李超线段树合集
李超线段树
用来维护二维平面上很多条线段(直线)在x = x0上的最值问题
定义：
1.永久化标记：即线段树标记不删除，每个结点维护的也不一定是最优的信息，需要查询的时候一路统计标记达到最优
2.优势线段：李超树每个节点含有一个优势线段，意为完全覆盖当前区间且在当前区间mid处相比于其他该位置线段最大(小)的线段，李超树的id记录的即为当前的优势线段
3.核心思想：假设维护最大值，每次插入线段时，如果斜率较大的线段为优势线段，则斜率较小的线段只有在左子树才有机会比优势线段大
如果斜率较小的线段为优势线段，则较大的线段只有在右子树才能翻盘
修改(log(n))²(但是常数不大),查询log(n)
1.luogu4254
操作1.增加一条斜率为P,起始点为S的直线
操作2.询问k点上所有直线的最大值
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 5e4 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; struct Tree{ int l,r; double S,P; //P为斜率,S为初始值 }tree[maxn << 2]; void Build(int t,int l,int r){ tree[t].l = l; tree[t].r = r; tree[t].S = tree[t].P = 0; if(l == r) return; int m = l + r >> 1; Build(t << 1,l,m); Build(t << 1 | 1,m + 1,r); } void update(int t,double S,double P){ int mid = tree[t].l + tree[t].r >> 1; if(P > tree[t].P){ swap(tree[t].P,P); swap(tree[t].S,S); } if(S + P * mid < tree[t].S + tree[t].P * mid){ if(tree[t].l == tree[t].r) return; update(t << 1,S,P); }else{ swap(tree[t].P,P); swap(tree[t].S,S); if(tree[t].l == tree[t].r) return; update(t << 1 | 1,S,P); } } double ans; void query(int t,int k){ int mid = tree[t].l + tree[t].r >> 1; ans = max(ans,k * tree[t].P + tree[t].S); if(tree[t].l == tree[t].r) return ; if(k <= mid) query(t << 1,k); else query(t << 1 | 1,k); } char op[10]; int main(){ Sca(N); Build(1,1,50000); for(int i = 1; i <= N ; i ++){ scanf("%s",op); if(op[0] == 'Q'){ ans = 0; query(1,read()); Pri((int)(ans / 100)); }else{ double S,P; scanf("%lf%lf",&S,&P); S -= P; update(1,S,P); } } return 0; }
2.luogu4097
- 在平面上加入一条线段。记第 i 条被插入的线段的标号为 i
- 给定一个数 k,询问与直线 x = k 相交的线段中，交点最靠上的线段的编号。
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 4e4 + 10; const int maxm = 1e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; double S[maxm],P[maxm]; struct Tree{ int l,r; int id; }tree[maxn << 2]; void Build(int t,int l,int r){ tree[t].l = l; tree[t].r = r; tree[t].id = 0; if(l == r) return; int m = l + r >> 1; Build(t << 1,l,m); Build(t << 1 | 1,m + 1,r); } bool dcmp(double a){ return fabs(a) > eps; } void update(int t,int l,int r,int id){ int m = (tree[t].l + tree[t].r) >> 1; if(l <= tree[t].l && tree[t].r <= r){ int &a = id; int &b = tree[t].id; if(P[a] > P[b]) swap(a,b); if(m * P[a] + S[a] > m * P[b] + S[b] || (!dcmp(m * P[a] + S[a] - m * P[b] + S[b]) && a < b)){ swap(a,b); if(tree[t].l == tree[t].r) return; update(t << 1 | 1,l,r,id); }else{ if(tree[t].l == tree[t].r) return; update(t << 1,l,r,id); } return; } if(r <= m) update(t << 1,l,r,id); else if(l > m) update(t << 1 | 1,l,r,id); else{ update(t << 1,l,m,id); update(t << 1 | 1,m + 1,r,id); } } int Id; void query(int t,int k){ int a = tree[t].id; double Ans = P[Id] * k + S[Id]; if(Ans < P[a] * k + S[a] || (!dcmp(Ans - P[a] * k - S[a]) && Id > a)){ Id = a; } if(tree[t].l == tree[t].r) return; int m = (tree[t].l + tree[t].r) >> 1; if(k <= m) query(t << 1,k); else query(t << 1 | 1,k); } int main(){ Sca(N); Id = 0; int tot = 0; Build(1,1,40000); for(int i = 1; i <= N ; i ++){ int op = read(); if(!op){ int k = (read() + Id - 1) % 39989 + 1; Id = 0; query(1,k); Pri(Id); }else{ int x0 = (read() + Id - 1) % 39989 + 1; int y0 = (read() + Id - 1) % 1000000000 + 1; int x1 = (read() + Id - 1) % 39989 + 1; int y1 = (read() + Id - 1) % 1000000000 + 1; tot++; if(x0 == x1){ P[tot] = 0; S[tot] = max(y1,y0); }else{ P[tot] = 1.0 * (y1 - y0) / (x1 - x0); S[tot] = y1 - x1 * P[tot]; } update(1,min(x0,x1),max(x0,x1),tot); } } return 0; }
3.luogu p4069
题意：树链上每个点添加形如a * dis + b的值的一个点，每次查询树链上最小点
树链剖分 + 李超树
关于李超树的区间查询：
用一个Min维护区间最小值,Min的组成来源为两个子树的Min的较小值和本结点优势线段在两个端点的较小值
如果查询包含整个区间则直接返回Min，否则来源为正常区间最小值查找 + 本结点优势线段在查询两端的较小值
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 1e5 + 10; const LL INF = 123456789123456789; const int mod = 1e9 + 7; int N,M,K; struct Edge{ int to,next; LL dis; }edge[maxn * 2]; int head[maxn],tot; void init(){ for(int i = 0 ; i <= N ; i ++) head[i] = -1; tot = 0; } void add(int u,int v,LL w){ edge[tot].to = v; edge[tot].next = head[u]; edge[tot].dis = w; head[u] = tot++; } int dep[maxn],top[maxn],fa[maxn],pos[maxn]; int sz[maxn],son[maxn],to[maxn]; LL dis[maxn]; void dfs1(int t,int la){ sz[t] = 1; son[t] = t; int heavy = 0; for(int i = head[t]; ~i ; i = edge[i].next){ int v = edge[i].to; if(v == la) continue; dis[v] = dis[t] + edge[i].dis; dep[v] = dep[t] + 1; fa[v] = t; dfs1(v,t); if(sz[v] > heavy){ heavy = sz[v]; son[t] = v; } sz[t] += sz[v]; } } int cnt; void dfs2(int t,int la){ top[t] = la; pos[t] = ++cnt; to[cnt] = t; if(son[t] == t) return; dfs2(son[t],la); for(int i = head[t]; ~i ; i = edge[i].next){ int v = edge[i].to; if((fa[t] == v) || (v == son[t])) continue; dfs2(v,v); } } LL S[maxn * 2],P[maxn * 2]; int ttt; struct Tree{ int l,r,id; LL Min; }tree[maxn << 2]; void Pushup(int t){ if(tree[t].l == tree[t].r) return; tree[t].Min = min(tree[t].Min,min(tree[t << 1].Min,tree[t << 1 | 1].Min)); } void Build(int t,int l,int r){ tree[t].l = l; tree[t].r = r; tree[t].id = 0; tree[t].Min = INF; if(l == r) return; int m = l + r >> 1; Build(t << 1,l,m); Build(t << 1 | 1,m + 1,r); } void update(int t,int l,int r,int id){ int m = tree[t].l + tree[t].r >> 1; if(l <= tree[t].l && tree[t].r <= r){ int &a = id; int &b = tree[t].id; if(P[a] > P[b]) swap(a,b); if(dis[to[m]] * P[a] + S[a] < dis[to[m]] * P[b] + S[b]){ swap(a,b); if(tree[t].l != tree[t].r) update(t << 1,l,r,id); }else{ if(tree[t].l != tree[t].r) update(t << 1 | 1,l,r,id); } Pushup(t); tree[t].Min = min(tree[t].Min,dis[to[tree[t].l]] * P[b] + S[b]); tree[t].Min = min(tree[t].Min,dis[to[tree[t].r]] * P[b] + S[b]); return; } if(r <= m) update(t << 1,l,r,id); else if(l > m) update(t << 1 | 1,l,r,id); else{ update(t << 1,l,m,id); update(t << 1 | 1,m + 1,r,id); } Pushup(t); } LL query(int t,int l,int r){ if(l <= tree[t].l && tree[t].r <= r) return tree[t].Min; int m = (tree[t].l + tree[t].r) >> 1; LL ans = INF; ans = min(ans,dis[to[l]] * P[tree[t].id] + S[tree[t].id]); ans = min(ans,dis[to[r]] * P[tree[t].id] + S[tree[t].id]); if(r <= m) return min(ans,query(t << 1,l,r)); else if(l > m) return min(ans,query(t << 1 | 1,l,r)); else{ return min(ans,min(query(t << 1,l,m),query(t << 1 | 1,m + 1,r))); } } int lca(int x,int y){ while(top[x] != top[y]) dep[top[x]] > dep[top[y]]?x = fa[top[x]]:y = fa[top[y]]; return dep[x] < dep[y]?x:y; } void update(int u,int v,int id){ while(top[u] != top[v]){ if(dep[top[u]] < dep[top[v]]) swap(u,v); update(1,pos[top[u]],pos[u],id); u = fa[top[u]]; } if(dep[u] > dep[v]) swap(u,v); update(1,pos[u],pos[v],id); } LL query(int u,int v){ LL ans = INF; while(top[u] != top[v]){ if(dep[top[u]] < dep[top[v]]) swap(u,v); ans = min(ans,query(1,pos[top[u]],pos[u])); u = fa[top[u]]; } if(dep[u] > dep[v]) swap(u,v); ans = min(ans,query(1,pos[u],pos[v])); return ans; } int main(){ Sca2(N,M); init(); P[0] = 0; S[0] = INF; for(int i = 1; i <= N - 1; i ++){ int u = read(),v = read(); LL w = read(); add(u,v,w); add(v,u,w); } int root = 1; dfs1(root,-1); cnt = 0; dfs2(root,root); Build(1,1,N); ttt = 0; while(M--){ int op = read(); if(op == 1){ int s = read(),t = read(); LL a = read(),b = read(); int x = lca(s,t); ttt++; P[ttt] = -a; S[ttt] = a * dis[s] + b; update(s,x,ttt); ttt++; P[ttt] = a; S[ttt] = a * (dis[s] - 2 * dis[x]) + b; update(x,t,ttt); }else{ int s = read(),t = read(); Prl(query(s,t)); } } return 0; }
4.2019ICPC南京网络赛I
比赛的时候一直想寻找一个维护斜率最大值的数据结构，结果没有想到
看了题解才知道真有直接维护这个的数据结构
李超树的下标代表的是洗衣机花费的时间，每个x上的最大值是当前后缀需要花费的最大值时间
动态加入后缀 维护最大值即可
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 1e6 + 10; const LL INF = 1e18; const int mod = 1e9 + 7; int N,M,K; LL S[maxn],P[maxn]; LL a[maxn]; int cnt; struct Tree{ int l,r,id; }tree[maxn << 2]; void Build(int t,int l,int r){ tree[t].l = l; tree[t].r = r; tree[t].id = 0; if(l == r) return; int m = l + r >> 1; Build(t << 1,l,m); Build(t << 1 | 1,m + 1,r); } void update(int t,int id){ int &a = id;int &b = tree[t].id; if(P[a] > P[b]) swap(a,b); int m = tree[t].l + tree[t].r >> 1; if(P[a] * m + S[a] > P[b] * m + S[b]){ swap(a,b); if(tree[t].l != tree[t].r)update(t << 1 | 1,id); }else{ if(tree[t].l != tree[t].r)update(t << 1,id); } } bool cmp(LL a,LL b){ return a > b; } LL query(int t,int k){ int a = tree[t].id; LL ans = P[a] * k + S[a]; if(tree[t].l == tree[t].r) return ans; int m = tree[t].l + tree[t].r >> 1; if(k <= m) return max(query(t << 1,k),ans); else return max(query(t << 1 | 1,k),ans); } void add(int t){ cnt++; S[cnt] = a[t]; P[cnt] = t; update(1,cnt); } LL ans[maxn]; int main(){ while(~Sca2(N,M)){ for(int i = 1; i <= N; i ++) Scl(a[i]); S[0] = 0; P[0] = 0; sort(a + 1,a + 1 + N,cmp); Build(1,1,M); cnt = 0; int f = 1; for(int i = M; i >= 1; i --){ while(f <= N && i * f <= M) add(f++); LL sum = query(1,i); if(f <= N) sum = max(sum,a[f] + M); ans[i] = sum; } for(int i = 1; i <= M ; i ++){ printf("%lld",ans[i]); if(i != M) printf(" "); } puts(""); } return 0; }
