(轻儿子为根的子树)的答案,统计完后删除信息
②. 再统计它重子树(重儿子为根的子树)的答案 ,统计完后保留信息
③. 然后再将重子树的信息合并到 u上
④. 再去遍历 u 的轻子树,然后把轻子树的信息合并到 u 上
⑤. 判断 u 的信息是否需要传递给它的父节点(u 是否是它父节点的重儿子)
dsu on tree 暴力的操作体现于统计答案上(不同的题目统计方式不一样)
例题
Ⅰ.CF570D Tree Requests
重排之后能否构成回文串,即判断出现次数为奇数的字符个数是否小于等于1
因为要多次询问以某个点为根深度为 \(b\) 的节点上的字母
所以可以对每个节点开个 vector,再把和它有关的询问保存在vector
然后在 dsu on tree 的过程进行到该节点时遍历这个节点的 vector (相关询问)
#include<bits/stdc++.h>
#define pb push_back
#define mp make_pair
using namespace std;
const int N = 1e6 + 67;
int read(){
int x = 0, f = 1; char ch = getchar();
while(ch < '0' || ch > '9'){if(ch == '-') f = -f; ch = getchar();}
while(ch >= '0' && ch <= '9'){x = (x << 1) + (x << 3) + (ch ^ 48); ch = getchar();}
return x * f;
}
int n, m, tot;
int Head[N], to[N], Next[N];
int a[N], son[N], sz[N], d[N];
vector<pair<int, int> > Q[N];
bool vis[N];
int b[N][26], ct[N], ans[N];
void add(int u, int v){
to[++tot] = v, Next[tot] = Head[u], Head[u] = tot;
}
void dfs1(int x, int fa){
d[x] = d[fa] + 1, sz[x] = 1;
for(int i = Head[x]; i; i = Next[i]){
int y = to[i]; if(y == fa) continue;
dfs1(y, x), sz[x] += sz[y];
if(sz[y] > sz[son[x]]) son[x] = y;
}
}
void calc(int x, int fa){
if(b[d[x]][a[x]]) ct[d[x]]--;
else ct[d[x]]++;
b[d[x]][a[x]] ^= 1;
for(int i = Head[x]; i; i = Next[i]){
int y = to[i]; if(y == fa || vis[y]) continue;
calc(y, x);
}
}
void dfs2(int x, int fa, int opt){
for(int i = Head[x]; i; i = Next[i]){
int y = to[i]; if(y == fa || y == son[x]) continue;
dfs2(y, x, 0);
}
if(son[x]) dfs2(son[x], x, 1), vis[son[x]] = 1;
calc(x, fa), vis[son[x]] = 0;
for(auto it : Q[x])
ans[it.second] = (ct[it.first] <= 1);
if(!opt) calc(x, fa);
}
int main(){
n = read(), m = read();
for(int i = 2; i <= n; ++i){
int u = read();
add(u, i), add(i, u);
}
for(int i = 1; i <= n; ++i) {
char ch; cin >> ch;
a[i] = ch - 'a';
}
for(int i = 1; i <= m; ++i){
int u = read(), k = read();
Q[u].pb(mp(k, i));
}
dfs1(1, 0), dfs2(1, 0, 0);
for(int i = 1; i <= m; ++i)
printf("%s\n", ans[i] ? "Yes" : "No");
return 0;
}
Ⅱ.CF375D Tree and Queries
开个数组 \(cnt_i\) 表示 \(i\) 种颜色有多少个,\(f_i\) 表示数量大于 \(i\) 的有多少种。
#include<bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define pii pair<int,int>
using namespace std;
const int N = 2e5 + 67;
int read(){
int x = 0, f = 1; char ch = getchar();
while(ch < '0' || ch > '9'){if(ch == '-') f = -f; ch = getchar();}
while(ch >= '0' && ch <= '9'){x = (x << 1) + (x << 3) + (ch ^ 48); ch = getchar();}
return x * f;
}
int n, m, tot;
int fa[N], sz[N], son[N];
int c[N], cnt[N], ans[N], f[N];
int Head[N], to[N], Next[N];
vector<pii> Q[N];
bool vis[N];
void add(int u, int v){
to[++tot] = v, Next[tot] = Head[u], Head[u] = tot;
}
void dfs1(int x, int f){
fa[x] = f, sz[x] = 1;
for(int i = Head[x]; i; i = Next[i]){
int y = to[i]; if(y == fa[x]) continue;
dfs1(y, x); sz[x] += sz[y];
if(sz[y] > sz[son[x]]) son[x] = y;
}
}
void calc(int x, int val){
if(val == 1) f[cnt[c[x]] += val]++;
else --f[cnt[c[x]]], cnt[c[x]] += val;
for(int i = Head[x]; i; i = Next[i]){
int y = to[i]; if(y == fa[x] || vis[y]) continue;
calc(y, val);
}
}
void dfs2(int x, int opt){
for(int i = Head[x]; i; i = Next[i]){
int y = to[i]; if(y == fa[x] || y == son[x]) continue;
dfs2(y, 0);
}
if(son[x]) dfs2(son[x], 1), vis[son[x]] = 1;
calc(x, 1), vis[son[x]] = 0;
for(auto it : Q[x])
ans[it.second] = f[it.first];
if(!opt) calc(x, -1);
}
int main(){
n = read(), m = read();
for(int i = 1; i <= n; ++i) c[i] = read();
for(int i = 1; i < n; ++i){
int u = read(), v = read();
add(u, v), add(v, u);
}
for(int i = 1; i <= m; ++i){
int u = read(), k = read();
Q[u].pb(mp(k, i));
}
dfs1(1, 0), dfs2(1, 0);
for(int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
return 0;
}
Ⅲ. P4149 [IOI2011] Race
树上两点 \(u,v\) 的距离为 \(dis_u + dis_v - 2 \times dis_{lca(u,v)}\),边的数量为 \(d_u + d_v - 2 \times d_{lca(u,v}\) 。
那么问题就转换成在树上找到两点 \(u,v\) 使得 \(dis_u+dis_v−2×dis_{lca(u,v)}=k\) 且 \(d_u+d_v−2×d_{lca(u,v)}\) 尽可能小。
于是我们可以定义 \(minn_d\) 表示当前距离 \(1\) 号点距离为 \(d\) 的节点的最小深度。
因为我们计算两点的简单路径权值和、两点简单路径的边的个数是根据它们的lca ,所以一个分支内的节点不能相互影响
所以需要先对一个分支统计完贡献后,再添加它的信息
#include<bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define pii pair<int,int>
#define int long long
using namespace std;
const int N = 5e5 + 67;
int read(){
int x = 0, f = 1; char ch = getchar();
while(ch < '0' || ch > '9'){if(ch == '-') f = -f; ch = getchar();}
while(ch >= '0' && ch <= '9'){x = (x << 1) + (x << 3) + (ch ^ 48); ch = getchar();}
return x * f;
}
int n, m, k, tot, ans = 0x3f3f3f3f3f3f3f3f;
int fa[N], sz[N], son[N], d[N], dis[N];
int Head[N], to[N], Next[N], edge[N];
map<int, int> minn;
void add(int u, int v, int w){
to[++tot] = v, Next[tot] = Head[u], Head[u] = tot, edge[tot] = w;
}
void dfs1(int x, int f){
fa[x] = f, sz[x] = 1, d[x] = d[f] + 1;
for(int i = Head[x]; i; i = Next[i]){
int y = to[i]; if(y == fa[x]) continue;
dis[y] = dis[x] + edge[i], dfs1(y, x), sz[x] += sz[y];
if(sz[y] > sz[son[x]]) son[x] = y;
}
}
void calc(int x, int rt){
int s = k + 2 * dis[rt] - dis[x];
if(minn.count(s)) ans = min(ans, minn[s] + d[x] - d[rt] * 2);
for(int i = Head[x]; i; i = Next[i]){
int y = to[i]; if(y == fa[x]) continue;
calc(y, rt);
}
}
void change(int x){
if(minn.count(dis[x])) minn[dis[x]] = min(minn[dis[x]], d[x]);
else minn[dis[x]] = d[x];
for(int i = Head[x]; i; i = Next[i]){
int y = to[i]; if(y == fa[x]) continue;
change(y);
}
}
void dfs2(int x, int opt){
for(int i = Head[x]; i; i = Next[i]){
int y = to[i]; if(y == fa[x] || y == son[x]) continue;
dfs2(y, 0);
}
if(son[x]) dfs2(son[x], 1);
int s = k + dis[x];
if(minn.count(s)) ans = min(ans, minn[s] - d[x]);minn[dis[x]] = d[x];
for(int i = Head[x]; i; i = Next[i]){
int y = to[i]; if(y == fa[x] || y == son[x]) continue;
calc(y, x), change(y);
}
if(!opt) minn.clear();
}
signed main(){
n = read(), k = read();
for(int i = 1; i < n; ++ i){
int u = read() + 1, v = read() + 1, w = read();
add(u, v, w), add(v, u, w);
}
dfs1(1, 0), dfs2(1, 0);
printf("%lld\n", ans == 0x3f3f3f3f3f3f3f3f ? -1 : ans);
return 0;
}
Ⅳ.CF741D Arpa’s letter-marked tree and Mehrdad’s Dokhtar-kosh paths
1.重排后构成回文的条件为:
①.每个字母出现的次数都为偶数 ②.一个字母出现次数为奇数,其余字母出现次数为偶数
2.字母的范围为 a ~ z , 把其转换成二进制状态(偶数为0,奇数为1)
那么满足回文条件的二进制为 : 00...000 , 00..001 , 00..010 , ... , 10..000
3.维护一个从根节点到子节点u的前缀异或和数组 X
那么 u 到 v 的简单路径的字母重排后的二进制形式为 X[u] ^ X[v]
4.节点 u 的答案有三种可能:
①.它的两个不同分支的节点构成的简单路径 ②.它的一个分支到它本身构成的简单路径 ③.它的子节点的 ans
定义 \(f_x\) 表示状态为 x 的节点的最大深度。
#include<bits/stdc++.h>
#define pb push_back
#define pii pair<int, int>
#define mp make_pair
using namespace std;
const int N = 5e5 + 67;
int read(){
int x = 0, f = 1; char ch = getchar();
while(ch < '0' || ch > '9'){if(ch == '-') f = -f; ch = getchar();}
while(ch >= '0' && ch <= '9'){x = (x << 1) + (x << 3) + (ch ^ 48); ch = getchar();}
return x * f;
}
bool _u;
int n, tot, maxn, XQ;
int hd[N], to[N << 1], nxt[N << 1];
int son[N], sz[N], dep[N], X[N];
int a[N], f[(1 << 22) + 67], ans[N];
void add(int u, int v){
to[++tot] = v, nxt[tot] = hd[u], hd[u] = tot;
}
void dfs1(int x, int ff, int cur){
dep[x] = dep[ff] + 1, sz[x] = 1; X[x] = cur ^ a[x];
for(int i = hd[x]; i; i = nxt[i]){
int y = to[i]; if(y == ff) continue;
dfs1(y, x, X[x]); sz[x] += sz[y];
if(sz[y] > sz[son[x]]) son[x] = y;
}
}
void calc(int x, int ff, int rt){
if(f[X[x]]) maxn = max(maxn, f[X[x]] + dep[x] - dep[rt] * 2);
if((X[x] ^ X[rt]) == 0) maxn = max(maxn, dep[x] - dep[rt]);
for(int i = 0; i <= 21; ++i){
if((X[x] ^ X[rt]) == (1 << i)) maxn = max(maxn, dep[x] - dep[rt]);
if(f[(X[x] ^ (1 << i))]) maxn = max(maxn, dep[x] + f[X[x] ^ (1 << i)] - dep[rt] * 2);
}
for(int i = hd[x]; i; i = nxt[i]){
int y = to[i]; if(y == ff || y == XQ) continue;
calc(y, x, rt);
}
}
void change(int x, int ff, int v){
if(v == 1) f[X[x]] = max(f[X[x]], dep[x]);
else f[X[x]] = 0;
for(int i = hd[x]; i; i = nxt[i]){
int y = to[i]; if(y == ff) continue;
change(y, x, v);
}
}
void dfs2(int x, int ff, int opt){
for(int i = hd[x]; i; i = nxt[i]){
int y = to[i]; if(y == ff || y == son[x]) continue;
dfs2(y, x, 0); ans[x] = max(ans[x], ans[y]);
}
if(son[x]) XQ = son[x], dfs2(son[x], x, 1), ans[x] = max(ans[x], ans[son[x]]);
if(f[X[x]]) maxn = max(maxn, f[X[x]] - dep[x]);
for(int i = 0; i <= 21; ++i)
if(f[X[x] ^ (1 << i)]) maxn = max(maxn, f[X[x] ^ (1 << i)] - dep[x]);
for(int i = hd[x]; i; i = nxt[i]){
int y = to[i]; if(y == ff || y == son[x]) continue;
calc(y, x, x), change(y, x, 1);
}
ans[x] = max(ans[x], maxn), f[X[x]] = max(f[X[x]], dep[x]), XQ = 0;
if(!opt) maxn = 0, change(x, ff, -1);
}
bool _v;
int main(){
cerr << abs(&_u - &_v) / 1048576.0 << " MB\n";
n = read();
for(int i = 2; i <= n; ++i){
int p = read(); char s;
cin >> s;
a[i] = 1 << (s - 'a');
add(p, i), add(i, p);
}
dfs1(1, 0, 0);
dfs2(1, 0, 0);
for(int i = 1; i <= n; ++i) printf("%d%c", ans[i], i == n ? '\n' : ' ');
return 0;
}
Ⅴ. P5290 [十二省联考 2019] 春节十二响
考虑有两条链的情况:将两条链弄成两个堆,每次取出两个堆的堆顶,取 \(max\) 加入答案,当一个堆取尽后,把另一个堆里的所有元素加入答案。最后再加入 \(a_x\)。
然后就可以用类似的方法合并子树,复杂度 \(O(n \log n)\)
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 2e5 + 67;
int read(){
int x = 0, f = 1; char ch = getchar();
while(ch < '0' || ch > '9'){if(ch == '-') f = -f; ch = getchar();}
while(ch >= '0' && ch <= '9'){x = (x << 1) + (x << 3) + (ch ^ 48); ch = getchar();}
return x * f;
}
int n;
ll ans;
int a[N];
vector<int> e[N], cur;
priority_queue<int> pq[N];
void merge(int x, int y){
if(pq[x].size() < pq[y].size()) swap(pq[x], pq[y]);
while(!pq[y].empty()){
cur.push_back(max(pq[x].top(), pq[y].top()));
pq[x].pop(), pq[y].pop();
}
while(!cur.empty()) pq[x].push(cur.back()), cur.pop_back();
}
void dfs(int x){
for(auto y : e[x]) dfs(y), merge(x, y);
pq[x].push(a[x]);
}
int main(){
n = read();
for(int i = 1; i <= n; ++i) a[i] = read();
for(int i = 2, f; i <= n; ++i) f = read(), e[f].push_back(i);
dfs(1);
while(!pq[1].empty()) ans += pq[1].top(), pq[1].pop();
printf("%lld\n", ans);
return 0;
}