权值线段树
例题
【模板】普通平衡树
#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 rankint 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 rankint 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];}// 以下为离散化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有约会] 雨天的尾巴
考虑对每个点进行动态开点权值线段树。对于树链修改，采用树上差分，最后把每个节点的子树和自己合并起来就是最终答案(树上差分不会可以看我博客)。注意，需要特判这个节点的最大救济粮数量，如果数量为 0，那么救济粮编号也为 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].rsint 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); // 注意必须这时候统计答案
}
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;
}
蒟蒻的数列
服了，想复杂了。刚开始时在 modify 的同时统计 sum，于是就乱了。这道题需要动态开点维护 lazy_tag，最后再统计 sum 就非常简单了。
#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 提高组] 天天爱跑步
集树上差分与线段树合并一体的好题！看题解看的
首先，我们只需要记录这个点在某个时间点上有没有人即可，权值线段树即可解决。但是又有新的问题：这需要沿着某条树链 \(0:+1\)，\(1:+1\)，\(2:+1\) 等等，无法差分，并且最后合并的时候子节点的时间数据会加到父节点上，这是错误的。
可以发现，随着时间的推移，节点的深度也在随之变化，那么就可以把时间信息与深度挂钩。对于 \(u -> lca\) 的节点，把他们的 \(dep[u]:+1\)；对于 \(lca -> u\) 的节点，把他们的 \(dep[lca]*2-dep[a]:+1\)，最后统计的时候统计每个节点的 \(dep[u]+val[u]\) 和 \(dep[u]-val[u]\) 的权值即可。最后注意判断 \(val[u]\) 是否等于 0，若等于 0，ans 就会加两次。
#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
不说了，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
\(c<=7\)
好恶心啊，调了我一晚上 + 一下午
对于需要维护 tag 的动态开点线段树，对于这道题了来说，涉及区间修改的只有 把某一区间修改为 0。所以不开点不打 tag 不影响接续的修改，所以没开过的点就不要打 tag 了，会 T。另外开过的点一定打 tag。对于合并操作，不要试图维护 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; //因为修改为0，所以不开点不打tag不影响接续的修改 if(l <= x && y <= r) return (void)(addtag(id)); //开过的点一定打 tagint 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 ûÓÐÓà 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;
}
