Hdu 4010-Query on The Trees LCT,动态树

Posted 2020-06-25 微弱的世界

Query on The Trees

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 4091    Accepted Submission(s): 1774

Problem Description

We have met so many problems on the tree, so today we will have a query problem on a set of trees. 
There are N nodes, each node will have a unique weight Wi. We will have four kinds of operations on it and you should solve them efficiently. Wish you have fun! 

 

 

Input

There are multiple test cases in our dataset. 
For each case, the first line contains only one integer N.(1 ≤ N ≤ 300000) The next N‐1 lines each contains two integers x, y which means there is an edge between them. It also means we will give you one tree initially. 
The next line will contains N integers which means the weight Wi of each node. (0 ≤ Wi ≤ 3000) 
The next line will contains an integer Q. (1 ≤ Q ≤ 300000) The next Q lines will start with an integer 1, 2, 3 or 4 means the kind of this operation. 
1. Given two integer x, y, you should make a new edge between these two node x and y. So after this operation, two trees will be connected to a new one. 
2. Given two integer x, y, you should find the tree in the tree set who contain node x, and you should make the node x be the root of this tree, and then you should cut the edge between node y and its parent. So after this operation, a tree will be separate into two parts. 
3. Given three integer w, x, y, for the x, y and all nodes between the path from x to y, you should increase their weight by w. 
4. Given two integer x, y, you should check the node weights on the path between x and y, and you should output the maximum weight on it. 

 

 

Output

For each query you should output the correct answer of it. If you find this query is an illegal operation, you should output ‐1. 
You should output a blank line after each test case. 

 

 

Sample Input

5 1 2 2 4 2 5 1 3 1 2 3 4 5 6 4 2 3 2 1 2 4 2 3 1 3 5 3 2 1 4 4 1 4

 

 

Sample Output

3 -1 7

Hint

We define the illegal situation of different operations: In first operation: if node x and y belong to a same tree, we think it‘s illegal. In second operation: if x = y or x and y not belong to a same tree, we think it‘s illegal. In third operation: if x and y not belong to a same tree, we think it‘s illegal. In fourth operation: if x and y not belong to a same tree, we think it‘s illegal.

 

 

Source

The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest 

 

 

Recommend

lcy

 

 题解：

LCT的子树问题。

找到每个点所在的原始树(不是Splay树)的根。

又是子树判断最麻烦。。。

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 #define MAXN 300010
  4 #define INF 1e9
  5 struct node
  6 {
  7     int left,right,val,mx;
  8 }tree[MAXN];
  9 struct NODE
 10 {
 11     int begin,end,next;
 12 }edge[MAXN*2];
 13 int father[MAXN],rev[MAXN],tag[MAXN],U[MAXN],V[MAXN],Stack[MAXN],Head[MAXN],cnt;
 14 void addedge(int bb,int ee)
 15 {
 16     edge[++cnt].begin=bb;edge[cnt].end=ee;edge[cnt].next=Head[bb];Head[bb]=cnt;
 17 }
 18 void addedge1(int bb,int ee)
 19 {
 20     addedge(bb,ee);addedge(ee,bb);
 21 }
 22 int read()
 23 {
 24     int s=0,fh=1;char ch=getchar();
 25     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)fh=-1;ch=getchar();}
 26     while(ch>=‘0‘&&ch<=‘9‘){s=s*10+(ch-‘0‘);ch=getchar();}
 27     return s*fh;
 28 }
 29 int isroot(int x)
 30 {
 31     return tree[father[x]].left!=x&&tree[father[x]].right!=x;
 32 }
 33 void pushdown(int x)
 34 {
 35     int l=tree[x].left,r=tree[x].right;
 36     if(rev[x]!=0)
 37     {
 38         rev[x]^=1;rev[l]^=1;rev[r]^=1;
 39         swap(tree[x].left,tree[x].right);
 40     }
 41     if(tag[x]!=0)
 42     {
 43         /*tag[l]+=tag[x];tag[r]+=tag[x];
 44         tree[l].val+=tag[x];tree[r].val+=tag[x];
 45         tree[l].mx+=tag[x];tree[r].mx+=tag[x];*/
 46         if(l!=0){tag[l]+=tag[x];tree[l].val+=tag[x];tree[l].mx+=tag[x];}
 47         if(r!=0){tag[r]+=tag[x];tree[r].val+=tag[x];tree[r].mx+=tag[x];}
 48         tag[x]=0;
 49     }
 50 }
 51 void Pushup(int x)
 52 {
 53     int l=tree[x].left,r=tree[x].right;
 54     tree[x].mx=max(max(tree[l].mx,tree[r].mx),tree[x].val);
 55 }
 56 void rotate(int x)
 57 {
 58     int y=father[x],z=father[y];
 59     if(!isroot(y))
 60     {
 61         if(tree[z].left==y)tree[z].left=x;
 62         else tree[z].right=x;
 63     }
 64     if(tree[y].left==x)
 65     {
 66         father[x]=z;father[y]=x;tree[y].left=tree[x].right;tree[x].right=y;father[tree[y].left]=y;
 67     }
 68     else
 69     {
 70         father[x]=z;father[y]=x;tree[y].right=tree[x].left;tree[x].left=y;father[tree[y].right]=y;
 71     }
 72     Pushup(y);Pushup(x);
 73 }
 74 void splay(int x)
 75 {
 76     int top=0,i,y,z;Stack[++top]=x;
 77     for(i=x;!isroot(i);i=father[i])Stack[++top]=father[i];
 78     for(i=top;i>=1;i--)pushdown(Stack[i]);
 79     while(!isroot(x))
 80     {
 81         y=father[x];z=father[y];
 82         if(!isroot(y))
 83         {
 84             if((tree[y].left==x)^(tree[z].left==y))rotate(x);
 85             else rotate(y);
 86         }
 87         rotate(x);
 88     }
 89 }
 90 void access(int x)
 91 {
 92     int last=0;
 93     while(x!=0)
 94     {
 95         splay(x);
 96         tree[x].right=last;Pushup(x);
 97         last=x;x=father[x];
 98     }
 99 }
100 void makeroot(int x)
101 {
102     access(x);splay(x);rev[x]^=1;
103 }
104 void link(int u,int v)
105 {
106     /*access(u);*/makeroot(u);father[u]=v;//splay(u);
107 }
108 void cut(int u,int v)
109 {
110     /*access(u);*/makeroot(u);access(v);splay(v);/*father[u]=tree[v].left=0;*/father[tree[v].left]=0;tree[v].left=0;Pushup(v);
111 }
112 int findroot(int x)
113 {
114     access(x);splay(x);
115     while(tree[x].left!=0)x=tree[x].left;
116     return x;
117 }
118 int main()
119 {
120     int n,i,w,x,y,fh,Q,top=0,u,j,v;
121     while(scanf("%d",&n)!=EOF)
122     {
123         top=0;
124         for(i=0;i<=n;i++)tree[i].val=tree[i].mx=tree[i].left=tree[i].right=rev[i]=tag[i]=father[i]=0;
125         tree[0].mx=-INF;
126         memset(Head,-1,sizeof(Head));cnt=1;
127         for(i=1;i<n;i++)
128         {
129             U[i]=read();V[i]=read();
130             addedge1(U[i],V[i]);
131         }
132         Stack[++top]=1;
133         for(i=1;i<=top;i++)
134         {
135             u=Stack[i];
136             for(j=Head[u];j!=-1;j=edge[j].next)
137             {
138                 v=edge[j].end;
139                 if(v!=father[u])
140                 {
141                     father[v]=u;
142                     Stack[++top]=v;
143                 }
144             }
145         }
146         for(i=1;i<=n;i++)tree[i].mx=tree[i].val=read();
147         //for(i=1;i<n;i++)link(U[i],V[i]);
148         Q=read();
149         for(i=1;i<=Q;i++)
150         {
151             fh=read();
152             if(fh==1)
153             {
154                 x=read();y=read();
155                 if(findroot(x)!=findroot(y))link(x,y);
156                 else {printf("-1\n");continue;}
157             }
158             else if(fh==2)
159             {
160                 x=read();y=read();
161                 if(findroot(x)==findroot(y)&&x!=y)
162                 {
163                     /*makeroot(x);*/cut(x,y);
164                     //access(y);access(father[y]);splay(father[y]);father[y]=tree[father[y]].left=0;
165                 }
166                 else {printf("-1\n");continue;}
167             }
168             else if(fh==3)
169             {
170                 w=read();x=read();y=read();
171                 if(findroot(x)==findroot(y))
172                 {
173                     makeroot(x);access(y);splay(y);
174                     tag[y]+=w;tree[y].mx+=w;tree[y].val+=w;
175                 }
176                 else {printf("-1\n");continue;}
177             }
178             else
179             {
180                 x=read();y=read();
181                 makeroot(x);access(y);splay(y);
182                 if(findroot(x)!=findroot(y)){printf("-1\n");continue;}
183                 printf("%d\n",tree[y].mx);
184             }
185         }
186         printf("\n");
187     }
188     return 0;
189 }

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