Jeanny
寂兮,寥兮,独立不改,周行而不殆
树链剖分

昨夜雨疏风骤，浓睡不消残酒，试问卷帘人，却道海棠依旧。知否知否，应是绿肥红瘦。  --《如梦令》

 

O 树链剖分的本质是把一棵树映射到线段上，且树被剖出来的链是连续的一段。看下图：

 

 

树剖的方法是先剖重的，这样dfs,并记录时间戳，也就是dfs序的序号。上图映射到线段上就是：

 

 

其中加括号的区间是重链部分。

O 那么这样做有什么好处？

比如我们要将树上某一路径x到y（比如7点-13点）上的点权值都+z，可以转化为在序列上进行。方法是：首先判断两个点的top[]是否相同，如果相同说明就在一个重链上，直接用线段树去处理。如若不然，判断dep(top[7])和dep(top[13])，即两个点的top[]值的深度。深的为7,浅的是13，那么先计算7-top[7]，将该区间+z，区间更新使用线段树。走到top[7]后，将x = fa[top[7]]，跳到1号点。重复判断两个点的top[]是否相同，如果不相同，继续判断dep[top[x]]和dep[top[13]]，以此类推。

可以看出树上的问题转到序列上就是处理区间[8,9],[1,4]的问题，使用线段树即可。

O 又如将x（例如4）的子树节点权值都+z,在序列上就是区间[2,6]的区域。我们可以在 dfs过程中记录子树大小size[x],那么就用线段树处理学序列上的区间[x,x+size[x]-1]了。

O 其余的树链剖分方法看博客：

https://www.cnblogs.com/ivanovcraft/p/9019090.html

 

1.写回顾了一下线段树的单点修改，区间查询和与最大值：

洛谷2590浙江OI2008树的统计:

输入：

4
1 2
2 3
4 1
4 2 1 3
5
QMAX 1 3
QSUM 2 3
CHANGE 2 5
QMAX 1 3
QSUM 2 3

输出：
4
3
5
6
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
int const MXN=100005;
char str[10]; int n, w[3*10000+5],x,y,cnt,hd[3*10000+5],q;
struct Tree{
    int l,r,sum,mx;
}tree[12*10000];
struct Edge{
    int to,nxt;
}edge[6*10000+5];
void add(int u,int v){
    cnt++;
    edge[cnt].to = v;
    edge[cnt].nxt = hd[u];
    hd[u] = cnt;
}
void pushup(int rt){
    tree[rt].sum = tree[rt<<1].sum + tree[rt<<1|1].sum;
    tree[rt].mx = max(tree[rt<<1].mx, tree[rt<<1|1].mx);
}
void build(int rt,int L,int R){
    tree[rt].l = L; tree[rt].r = R;
    if(L == R){
        tree[rt].sum = w[L];//写成了小写w[l]
        tree[rt].mx = w[L];
        return ;
    }
    int mid = (L+R)>>1;
    build(rt<<1,L,mid);
    build(rt<<1|1,mid+1,R);
    pushup(rt);
}
void change(int rt,int L,int t){
    if(L < tree[rt].l || L >tree[rt].r) return;
    if(tree[rt].l == tree[rt].r){
        tree[rt].sum = t;
        tree[rt].mx = t;
        return;
    }
    int mid = (tree[rt].l + tree[rt].r)>>1;
    if(L <= mid)
        change(rt<<1,L,t);
    else
        change(rt<<1|1,L,t);
    pushup(rt);
}
int querymax(int rt, int L ,int R){
    if(L <= tree[rt].l && R >= tree[rt].r){//err: L >=... R<=..
        return tree[rt].mx;
    }
    int mid = (tree[rt].l + tree[rt].r)>>1;
    int tmp = -0x7fffffff;
    if(L <= mid)
        tmp = max(tmp,querymax(rt<<1,L,R));
    if(R >= mid+1)//err:else
        tmp = max(tmp,querymax(rt<<1|1,L,R));
    return tmp;
}
int querysum(int rt, int L, int R){
    if(L <= tree[rt].l && R >= tree[rt].r){
        return tree[rt].sum;
    }
    int mid = (tree[rt].l + tree[rt].r)>>1;
    int sum = 0;
    if(L <= mid)
        sum += querysum(rt<<1,L,R);
    if(R >= mid+1)
        sum += querysum(rt<<1|1,L,R);
    return sum;
}
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n-1;i++){
        scanf("%d%d",&x,&y);
        add(x,y);add(y,x);
    }
    for(int i=1;i<=n;i++) scanf("%d",&w[i]);
    build(1,1,n);
    scanf("%d",&q); int l,r;
    for(int i = 1; i <= q; i++){
        scanf("%s%d%d",str,&l,&r);
        if(str[1]=='H'){
            change(1,l,r);
        }else if(str[1]=='M'){
            printf("%d\n",querymax(1,l,r));
        }else if(str[1]=='S'){
            printf("%d\n", querysum(1,l,r));
        }
    }
}
View Code

 

完整代码如下：

 

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
char str[10]; int n, w[300005],x,y,cnt,hd[300005],q,size[300005],son[300005],dfn,id[300005],a[300005],tp[300005];
int dep[300005],fa[300005];
struct Tree{
    int l,r,sum,mx;
}tree[12*10000];
struct Edge{
    int to,nxt;
}edge[6*10000+5];
void add(int u,int v){
    cnt++;
    edge[cnt].to = v;
    edge[cnt].nxt = hd[u];
    hd[u] = cnt;
}
void dfs1(int u,int f){//第一步，主要求出son[]
    size[u] = 1;
    int mx = 0;
    for(int i = hd[u]; i; i = edge[i].nxt){
        int v = edge[i].to;
        if(f == v) continue;
        fa[v] = u;
        dep[v] = dep[u] + 1;
        dfs1(v,u);
        if(size[v] > mx){
            mx = size[v];
            son[u] = v;
        }
        size[u] += size[v];
    }
}
void dfs2(int u,int f){//第二步，主要将树映射到序列上
    ++dfn;
    a[dfn] = w[u];//a[1] a[2]...
    id[u] = dfn;
    if(son[u]){
        tp[son[u]] = tp[u];
        dfs2(son[u],u);
    }
    for(int i = hd[u]; i; i = edge[i].nxt){
        int v = edge[i].to;
        if(v == f) continue;
        if(v != son[u]){
            tp[v] = v;
            dfs2(v,u);
        }
    }
}
void pushup(int rt){
    tree[rt].sum = tree[rt<<1].sum + tree[rt<<1|1].sum;
    tree[rt].mx = max(tree[rt<<1].mx, tree[rt<<1|1].mx);
}
void build(int rt,int L,int R){
    tree[rt].l = L; tree[rt].r = R;
    if(L == R){
        tree[rt].sum = a[L];//写成了小写w[l]
        tree[rt].mx = a[L];
        return ;
    }
    int mid = (L+R)>>1;
    build(rt<<1,L,mid);
    build(rt<<1|1,mid+1,R);
    pushup(rt);
}
void change(int rt,int L,int t){
    if(L < tree[rt].l || L >tree[rt].r) return;
    if(tree[rt].l == tree[rt].r){
        tree[rt].sum = t;
        tree[rt].mx = t;
        return;
    }
    int mid = (tree[rt].l + tree[rt].r)>>1;
    if(L <= mid)
        change(rt<<1,L,t);
    else
        change(rt<<1|1,L,t);
    pushup(rt);
}
int querymax(int rt, int L ,int R){
    if(L <= tree[rt].l && R >= tree[rt].r){//err: L >=... R<=..
        return tree[rt].mx;
    }
    int mid = (tree[rt].l + tree[rt].r)>>1;
    int tmp = -0x7fffffff;
    if(L <= mid)
        tmp = max(tmp,querymax(rt<<1,L,R));
    if(R >= mid+1)//err:else
        tmp = max(tmp,querymax(rt<<1|1,L,R));
    return tmp;
}
int querysum(int rt, int L, int R){
    if(L <= tree[rt].l && R >= tree[rt].r){
        return tree[rt].sum;
    }
    int mid = (tree[rt].l + tree[rt].r)>>1;
    int sum = 0;
    if(L <= mid)
        sum += querysum(rt<<1,L,R);
    if(R >= mid+1)
        sum += querysum(rt<<1|1,L,R);
    return sum;
}
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n-1;i++){
        scanf("%d%d",&x,&y);
        add(x,y);add(y,x);
    }
    for(int i=1;i<=n;i++) scanf("%d",&w[i]);
    dep[1] = 1; dfs1(1,0);
    tp[1] = 1; dfs2(1,0);
    // for(int i = 1; i <= n; i++)
    //     cout<<i<<" "<<dep[i]<<" "<<tp[i]<<endl;
    build(1,1,n);
    // cout<<querymax(1,6,6)<<endl;
    scanf("%d",&q); int l,r;
    for(int i = 1; i <= q; i++){
        scanf("%s%d%d",str,&l,&r);
        if(str[1]=='H'){
            change(1,id[l],r);
        }else if(str[1]=='M'){
            // printf("%d\n",querymax(1,l,r));
            int mx = -0x7fffffff;
            while(tp[l] != tp[r]){
                if(dep[tp[l]] < dep[tp[r]]) swap(l,r);
                // cout<<l<<" "<<tp[l]<<" "<<r<<" "<<tp[r]<<endl;
                // cout<<id[l]<<" "<<id[tp[l]]<<endl;
                mx = max(mx,querymax(1,id[tp[l]],id[l]));//err:深的在左
                // cout<<"mx:"<<mx<<endl;
                l = fa[tp[l]];
            }
            // cout<<l<<" "<<r<<endl;
            if(dep[l] > dep[r]) swap(l,r);
            // cout<<dep[l]<<" "<<dep[r]<<endl;
            mx = max(mx,querymax(1,id[l],id[r]));//err:深的在左边
            printf("%d\n",mx);
        }else if(str[1]=='S'){
            // printf("%d\n", querysum(1,l,r));
            int sum = 0;
            while(tp[l] != tp[r]){
                if(dep[tp[l]] < dep[tp[r]]) swap(l,r);
                sum += querysum(1,id[tp[l]],id[l]);
                l = fa[tp[l]];
            }
            if(dep[l] > dep[r]) swap(l,r);
            sum += querysum(1,id[l],id[r]);
            printf("%d\n",sum);
        }
    }
}
/*
8
1 2
1 3
2 4
2 5
3 6
5 7
5 8
1 2 3 4 5 6 7 8
3
QMAX 4 8
CHANGE 5 10
QMAX 4 8
*/
View Code

 

P2146软件包管理系统

题解：一开始将所有数据都置1，flag=-1,如果安装就变成0，卸载就变成1。如果安装则计算从0到该节点的和，如果卸载就计算子树大小-子树的和就是需要卸载的软件个数。

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
int n,x,m,a[100005],cnt,hd[100005],dfn,id[100005],size[100005],son[100005],tp[100005],fa[100005];
char str[15];
struct Edge{
    int to,nxt;
}edge[100005];
void add(int u,int v){
    cnt++;
    edge[cnt].to = v;
    edge[cnt].nxt = hd[u];
    hd[u] = cnt;
}
struct Tree{
    int l,r,sum,one;
}tree[400005];

//0表示已经安装的，1表示没有安装的
void dfs1(int u){
    size[u] = 1;
    int mx = 0;
    for(int i = hd[u]; i; i = edge[i].nxt){
        int v = edge[i].to;
        fa[v] = u;
        dfs1(v);
        if(size[v] > mx){
            mx = size[v];
            son[u] = v;
        }
        size[u] += size[v];
    }
}
void dfs2(int u){
    dfn++;
    a[dfn] = 1;
    id[u] = dfn;
    if(son[u]){
        tp[son[u]] = tp[u];
        dfs2(son[u]);
    }
    for(int i = hd[u]; i ; i = edge[i].nxt){
        int v = edge[i].to;
        if(son[u] != v){
            tp[v] = v;
            dfs2(v);
        }
    }
}

void pushup(int rt){
    tree[rt].sum = tree[rt<<1].sum + tree[rt<<1|1].sum;
    // tree[rt].one = tree[rt<<1].one || tree[rt<<1|1].one;
}
void pushdown(int rt){//标记下放不仅下放标记，还有sum值
    int one = tree[rt].one ;
    if(one != -1){
        tree[rt<<1].one = one;
        tree[rt<<1].sum = (tree[rt<<1].r - tree[rt<<1].l + 1) * one;
        tree[rt<<1|1].one = one;
        tree[rt<<1|1].sum = (tree[rt<<1|1].r - tree[rt<<1|1].l + 1) * one;
        tree[rt].one = -1;
    }
}
void build(int rt, int L, int R){
    tree[rt].l = L, tree[rt].r = R;
    tree[rt].one = -1;//
    if(L == R){
        tree[rt].sum = a[L];
        return ;
    }
    int mid = (tree[rt].l + tree[rt].r)>>1;
    build(rt<<1,L,mid);
    build(rt<<1|1,mid+1,R);
    pushup(rt);
}
void update(int rt,int L,int R,int t){//改变就要上升
    if(L <= tree[rt].l && R >= tree[rt].r){
        tree[rt].sum = t * (tree[rt].r - tree[rt].l + 1);
        tree[rt].one = t;
        return;
    }
    if(tree[rt].one != -1)pushdown(rt);
    int mid = (tree[rt].l + tree[rt].r)>>1;
    if(L <= mid)
        update(rt<<1,L,R,t);
    if(R > mid)
        update(rt<<1|1,L,R,t);
    pushup(rt);//和的时候是对的
}
int query(int rt, int L, int R){//询问就要下放
    // cout<<L<<" "<<R<<" "<<tree[rt].sum<<endl;
    if(L <= tree[rt].l && R >= tree[rt].r){
        return tree[rt].sum;//err:sum在update已经算过了
    }
    if(tree[rt].one != -1) pushdown(rt);
    int mid = (tree[rt].l + tree[rt].r)>>1;
    int sum = 0;
    if(L <= mid)
        sum += query(rt<<1,L,R);
    if(R > mid)
        sum += query(rt<<1|1,L,R);
    return sum;
}
int main(){
    scanf("%d",&n);
    for(int i = 1; i <= n-1; i++){
        scanf("%d",&x);
        add(x,i);
    }
    dfs1(0);
    tp[0] = 0; dfs2(0);
    build(1,1,n);
    scanf("%d",&m);
    for(int i = 1; i <= m; i++){
        scanf("%s%d",str,&x);
        int sum = 0;
        if(str[0] == 'i'){
            while(tp[0] != tp[x]){
                sum += query(1,id[tp[x]],id[x]);
                update(1,id[tp[x]],id[x],0);
                x = fa[tp[x]];
            }
            sum += query(1,id[0],id[x]);
            printf("%d\n",sum);
            update(1,id[0],id[x],0);
            // for(int i = 1; i <= 13; i++)
            //     cout<<i<<" "<<tree[i].one<<endl;
        }else if(str[0]=='u'){//卸载就是对子树进行操作，在序列上一定是个子段
            sum = size[x] - query(1,id[x],id[x]+size[x]-1);
            printf("%d\n",sum);
            update(1,id[x],id[x]+size[x]-1,1);
        }
    }

    return 0;
}
/*
9
0 1 2 0 1 5 2 5
1
install 3
*/

 这里少了一个加号，惨不忍睹。

3. 使用数链剖分求lca

5 3
1 2
3 1
3 4
5 3
2 3  -> 1
1 5  -> 1
3 3  -> 3
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int n,m,dep[100005],f[100005],size[100005],cnt,hd[100005],son[100005],tp[100005];
struct Edge{
    int nxt, to;
}edge[200005];
void add(int u, int v){
    cnt++; 
    edge[cnt].to = v;
    edge[cnt].nxt = hd[u];
    hd[u] = cnt;
}
void dfs1(int u, int fa){
    int mx = -1; size[u] = 1;
    for(int i = hd[u]; i; i = edge[i].nxt){
        int v = edge[i].to;
        if(v == fa) continue;
        dep[v] = dep[u] + 1; f[v] = u;
        dfs1(v, u);
        size[u] += size[v];
        if(size[v] > mx){
            mx = size[v], son[u] = v;
        }
    }
}
void dfs2(int u, int top, int fa){
    tp[u] = top;
    if(son[u]) dfs2(son[u], tp[u], u);
    for(int i = hd[u]; i; i = edge[i].nxt){
        int v = edge[i].to;
        if(v == fa) continue;
        if(v != son[u]) dfs2(v, v, u);
    }
}
int LCA(int x, int y){
    while(tp[x] != tp[y]){
        if(dep[tp[x]] < dep[tp[y]]) swap(x,y);
        x = f[tp[x]];
    }
    if(dep[x] < dep[y]) return x;
    else return y;
}
int main(){
    int x,y,z;
    scanf("%d%d",&n,&m);
    for(int i = 1; i <= n-1; i++){
        scanf("%d%d",&x,&y);
        add(x,y); add(y,x);
    }
     dep[1] = 1; dfs1(1,0);
     dfs2(1,1,1);
     for(int i = 1; i <= m; i++){
         scanf("%d%d%d",&x,&y);
     }
    return 0;
}
二刷
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define N 300005
using namespace std;
int n, m, x, y, siz[N], top[N], hd[N], cnt, dep[N], son[N], f[N];
struct Edge{
    int nxt, to;
}edge[N*2];
void add(int u, int v){
    edge[++cnt].to = v;
    edge[cnt].nxt = hd[u];
    hd[u] = cnt;
}
void dfs1(int u, int fa){
    f[u] = fa; siz[u] = 1;
    dep[u] = dep[fa] + 1;
    for(int i = hd[u]; i; i = edge[i].nxt){
        int v = edge[i].to;
        if(v == fa) continue;
        dfs1(v, u);
        siz[u] += siz[v];
        if(siz[v] > siz[son[u]]) son[u] = v;
    }
}
void dfs2(int u, int fa, int tp){
    top[u] = tp;
    if(son[u]) dfs2(son[u], u, tp);
    for(int i = hd[u]; i; i = edge[i].nxt){
        int v = edge[i].to;   //nnd,写成edge[i].nxt
        if(v == fa || v == son[u]) continue;
        dfs2(v, u, v);
    }
}
int lca(int x, int y){
    while(top[x] != top[y]){
        if(dep[top[x]] < dep[top[y]]) swap(x,y);
        x = f[top[x]];
    }
    return dep[x] < dep[y] ? x:y;//error dep,top搞混
}
int main(){
    scanf("%d%d",&n,&m);
    for(int i = 1; i <= n-1; i++){
        scanf("%d%d",&x,&y);
        add(x,y); add(y,x);
    }
    dfs1(1,0);
    // for(int i = 1; i <= n; i++)
    //     cout<<"son: "<<i<<" "<<son[i]<<endl;
    dfs2(1,0,1);
    // for(int i = 1; i <= n; i++)
        // cout<<i<<" "<<f[i]<<" "<<siz[i]<<" "<<dep[i]<<endl;
        // cout<<"hhh: "<<i<<" "<<top[i]<<endl;
    for(int i = 1;i <= m; i++){
        scanf("%d%d",&x,&y);
        printf("%d\n",lca(x,y));
    }
    return 0;
}

 

posted on 2020-06-08 10:09  Jeanny  阅读(247)  评论(0)    收藏  举报

博客园  ©  2004-2025
浙公网安备 33010602011771号 浙ICP备2021040463号-3 Powered By博客园