權值線段樹
例題
【模板】普通平衡樹
#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 1;
int n, val[N], opt[N], num[N], cnt, len, san[N], m[N], rnk[N];
unordered_map<int, int> dfn;
struct WeightedSegmentTree{
#define ls (id << 1)
#define rs (id << 1 | 1)
struct node{ int l, r, num; }t[N<<2];
inline void pushup(int id){ t[id].num = t[ls].num + t[rs].num; }
inline void build(int id, int l, int r){
t[id].l = l, t[id].r = r;
if(l == r) return;
int mid = (l + r) >> 1;
build(ls, l, mid), build(rs, mid+1, r);
}
inline void modify(int id, int u, int v){
if(t[id].l == t[id].r) return (void)(t[id].num += v);
int mid = (t[id].l + t[id].r) >> 1;
modify(u<=mid?ls:rs, u, v);
pushup(id);
}
inline int query_rank(int id, int th){
if(t[id].l == t[id].r) return rnk[t[id].l];
if(th <= t[ls].num) return query_rank(ls, th);
else return query_rank(rs, th-t[ls].num);
}
inline int query_minnum(int id, int u){
if(t[id].l == t[id].r) return 1; // output the smallest rank
int mid = (t[id].l + t[id].r) >> 1;
if(u <= mid) return query_minnum(ls, u);
else return query_minnum(rs, u) + t[ls].num;
}
inline int query_maxnum(int id, int u){
if(t[id].l == t[id].r) return t[id].num; // output the biggest rank
int mid = (t[id].l + t[id].r) >> 1;
if(u <= mid) return query_maxnum(ls, u);
else return query_maxnum(rs, u) + t[ls].num;
}
inline int query_nxt(int id, int u){
return query_rank(1, query_minnum(1, u)-1);
}
inline int query_lst(int id, int u){
return query_rank(1, query_maxnum(1, u)+1);
}
}WST;
int main(){
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin>>n;
for(int i=1, op; i<=n; ++i){
cin>>op;
opt[i] = op, cin>>num[i];
if(op != 4 && op != 2)
val[++cnt] = num[i], san[cnt] = m[cnt] = val[cnt];
}
// 以下為離散化
sort(m+1, m+1+cnt);
len = unique(m+1, m+1+cnt) - (m+1);
for(int i=1; i<=cnt; ++i){
san[i] = lower_bound(m+1, m+1+len, san[i]) - m;
dfn[val[i]] = san[i], rnk[san[i]] = val[i];
}
WST.build(1, 1, cnt);
for(int i=1; i<=n; ++i){
switch(opt[i]){
case 1: WST.modify(1, dfn[num[i]], 1); break;
case 2: WST.modify(1, dfn[num[i]], -1); break;
case 3: cout<<WST.query_minnum(1, dfn[num[i]])<<'\n'; break;
case 4: cout<<WST.query_rank(1, num[i])<<'\n'; break;
case 5: cout<<WST.query_nxt(1, dfn[num[i]])<<'\n'; break;
case 6: cout<<WST.query_lst(1, dfn[num[i]])<<'\n'; break;
}
} return 0;
}
#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 1;
int n, m, f[N], rt[N], q, rnk[N];
inline int find(int k){
if(!f[k]) return k;
return f[k] = find(f[k]);
}
struct WeightedSegmentTree{
int ls[N<<6], rs[N<<6], sum[N<<6], cnt;
inline void pushup(int id){ sum[id] = sum[ls[id]] + sum[rs[id]]; }
inline void modify(int& id, int l, int r, int u){
if(!id) id = ++cnt;
if(l == r) return (void)(sum[id] = 1);
int mid = (l + r) >> 1;
if(u <= mid) modify(ls[id], l, mid, u);
else modify(rs[id], mid+1, r, u);
pushup(id);
}
inline int query(int id, int l, int r, int rk){
if(rk > sum[id]) return -1;
if(l == r) return rnk[l];
int mid = (l + r) >> 1;
if(rk <= sum[ls[id]]) return query(ls[id], l, mid, rk);
else return query(rs[id], mid+1, r, rk-sum[ls[id]]);
}
inline int merge(int a, int b, int l, int r){
if(!a && !b) return 0;
if(!a) return b; if(!b) return a;
if(l == r) return a;
int mid = (l + r) >> 1;
ls[a] = merge(ls[a], ls[b], l, mid);
rs[a] = merge(rs[a], rs[b], mid+1, r);
pushup(a);
return a;
}
}MST;
int main(){
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin>>n>>m;
for(int i=1, val; i<=n; ++i){
cin>>val, rnk[val] = i;
MST.modify(rt[i], 1, n, val);
}
for(int i=1, a, b; i<=m; ++i){
cin>>a>>b;
a = find(a), b = find(b);
// cout<<a<<' '<<b<<'\n';
rt[b] = MST.merge(rt[a], rt[b], 1, n);
f[b] = a;
}
// for(int i=1; i<=n; ++i) cout<<rt[i]<<' '; printf("\n");
// printf("%d", rt[find(2)]);
// return 0;
cin>>q; char opt;
for(int i=1, a, b; i<=q; ++i){
cin>>opt>>a>>b;
if(opt == 'Q') cout<<MST.query(rt[find(a)], 1, n, b)<<'\n';
else{
a = find(a), b = find(b);
if(a == b) continue;
f[b] = a;
rt[a] = MST.merge(rt[a], rt[b], 1, n);
}
} return 0;
}
[Vani有約會] 雨天的尾巴
考慮對每個點進行動態開點權值線段樹。對於樹鏈修改，採用樹上差分，最後把每個節點的子樹和自己合併起來就是最終答案(樹上差分不會可以看我部落格)。注意，需要特判這個節點的最大救濟糧數量，如果數量為 0，那麼救濟糧編號也為 0。
#include<bits/stdc++.h>
using namespace std;
#define min(x,y) (dfn[x] < dfn[y]) ? x : y
constexpr int N = 1e5 + 1;
int n, m, rt[N], dcnt, st[N][21], dfn[N], ans[N];
struct WeightedSegmentTree{
#define tls t[id].ls
#define trs t[id].rs
int cnt;
struct node{ int ls, rs, mx; }t[N<<6];
inline void pushup(int id){ t[id].mx = max(t[tls].mx, t[trs].mx); }
inline void modify(int& id, int l, int r, int u, int k){
if(!id) id = ++cnt;
if(l == r) return (void)(t[id].mx += k);
int mid = (l + r) >> 1;
if(u <= mid) modify(tls, l, mid, u, k);
else modify(trs, mid+1, r, u, k);
pushup(id);
}
inline int query(int id, int l, int r){
if(!id) return 0;
if(l == r){
if(t[id].mx == 0) return 0; // spj!!!
else return l;
}
int mid = (l + r) >> 1;
if(t[tls].mx >= t[trs].mx) return query(tls, l, mid);
else return query(trs, mid+1, r);
}
inline int merge(int a, int b, int l, int r){
if(!a || !b) return a+b;
if(l == r){ t[a].mx += t[b].mx; 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);
pushup(a);
return a;
}
}MST;
vector<int> G[N];
inline void dfs1(int u, int f){
st[dfn[u] = ++dcnt][0] = f;
for(int v : G[u]) if(!dfn[v]) dfs1(v, u);
}
inline int Lca(int u, int v){
if(u == v) return u;
if((u = dfn[u]) > (v = dfn[v])) swap(u, v);
int k = __lg(v - u++);
return min(st[u][k], st[v-(1<<k)+1][k]);
}
bitset<N> vis;
inline void dfs2(int u){
vis[u] = 1;
for(int v : G[u])
if(!vis[v]) dfs2(v), rt[u] = MST.merge(rt[u], rt[v], 1, N);
ans[u] = MST.query(rt[u], 1, N); // 注意必須這時候統計答案
}
int main(){
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin>>n>>m;
for(int i=1, a, b; i<n; ++i){
cin>>a>>b;
G[a].push_back(b), G[b].push_back(a);
}
dfs1(1, 0);
for(int j=1; j<=__lg(n); ++j)
for(int i=1; i<=n-(1<<j)+1; ++i)
st[i][j] = min(st[i][j-1], st[i+(1<<(j-1))][j-1]);
for(int i=1, a, b, c, lca; i<=m; ++i){
cin>>a>>b>>c; lca = Lca(a, b);
MST.modify(rt[a], 1, N, c, 1);
MST.modify(rt[b], 1, N, c, 1);
MST.modify(rt[lca], 1, N, c, -1);
MST.modify(rt[st[dfn[lca]][0]], 1, N, c, -1);
}
dfs2(1);
for(int i=1; i<=n; ++i) cout<<ans[i]<<'\n';
return 0;
}
蒟蒻的數列
服了，想複雜了。剛開始時在 modify 的同時統計 sum，於是就亂了。這道題需要動態開點維護 lazy_tag，最後再統計 sum 就非常簡單了。
#include<bits/stdc++.h>
using namespace std;
#define int long long
constexpr int N = 1e7 + 1, MAX = 1e9;
int n, rt;
struct SegmentTree{
int ls[N], rs[N], tag[N], cnt;
inline void addtag(int& id, int v){ if(!id) id = ++cnt; tag[id] = max(tag[id], v); }
inline void pushdown(int id){ if(tag[id]) addtag(ls[id], tag[id]), addtag(rs[id], tag[id]); tag[id] = 0; }
inline void modify(int& id, int x, int y, int l, int r, int k){
if(!id) id = ++cnt;
if(l <= x && y <= r){ addtag(id, k); return; }
pushdown(id);
int mid = (x + y) >> 1;
if(l <= mid) modify(ls[id], x, mid, l, r, k);
if(r > mid ) modify(rs[id], mid+1, y, l, r, k);
}
inline int query(int id, int l, int r){
if(!ls[id] && !rs[id]) return tag[id]*(r-l+1);
pushdown(id);
int mid = (l + r) >> 1, ans = 0;
if(ls[id]) ans += query(ls[id], l, mid);
if(rs[id]) ans += query(rs[id], mid+1, r);
return ans;
}
}ST;
signed main(){
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin>>n;
for(int i=1, a, b, c; i<=n; ++i){
cin>>a>>b>>c;
if(a <= b-1) ST.modify(rt, 1, MAX, a, b-1, c);
}
return cout<<ST.query(1, 1, MAX), 0;
}
[NOIP2016 提高組] 天天愛跑步
集樹上差分與線段樹合併一體的好題！看題解看的
首先，我們只需要記錄這個點在某個時間點上有沒有人即可，權值線段樹即可解決。但是又有新的問題：這需要沿著某條樹鏈 \(0:+1\)，\(1:+1\)，\(2:+1\) 等等，無法差分，並且最後合併的時候子節點的時間資料會加到父節點上，這是錯誤的。
可以發現，隨著時間的推移，節點的深度也在隨之變化，那麼就可以把時間資訊與深度掛鉤。對於 \(u -> lca\) 的節點，把他們的 \(dep[u]:+1\)；對於 \(lca -> u\) 的節點，把他們的 \(dep[lca]*2-dep[a]:+1\)，最後統計的時候統計每個節點的 \(dep[u]+val[u]\) 和 \(dep[u]-val[u]\) 的權值即可。最後注意判斷 \(val[u]\) 是否等於 0，若等於 0，ans 就會加兩次。
#include<bits/stdc++.h>
using namespace std;
constexpr int N = 3e5, M = 6e5;
int n, m, val[N], fa[N], size[N], son[N], dep[N], rt[N], ans[N], top[N];
vector<int> G[N];
inline void dfs1(int u){
size[u] = 1, son[u] = -1;
for(int v : G[u]){
if(!dep[v]){
dep[v] = dep[fa[v]=u] + 1;
dfs1(v);
size[u] += size[v];
if(son[u] == -1 || size[v] > size[son[u]]) son[u] = v;
}
}
}
inline void dfs2(int u, int t){
top[u] = t;
if(son[u] == -1) return;
dfs2(son[u], t);
for(int v : G[u]) if(v != son[u] && v != fa[u]) dfs2(v, v);
}
inline int Lca(int u, int v){
int tu = top[u], tv = top[v];
while(tu != tv){
if(dep[tu] > dep[tv]) tu = top[u=fa[tu]];
else tv = top[v=fa[tv]];
} return dep[u]>dep[v]?v:u;
}
struct WeightedSegmentTree{
int ls[N<<5], rs[N<<5], sum[N<<5], cnt;
inline void modify(int &id, int l, int r, int u, int k){
if(!id) id = ++cnt;
if(l == r) return (void)(sum[id] += k);
int mid = (l + r) >> 1;
if(u <= mid) modify(ls[id], l, mid, u, k);
else modify(rs[id], mid+1, r, u, k);
}
inline int query(int id, int l, int r, int u){
if(!id) return 0;
if(l == r) return sum[id];
int mid = (l + r) >> 1;
if(u <= mid) return query(ls[id], l, mid, u);
else return query(rs[id], mid+1, r, u);
}
inline int merge(int a, int b, int l, int r){
if(!a || !b) return a+b;
if(l == r){ sum[a] += sum[b]; return a; }
int mid = (l + r) >> 1;
ls[a] = merge(ls[a], ls[b], l, mid);
rs[a] = merge(rs[a], rs[b], mid+1, r);
sum[a] = sum[ls[a]] + sum[rs[a]];
return a;
}
}WST;
bitset<N> vis;
inline void dfs(int u){
vis[u] = 1;
for(int v : G[u])
if(!vis[v]) dfs(v), rt[u] = WST.merge(rt[u], rt[v], 1, M<<1);
if(val[u]) ans[u] += WST.query(rt[u], 1, M<<1, dep[u]+val[u]+M);
ans[u] += WST.query(rt[u], 1, M<<1, dep[u]-val[u]+M);
}
int main(){
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin>>n>>m;
for(int i=1, a, b; i<n; ++i) cin>>a>>b, G[a].push_back(b), G[b].push_back(a);
for(int i=1; i<=n; ++i) cin>>val[i];
dep[1] = 1, dfs1(1); dfs2(1, 1);
for(int i=1, a, b, lca; i<=m; ++i){
cin>>a>>b; lca = Lca(a, b);
WST.modify(rt[a], 1, M<<1, dep[a]+M, 1);
WST.modify(rt[fa[lca]], 1, M<<1, dep[a]+M, -1);
WST.modify(rt[b], 1, M<<1, dep[lca]*2-dep[a]+M, 1);
WST.modify(rt[lca], 1, M<<1, dep[lca]*2-dep[a]+M, -1);
} dfs(1);
for(int i=1; i<=n; ++i) cout<<ans[i]<<' ';
return 0;
}
[USACO17JAN] Promotion Counting P
不說了，板子。
#include<bits/stdc++.h>
using namespace std;
constexpr int N = 1e5 + 1, M = 9e6 + 1, MAX = 1e9+1;
int n, rt[N], ans[N], val[N];
vector<int> G[N];
struct WeightedSegmentTree{
int cnt, ls[M], rs[M], sum[M];
inline void modify(int &id, int l, int r, int u){
if(!id) id = ++cnt;
if(l == r) return (void)(++sum[id]);
int mid = (l + r) >> 1;
if(u <= mid) modify(ls[id], l, mid, u);
else modify(rs[id], mid+1, r, u);
sum[id] = sum[ls[id]] + sum[rs[id]];
}
inline int query(int id, int l, int r, int u){
if(!id) return 0;
if(l == r) return sum[id];
int mid = (l + r) >> 1, ans = 0;
if(u <= mid) ans += sum[rs[id]] + query(ls[id], l, mid, u);
else ans += query(rs[id], mid+1, r, u);
return ans;
}
inline int merge(int a, int b, int l, int r){
if(!a || !b) return a+b;
if(l == r){ sum[a] += sum[b]; return a; }
int mid = (l + r) >> 1;
ls[a] = merge(ls[a], ls[b], l, mid);
rs[a] = merge(rs[a], rs[b], mid+1, r);
sum[a] = sum[ls[a]] + sum[rs[a]];
return a;
}
}WST;
bitset<N> vis;
inline void dfs(int u){
vis[u] = 1;
for(int v : G[u]) if(!vis[v]) dfs(v), rt[u] = WST.merge(rt[u], rt[v], 1, MAX);
ans[u] = WST.query(rt[u], 1, MAX, val[u]+1);
}
int main(){
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin>>n;
for(int i=1; i<=n; ++i) cin>>val[i], WST.modify(rt[i], 1, MAX, val[i]);
for(int i=2, a; i<=n; ++i) cin>>a, G[i].push_back(a), G[a].push_back(i);
dfs(1); for(int i=1; i<=n; ++i) cout<<ans[i]<<'\n'; return 0;
}
魔法少女LJJ
\(c<=7\)
好惡心啊，調了我一晚上 + 一下午
對於需要維護 tag 的動態開點線段樹，對於這道題了來說，涉及區間修改的只有 把某一區間修改為 0。所以不開點不打 tag 不影響接續的修改，所以沒開過的點就不要打 tag 了，會 T。另外開過的點一定打 tag。對於合併操作，不要試圖維護 tag，直接 pushdown 即可。
#include<bits/stdc++.h>
using namespace std;
constexpr int N = 4e5 + 1, M = 1e7 + 1, MAX = 1e9 + 1;
int m, tot, num[N], rt[N], f[N];
inline int find(int k){
if(!f[k]) return k;
return f[k] = find(f[k]);
}
struct WeightedSegmentTree{
int ls[M], rs[M], sum[M], cnt=1;
double mul[M]; bitset<M> tag;
inline void addtag(int &id){
tag[id] = 0; sum[id] = mul[id] = 0;
}
inline void pushdown(int id){
if(!tag[id]) addtag(ls[id]), addtag(rs[id]);
tag[id] = 1;
}
inline void pushup(int id){
sum[id] = sum[ls[id]] + sum[rs[id]];
mul[id] = mul[ls[id]] + mul[rs[id]];
}
inline void modify_itv(int &id, int x, int y, int l, int r){
if(l > r || !id) return; //因為修改為0，所以不開點不打tag不影響接續的修改
if(l <= x && y <= r) return (void)(addtag(id)); //開過的點一定打 tag
int mid = (x + y) >> 1; pushdown(id);
if(l <= mid) modify_itv(ls[id], x, mid, l, r);
if(r > mid) modify_itv(rs[id], mid+1, y, l, r);
pushup(id);
}
inline void modify_nod(int &id, int l, int r, int u, int k){
if(!id) id = ++cnt, tag[id] = 1;
if(l == r) return (void)(sum[id]+=k, mul[id]+=log(u)*k);
int mid = (l + r) >> 1; pushdown(id);
if(u <= mid) modify_nod(ls[id], l, mid, u, k);
else modify_nod(rs[id], mid+1, r, u, k);
pushup(id);
}
inline int query_rnk(int id, int l, int r, int k){
if(!id || k > sum[id]) return 0;
if(l == r) return l;
int mid = (l + r) >> 1; pushdown(id);
if(k <= sum[ls[id]]) return query_rnk(ls[id], l, mid, k);
else return query_rnk(rs[id], mid+1, r, k-sum[ls[id]]);
}
inline int query_sum(int id, int x, int y, int l, int r){
if(!id || l > r) return 0;
if(l <= x && y <= r) return sum[id];
int mid = (x + y) >> 1, ans = 0; pushdown(id);
if(l <= mid) ans += query_sum(ls[id], x, mid, l, r);
if(r > mid) ans += query_sum(rs[id], mid+1, y, l, r);
return ans;
}
inline int merge(int a, int b, int l, int r){
if(!a || !b) return a+b; //²»ÒªºÏ²¢tag ûÓÐÓÃ
if(l == r){ sum[a] += sum[b]; return a; }
int mid = (l + r) >> 1; pushdown(a), pushdown(b);
sum[a] += sum[b], mul[a] += mul[b];
ls[a] = merge(ls[a], ls[b], l, mid);
rs[a] = merge(rs[a], rs[b], mid+1, r);
return a;
}
inline void change_max(int &id, int u){
int s = query_sum(id, 1, MAX, u+1, MAX);
if(!s) return;
modify_itv(id, 1, MAX, u+1, MAX);
modify_nod(id, 1, MAX, u, s);
}
inline void change_min(int &id, int u){
int s = query_sum(id, 1, MAX, 1, u-1);
if(!s) return;
modify_itv(id, 1, MAX, 1, u-1);
modify_nod(id, 1, MAX, u, s);
}
}WST;
int main(){
ios::sync_with_stdio(0), cout.tie(0), cout.tie(0);
cin>>m;
for(int i=1, opt, a, b, fa, fb; i<=m; ++i){
cin>>opt>>a;
switch(opt){
case 1:
num[++tot] = 1; WST.modify_nod(rt[tot], 1, MAX, a, 1); break;
case 2:
cin>>b; fa = find(a), fb = find(b); if(fa == fb) break;
f[fb] = fa; rt[fa] = WST.merge(rt[fa], rt[fb], 1, MAX);
num[fa] += num[fb]; break;
case 3:
cin>>b; WST.change_min(rt[find(a)], b); break;
case 4:
cin>>b; WST.change_max(rt[find(a)], b); break;
case 5:
cin>>b; fa = find(a); if(b > num[fa]){ cout<<"0\n"; break; }
cout<<WST.query_rnk(rt[fa], 1, MAX, b)<<'\n'; break;
case 6:
cin>>b; fa = find(a), fb = find(b);
if(fa == fb){ cout<<"0\n"; break; }
cout<<(WST.mul[rt[fa]]>WST.mul[rt[fb]]?1:0)<<'\n'; break;
default: cout<<num[find(a)]<<'\n';
}
} return 0;
}