线段树进阶应用学习笔记（二）：动态开点线段树相关
动态开点线段树
回忆普通线段树，它的空间一般要开到数组长度的 (因此经常MLE)，考虑如何优化它。
权值线段树
线段树合并
算法流程
复杂度分析
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 9, LOGN = 22;
struct Edge{
int v, nex;
} e[N << 1];
int head[N], ecnt;
void addEdge(int u, int v){
e[++ecnt] = Edge{v, head[u]};
head[u] = ecnt;
}
struct Node{
int ls, rs, sum, res;
} t[50 * N];
int ans[N], cnt;
void update(int x){
if(t[t[x].ls].sum < t[t[x].rs].sum) {
t[x].res = t[t[x].rs].res;
t[x].sum = t[t[x].rs].sum;
} else {
t[x].res = t[t[x].ls].res;
t[x].sum = t[t[x].ls].sum;
}
}
int modify(int a, int b, int l, int r) {
if(!a)
return b;
if(!b)
return a;
if(l == r) {
t[a].sum += t[b].sum;
return a;
}
int mid = (l + r) >> 1;
t[a].ls = modify(t[a].ls, t[b].ls, l, mid);
t[a].rs = modify(t[a].rs, t[b].rs, mid + 1, r);
update(a);
return a;
}
int Build(int id, int l, int r, int co, int z) {
if(!id)
id = ++cnt;
if(l == r) {
t[id].sum += z;
t[id].res = co;
return id;
}
int mid = (l + r) >> 1;
if(co <= mid)
t[id].ls = Build(t[id].ls, l, mid, co, z);
else
t[id].rs = Build(t[id].rs, mid + 1, r, co, z);
update(id);
return id;
}
int fa[N][LOGN], dep[N], rt[N];
void dfs1(int u){
for(int i = head[u]; i; i = e[i].nex){
int v = e[i].v;
if(v != fa[u][0]){
fa[v][0] = u;
dep[v] = dep[u] + 1;
dfs1(v);
}
}
}
int query(int x, int y){
if(dep[x] > dep[y])
swap(x, y);
int d = dep[y] - dep[x];
int t = 0;
while(d){
if(d & 1)
y = fa[y][t];
++t;
d >>= 1;
}
t = 0;
while(fa[x][t] != fa[y][t])
++t;
--t;
while(t >= 0){
if(fa[x][t] != fa[y][t]){
x = fa[x][t];
y = fa[y][t];
}
--t;
}
if(x != y){
x = fa[x][0];
y = fa[y][0];
}
return x;
}
void dfs2(int u) {
for(int i = head[u]; i; i = e[i].nex) {
int v = e[i].v;
if(v == fa[u][0])
continue;
dfs2(v);
rt[u] = modify(rt[u], rt[v], 1, 100000);
}
ans[u] = t[rt[u]].res;
if(t[rt[u]].sum == 0)
ans[u] = 0;
}
int n, m;
int main(){
scanf("%d%d", &n, &m);
for(int i = 1; i < n; i++){
int u, v;
scanf("%d%d", &u, &v);
addEdge(u, v);
addEdge(v, u);
}
dfs1(1);
for(int j = 1, j2 = 2; j2 <= n; ++j, j2 <<= 1)
for(int i = 1; i <= n; ++i)
fa[i][j] = fa[fa[i][j - 1]][j - 1];
for(int i = 1; i <= m; i++){
int x, y, z;
scanf("%d%d%d", &x, &y, &z);
int lca = query(x, y);
rt[x] = Build(rt[x], 1, 100000, z, 1);
rt[y] = Build(rt[y], 1, 100000, z, 1);
rt[lca] = Build(rt[lca], 1, 100000, z, -1);
if(fa[lca][0])
rt[fa[lca][0]] = Build(rt[fa[lca][0]], 1, 100000, z, -1);
}
dfs2(1);
for(int i = 1; i <= n; i++)
printf("%d\n", ans[i]);
return 0;
}
线段树分裂
算法流程
复杂度分析
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N = 2e5 + 9;
struct Tree{
int ls, rs, sum;
} t[N << 5];
int rub[N << 5], rt[N << 2], cnt, tot, n, m, idx = 1;
int New(){
return cnt ? rub[cnt--] : ++tot;
}
void del(int &p){
t[p].ls = t[p].rs = t[p].sum = 0;
rub[++cnt] = p;
p = 0;
}
int build(int id, int L, int R) {
if(!id)
id = New();
if(L == R){
scanf("%lld", &t[id].sum);
return id;
}
int mid = (L + R) >> 1;
t[id].ls = build(t[id].ls, L, mid);
t[id].rs = build(t[id].rs, mid + 1, R);
t[id].sum = t[t[id].ls].sum + t[t[id].rs].sum;
return id;
}
int update(int id, int L, int R, int co, int z) {
if(!id)
id = New();
if(L == R) {
t[id].sum += z;
return id;
}
int mid = (L + R) >> 1;
if(co <= mid)
t[id].ls = update(t[id].ls, L, mid, co, z);
else
t[id].rs = update(t[id].rs, mid + 1, R, co, z);
t[id].sum = t[t[id].ls].sum + t[t[id].rs].sum;
return id;
}
int merge(int a, int b, int L, int R) {
if(!a)
return b;
if(!b)
return a;
if(L == R) {
t[a].sum += t[b].sum;
del(b);
return a;
}
int mid = (L + R) >> 1;
t[a].ls = merge(t[a].ls, t[b].ls, L, mid);
t[a].rs = merge(t[a].rs, t[b].rs, mid + 1, R);
t[a].sum = t[t[a].ls].sum + t[t[a].rs].sum;
del(b);
return a;
}
void split(int &p, int &q, int L, int R, int qL, int qR){
if(!p)
return;
if(L == qL && R == qR){
q = p;
p = 0;
return;
}
if(!q)
q = New();
int mid = (L + R) >> 1;
if(qR <= mid)
split(t[p].ls, t[q].ls, L, mid, qL, qR);
else if(qL > mid)
split(t[p].rs, t[q].rs, mid + 1, R, qL, qR);
else{
split(t[p].ls, t[q].ls, L, mid, qL, mid);
split(t[p].rs, t[q].rs, mid + 1, R, mid + 1, qR);
}
t[p].sum = t[t[p].ls].sum + t[t[p].rs].sum;
t[q].sum = t[t[q].ls].sum + t[t[q].rs].sum;
}
int query(int id, int L, int R, int qL, int qR){
if(!id)
return 0;
if(L == qL && R == qR)
return t[id].sum;
int ans = 0;
int mid = (L + R) >> 1;
if(qR <= mid)
ans += query(t[id].ls, L, mid, qL, qR);
else if(qL > mid)
ans += query(t[id].rs, mid + 1, R, qL, qR);
else
ans += query(t[id].ls, L, mid, qL, mid) + query(t[id].rs, mid + 1, R, mid + 1, qR);
return ans;
}
int kth(int id, int L, int R, int q){
if(L == R)
return L;
int mid = (L + R) >> 1;
int l = t[t[id].ls].sum;
if(q <= l)
return kth(t[id].ls, L, mid, q);
else
return kth(t[id].rs, mid + 1, R, q - l);
}
signed main(){
scanf("%lld%lld", &n, &m);
rt[1] = build(rt[1], 1, n);
while(m--){
int opt, p, x, y, tt, q, k;
scanf("%lld", &opt);
if(opt == 0){
scanf("%lld%lld%lld", &p, &x, &y);
split(rt[p], rt[++idx], 1, n, x, y);
} else if(opt == 1){
scanf("%lld%lld", &p, &tt);
rt[p] = merge(rt[p], rt[tt], 1, n);
} else if(opt == 2){
scanf("%lld%lld%lld", &p, &x, &q);
rt[p] = update(rt[p], 1, n, q, x);
} else if(opt == 3){
scanf("%lld%lld%lld", &p, &x, &y);
printf("%lld\n", query(rt[p], 1, n, x, y));
} else {
scanf("%lld%lld", &p, &k);
if(t[rt[p]].sum < k)
printf("-1\n");
else
printf("%lld\n", kth(rt[p], 1, n, k));
}
}
return 0;
}
本文来自博客园，作者：JPGOJCZX，转载请注明原文链接：https://www.cnblogs.com/JPGOJCZX/p/18555996
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