树链剖分——边上权值和

Queries On Tree Problem code: TAQTREE Tweet

All submissions for this problem are available.

Read problems statements in Mandarin Chinese and Russian.

You are given a tree of N nodes numbered from 1 to N.
The i^th edge connecting node u_i and v_i has a weight of w_i.

Your target is to handle the queries of the following two types:

• "1 i c" : Update the weight of i^th edge with the new weight c.
  (1 represents the query type).
• "2 u v" : Find the length of the path from node u to v. (2 represents the query type).

Input

• The first line contains a single integer N.
• Each of the next N - 1 lines contains three integers u v w, representing an edge
  between u and v with the weight of w.
• The next line contains a single integer Q representing the number of queries
• Each of the next Q lines contains three integers representing a query as described above.

Output

• For each query of type 2, output the answer in a single line.

Constraints

All test:

• 1 ≤ i < N
• 1 ≤ u, v ≤ N; u ≠ v
• 1 ≤ w, c ≤ 10⁴

40 points:

• 1 ≤ N, Q ≤ 10³

60 points:

• 1 ≤ N, Q ≤ 10⁵

Example

Input: 5 1 2 2 2 3 4 4 2 3 5 4 1 3 2 5 3 1 3 1 2 5 3  Output: 8 6  

Explanation

The path from 5 to 3 is 5 -> 4 -> 2 -> 3. Initially this path has the length of 1 + 3 + 4 = 8.
After the weight of the edge connect 4 and 2 was changed to 1, the new length of the path is 1 + 1 + 4 = 6.

 

比较基础的树链剖分题，算边上的权值和，单边修改

  1 #include<bits/stdc++.h>
  2 
  3 using namespace std;
  4 typedef long long ll;
  5 
  6 #define N 223456
  7 struct node
  8 {
  9     int v,next;
 10 }e[N<<2];
 11 int head[N],tot,top[N],fa[N],deep[N],num[N];
 12 int p[N],fp[N],son[N];
 13 int pos;
 14 
 15 void init()
 16 {
 17     tot=0;
 18     memset(head,-1,sizeof(head));
 19     pos=0;
 20     memset(son,-1,sizeof(son));
 21 }
 22 void add(int u,int v)
 23 {
 24     e[tot].v=v;
 25     e[tot].next=head[u];
 26     head[u]=tot++;
 27 }
 28 void dfs1(int u,int pre,int d)
 29 {
 30     deep[u]=d;
 31     fa[u]=pre;
 32     num[u]=1;
 33     for (int i=head[u];i!=-1;i=e[i].next)
 34     {
 35         int v=e[i].v;
 36         if (v!=pre)
 37         {
 38             dfs1(v,u,d+1);
 39             num[u]+=num[v];
 40             if (son[u]==-1||num[v]>num[son[u]]) son[u]=v;
 41         }
 42     }
 43 }
 44 void getpos(int u,int sp)
 45 {
 46     top[u]=sp;
 47     p[u]=pos++;
 48     fp[p[u]]=u;
 49     if (son[u]==-1) return;
 50     getpos(son[u],sp);
 51     for (int i=head[u];i!=-1;i=e[i].next)
 52     {
 53         int v=e[i].v;
 54         if (v!=son[u]&&v!=fa[u])
 55             getpos(v,v);
 56     }
 57 }
 58 
 59 struct nod
 60 {
 61     int l,r;
 62     ll sum;
 63 }tree[N<<4];
 64 
 65 void build(int rt,int l,int r)
 66 {
 67     tree[rt].l=l;
 68     tree[rt].r=r;
 69     tree[rt].sum=0;
 70     if (l==r) return;
 71     int mid=(l+r)/2;
 72     build(rt<<1,l,mid);
 73     build(rt<<1|1,mid+1,r);
 74 }
 75 
 76 void pushup(int rt)
 77 {
 78     tree[rt].sum=tree[rt<<1].sum+tree[rt<<1|1].sum;
 79 }
 80 void update(int rt,int k,int val)
 81 {
 82     if (tree[rt].l==k&&tree[rt].r==k)
 83     {
 84         tree[rt].sum=val;
 85         return;
 86     }
 87     int mid=(tree[rt].l+tree[rt].r)/2;
 88     if (k<=mid) update(rt<<1,k,val);
 89     else update(rt<<1|1,k,val);
 90     pushup(rt);
 91 }
 92 
 93 ll query(int rt,int l,int r)
 94 {
 95     if (tree[rt].l==l&&tree[rt].r==r) return tree[rt].sum;
 96 
 97     int mid=(tree[rt].l+tree[rt].r)/2;
 98     if (r<=mid) return query(rt<<1,l,r);
 99     else if (l>mid) return query(rt<<1|1,l,r);
100     else return query(rt<<1,l,mid)+query(rt<<1|1,mid+1,r);
101 }
102 
103 ll find(int u,int v)
104 {
105     int f1=top[u],f2=top[v];
106     ll tmp=0;
107     while (f1!=f2)
108     {
109         if (deep[f1]<deep[f2])
110         {
111             swap(f1,f2);
112             swap(u,v);
113         }
114         tmp+=query(1,p[f1],p[u]);
115         u=fa[f1];
116         f1=top[u];
117     }
118     if (u==v) return tmp;
119     if (deep[u]>deep[v]) swap(u,v);
120     return tmp+query(1,p[son[u]],p[v]);
121 }
122 
123 int x[N],y[N],z[N];
124 int main()
125 {
126     int n;
127     scanf("%d",&n);
128     init();
129     for (int i=1;i<n;i++)
130     {
131         scanf("%d%d%d",&x[i],&y[i],&z[i]);
132         add(x[i],y[i]);
133         add(y[i],x[i]);
134     }
135     dfs1(1,0,0);
136     getpos(1,1);
137     build(1,0,pos-1);
138     for (int i=1;i<n;i++)
139     {
140         if (deep[x[i]]>deep[y[i]])
141             swap(x[i],y[i]);
142         update(1,p[y[i]],z[i]);
143     }
144     int Q;
145     scanf("%d",&Q);
146     while (Q--)
147     {
148         int ty,a,b;
149         scanf("%d%d%d",&ty,&a,&b);
150         if (ty==1) update(1,p[y[a]],b);
151         else printf("%lld\n",find(a,b));
152     }
153 
154     return 0;
155 }