hdu 5044 树链剖分

Tree

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1329    Accepted Submission(s): 222

Problem Description

You are given a tree (an acyclic undirected connected graph) with N nodes. The tree nodes are numbered from 1 to N

There are N - 1 edges numbered from 1 to N - 1.

Each node has a value and each edge has a value. The initial value is 0.

There are two kind of operation as follows:

● ADD1 u v k: for nodes on the path from u to v, the value of these nodes increase by k.

● ADD2 u v k: for edges on the path from u to v, the value of these edges increase by k.

After finished M operation on the tree, please output the value of each node and edge.

 

Input

The first line of the input is T (1 ≤ T ≤ 20), which stands for the number of test cases you need to solve.

The first line of each case contains two integers N ,M (1 ≤ N, M ≤10⁵),denoting the number of nodes and operations, respectively.

The next N - 1 lines, each lines contains two integers u, v(1 ≤ u, v ≤ N ), denote there is an edge between u,v and its initial value is 0.

For the next M line, contain instructions “ADD1 u v k” or “ADD2 u v k”. (1 ≤ u, v ≤ N, -10⁵ ≤ k ≤ 10⁵)

 

Output

For each test case, print a line “Case #t:”(without quotes, t means the index of the test case) at the beginning.

The second line contains N integer which means the value of each node.

The third line contains N - 1 integer which means the value of each edge according to the input order.

 

Sample Input

24 21 22 32 4ADD1 1 4 1ADD2 3 4 24 21 22 31 4ADD1 1 4 5ADD2 3 2 4

 

Sample Output

Case #1:1 1 0 10 2 2Case #2:5 0 0 50 4 0

 

Source

2014 ACM/ICPC Asia Regional Shanghai Online

两种操作，一种更新结点值，一种更新路径值，最后输出更改后的结点值和路径值。

对于区间[a,b]，区间的每个值加上c，可以用一个数组标记，ans[a]+=c,ans[b+1]-=c;然后下标从a，遍历到b，把所有的ans[]值加上，就等于当前结点修改后的值。注意两点，一是手动扩栈，二是最终的结果用64位。

#pragma comment(linker, "/STACK:102400000,102400000")#include<stdio.h>#include<iostream>#include<string.h>using namespace std;#define N 100010struct pp{int u,v;}ed[N];struct node{int u,v,next;}bian[N*2];int e,id,dep[N],son[N],father[N],sz[N],ti[N],mark1[N],mark2[N],top[N],head[N];__int64 a[N],b[N],ans1[N],ans2[N];void add(int u,int v){bian[e].u=u;bian[e].v=v;bian[e].next=head[u];head[u]=e++;}void dfs1(int u,int fa){int i,v;     dep[u]=dep[fa]+1; son[u]=0;   father[u]=fa; sz[u]=1;  for(i=head[u];i!=-1;i=bian[i].next) { v=bian[i].v; if(v==fa) continue; dfs1(v,u); sz[u]+=sz[v]; if(sz[son[u]]<sz[v]) son[u]=v; }}void dfs2(int u,int fa){int i,v;ti[u]=id++;mark1[id-1]=u;top[u]=fa;if(son[u]!=0)dfs2(son[u],fa);for(i=head[u];i!=-1;i=bian[i].next){v=bian[i].v;if(v==father[u]||v==son[u])continue;dfs2(v,v);}}void getnode(int u,int v,int k){while(top[u]!=top[v]){if(dep[top[u]]>dep[top[v]])swap(u,v);a[ti[top[v]]]+=k;a[ti[v]+1]-=k;v=father[top[v]];}if(ti[u]>ti[v])swap(u,v);a[ti[u]]+=k;a[ti[v]+1]-=k;}void getedge(int u,int v,int k){while(top[u]!=top[v]){if(dep[top[u]]>dep[top[v]])swap(u,v);b[ti[top[v]]]+=k;b[ti[v]+1]-=k;v=father[top[v]];}if(ti[u]>ti[v])swap(u,v);if(u!=v){b[ti[u]+1]+=k;b[ti[v]+1]-=k;}}int main(){int t,cnt=1,n,m,i,u,v,k;__int64 s;char str[10];scanf("%d",&t);while(t--){      scanf("%d%d",&n,&m);  memset(a,0,sizeof(a));  memset(head,-1,sizeof(head));  memset(b,0,sizeof(b));  e=0;  for(i=1;i<n;i++)  {  scanf("%d%d",&ed[i].u,&ed[i].v);  add(ed[i].u,ed[i].v);  add(ed[i].v,ed[i].u);  }  sz[0]=0; id=1; dep[1]=0;  dfs1(1,1);  dfs2(1,1);  for(i=1;i<=m;i++)  {  scanf("%s%d%d%d",str,&u,&v,&k);  if(strcmp(str,"ADD1")==0)       getnode(u,v,k);  else  getedge(u,v,k);  }  for(i=1;i<n;i++)  {  if(dep[ed[i].u]<dep[ed[i].v])  mark2[ti[ed[i].v]]=i;  else  mark2[ti[ed[i].u]]=i;  }  printf("Case #%d:\n",cnt++);  s=0;  for(i=1;i<=n;i++)  {          s+=a[i];  ans1[mark1[i]]=s;  }  for(i=1;i<=n;i++)  {  if(i==1)  printf("%I64d",ans1[i]);  else  printf(" %I64d",ans1[i]);  }  printf("\n");  s=0;  for(i=2;i<=n;i++)  {  s+=b[i];          ans2[mark2[i]]=s;  }  for(i=1;i<n;i++)  {  if(i==1)  printf("%I64d",ans2[i]);  else  printf(" %I64d",ans2[i]);  }  printf("\n");}return 0;}