POJ3321——Apple Tree(DFS+树状数组)

Apple Tree

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 19110 Accepted: 5816

Description

There is an apple tree outside of kaka's house. Every autumn, a lot of apples will 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 are connected by branches. Kaka numbers the forks by 1 to N and the root is always numbered by 1. Apples will grow on the forks and two apple won't grow on the same fork. kaka wants to know how many apples are there in a sub-tree, for his study of the produce ability of the apple tree.

The trouble is that a new apple may grow on an empty fork some time and kaka may pick an apple from the tree for his dessert. Can you help kaka?

Input

The first line contains an integer N (N ≤ 100,000) , which is the number of the forks in the tree.
The following N - 1 lines each contain two 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 is either
"C x" which means the existence of the apple on fork x has been changed. i.e. if there is an apple on the fork, 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 fork x, including the apple (if exists) on the fork x
Note the tree is full of apples at the beginning

Output

For every inquiry, output the correspond answer per line.

Sample Input

31 21 33Q 1C 2Q 1

Sample Output

32

题意：
在一棵树上，每个分叉点及末梢可能有苹果（最多1个）， 每次可以摘掉一个苹果，或有一个苹果新长出来，随时查询某个分叉点往上的子树里，一共有多少个苹果。(分叉点数<100,000) 
根据题意，一开始时，所有能长苹果的地方都有苹果。


分析：
具体做法是做一次dfs，记下每个节点的开始时间Start[i]和结 束时间End[i]，那么对于i节点的所有子孙的开始时间和结束时间都应位于Start[i]和End[i]之间然后用树状数组统计Start[i]到End[i]之间的附加苹果总数。 
这里用树状数组统计区间可以用Sum(End[i])-Sum(Start[i]-1)来计算。 


但是这道题比较坑的地方是会卡Vector邻接表，但是换一种写法也能过，不知道为毛！

Vector邻接表版    938MS

#include <algorithm>#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <vector>#define INF 0x7fffffffusing namespace std;const int N = 100000 + 10;vector<vector<int> > G(N);//vector<int> G[N];  为毛这样写就TLE！int Start[N],End[N];int a[N],Time;bool Have[N];void init(){    Time=0;    for(int i=0; i<N; i++)    {        G[i].clear();        Have[i]=true;    }}int lowbit(int x){    return x&-x;}void add(int x,int v){    for(int i=x; i<N; i+=lowbit(i))        a[i]+=v;}int getsum(int x){    int ret=0;    for(int i=x; i>0; i-=lowbit(i))        ret+=a[i];    return ret;}void dfs(int u){    Time++;    Start[u]=Time;    for(int i=0; i<G[u].size(); i++)        dfs(G[u][i]);    End[u]=Time;}int main(){    int n,m;    while(~scanf("%d",&n))    {        init();        for(int i=0; i<n-1; i++)        {            int a,b;            scanf("%d%d",&a,&b);            G[a].push_back(b);        }                dfs(1);        for(int i=1; i<=n; i++) //初始化树上苹果            a[i]=lowbit(i);        scanf("%d",&m);        while(m--)        {            char s[3];            int x;            scanf("%s%d",s,&x);            if(s[0]=='Q')            {                int ans=getsum(End[x])-getsum(Start[x]-1);                printf("%d\n",ans);            }            else if(s[0]=='C')            {                if(Have[x])                    add(Start[x],-1);                else                    add(Start[x],1);                Have[x]^=1;            }        }    }        return 0;}

手写邻接表 391MS C++

#include <algorithm>#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <vector>#define INF 0x7fffffffusing namespace std;const int N = 100000 + 10;int Start[N],End[N];int a[N],Time;bool Have[N];struct Edge{    int v,next;}e[N];int head[N],top=0;void Add_edge(int u,int v){    e[top].v=v;    e[top].next=head[u];    head[u]=top++;}void dfs(int u){    Time++;    Start[u]=Time;    for(int i=head[u]; i!=-1; i=e[i].next)        dfs(e[i].v);    End[u]=Time;}int lowbit(int x){    return x&-x;}void add(int x,int v){    for(int i=x; i<N; i+=lowbit(i))        a[i]+=v;}int getsum(int x){    int ret=0;    for(int i=x; i>0; i-=lowbit(i))        ret+=a[i];    return ret;}void init(){    Time=0;    top=0;    memset(Have,1,sizeof(Have));    memset(head,-1,sizeof(head));}int main(){    int n,m;    while(~scanf("%d",&n))    {        init();        for(int i=0; i<n-1; i++)        {            int a,b;            scanf("%d%d",&a,&b);            Add_edge(a,b);        }        dfs(1);        for(int i=1; i<=n; i++) //初始化树上苹果            a[i]=lowbit(i);        scanf("%d",&m);        while(m--)        {            char s[3];            int x;            scanf("%s%d",s,&x);            if(s[0]=='Q')            {                int ans=getsum(End[x])-getsum(Start[x]-1);                printf("%d\n",ans);            }            else if(s[0]=='C')            {                if(Have[x])                    add(Start[x],-1);                else                    add(Start[x],1);                Have[x]^=1;            }        }    }    return 0;}