POJ 3321:apple trees (tree-like array, raising questions)

Apple Tree Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 18623 Accepted: 5629 Description There is a apple tree outside of Kaka ' s house. Every autumn, a lot of apples would grow in the tree. Kaka likes Apple very much, so he has been carefully nurturing the Big Apple tree. The tree has N forks which is connected by branches. Kaka numbers the forks by 1 to N and the root are always numbered by 1. Apples'll grow on the forks and both Apple won ' t grow on the same fork. Kaka wants to know how many apples is there in a sub-tree, for his study of the produce ability of the apple tree. The trouble is, a new Apple may grow in an empty fork some time and Kaka could pick an apple from the tree for his Desse Rt. Can you help Kaka? Input The first line contains a integer  n   ( n  ≤100,000), which is the number of the forks In the tree. the following  N  -1 lines each contain the integers  u  and  v , which means fork  u  and fork  v  are connected by a branch. The next line contains an integer  m   ( m  ≤100,000). the following  M  lines each contain a message which be either " c  x strong> "which means the existence of the apple on fork  x  has been changed. i.e. if there is a apple on the fork and then Kaka pick it; Otherwise a new Apple has grown on the empty fork. or " q  x " which means an inquiry for the number of apples in the sub-tree above the F ork  x , including the apple (if exists) on the fork x Note The tree was full of apples at the beginning Output For every inquiry, the output of the correspond answer per line. Sample Input 33Q 1C 2Q 1 Sample Output 32 Source POJ monthly--2007.08.05, Huang, Jinsong

　　tree-like array to improve the problem .

Some difficulty, to turn the bend.

　　Test Instructions :

At first, I gave you a n-1, an apple tree, by default there are apples on every fork. Then there are Q operations, these operations are divided into two, or Apple picked away apples, no apples grow apple, or a fork on the number of apples.

　　Ideas :

The key is to record a start time, and an end time at each fork, indicating the time at which the DFS went through the fork and back to the fork. The time here is actually the location of the DFS traversal order. This is equivalent to having an interval in which the number of all apples on the fork point can be calculated using a tree array.

　　Note :

To your n-1 edge, is bidirectional, that is, a point to b,b also points to a, is a non-directed graph.

　　test data (from POJ discussion edition):

Ten2 13 13 44 74 57 95 86 5Ten 6TenQ1C5Q3Q4C5Q6C6Q8C9Q1ans:Ten76218

　　Code :

1#include <iostream>2#include <stdio.h>3 using namespacestd;4 5 #defineMAXN 1000106 7 intN;8 intCnt=0;9 intC[MAXN];Ten intSTART[MAXN]; One intEND[MAXN]; A  - structnode{ -     intnum; theNode* Next;//child node - Node () -     { -Next =NULL; +     } -}TREE[MAXN];//Critical Table +  A intLowbit (intx) at { -     returnX & (-x); - } -  - voidAddintDintx) - { in      while(d<=N) { -C[D] + =x; toD + =Lowbit (d); +     } - } the  * intSumintd) $ {Panax Notoginseng     intAns =0; -      while(d>=1){ theAns + =C[d]; +D-=Lowbit (d); A     } the     returnans; + } -  $ voidDfsintV//DFS traversal with R as the root node, returning the time after the entire traversal $ { -START[V] = + +CNT; -node* p =Tree[v].next; the      while(p) { -         if(start[p->num]==0)WuyiDFS (p->num); thep = p->Next; -     } WuEND[V] =CNT; - } About  $ voidAddedge (intAintb//Add a branch to the apple tree - {     -node* p =NewNode; -P->num =b; AP->next =Tree[a].next; +Tree[a].next =p; the } -  $ intMain () the {     the     inti,q; thescanf"%d",&N); the      for(i=1; i<n;i++){ -         intb; inscanf"%d%d",&a,&b); the Addedge (A, b); the Addedge (b,a); About     } theDfs1); the  the      for(i=1; i<=n;i++)//Initialize c[] +Add (I,1); -  thescanf"%d",&q);Bayi      while(q--) {//Q-time Operation the         Charcmd[Ten]; the         intD; -scanf"%s%d",cmd,&d); -         if(cmd[0]=='C'){ the             if(SUM (start[d])-sum (start[d]-1)==1) theAdd (start[d],-1); the             Else theAdd (Start[d],1); -         } the         Else if(cmd[0]=='Q'){ theprintf"%d\n", SUM (End[d])-sum (start[d]-1)); the         }94     } the     return 0; the}

freecode:www.cnblogs.com/yym2013