Queries On TreeProblem code: TAQTREE |
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 ith edge connecting node ui and vi has a weight of wi.
Your target is to handle the queries of the following two types:
- "1 i c" : Update the weight of ith 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 }