poj 3321 单点更新 区间求和

Apple Tree

Time Limit: 2000 MS Memory Limit: 65536 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

[Submit] [Status] [Discuss]

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

3
1 2
1 3
3
Q 1
C 2
Q 1

Sample Output

3
2/*#include<cstdio>
#include<cstring>
using namespace std;
#define MAX 100005

int edge[MAX<<1];//表示第i条边的终点
int next[MAX<<1];//与第i条边同起点的下一条边的位置
int head[MAX<<1];//以i为起点的第一条边的储存位置
int low[MAX],high[MAX],c[MAX];
bool visit[MAX];
int cnt,n;

int lowbit(int x)
{
    return x&-x;
}

void update(int pos,int value) //更新pos的值
{
    int x=pos;
    while(x<=n)
    {
        c[x]+=value;
        x+=lowbit(x);
    }
}

int get_sum(int pos)//求1到pos位置的和
{
    int x=pos,sum=0;
    while(x>0)
    {
        sum+=c[x];
        x-=lowbit(x);
    }

    return sum;
}

void insert(int i,int a,int b)//a起点,b终点
{
    edge[i]=b;
    next[i]=head[a];
    head[a]=i;
}

void dfs(int x,int pre)
{
    low[x]=(++cnt);
    for(int i=head[x]; i!=-1; i=next[i])
        if(edge[i]!=pre)
            dfs(edge[i],x);
    high[x]=cnt;
}

int main()
{
    int m;
    int a,b;
    char op[5];
    for(; ~scanf("%d",&n);)
    {
        cnt=0;
        memset(head,-1,sizeof(head));
        memset(next,-1,sizeof(next));
        memset(c,0,sizeof(c));
        for(int i=1,j=1; i<n; ++i)
        {
            scanf("%d%d",&a,&b);
            insert(j++,a,b);
            insert(j++,b,a);
        }
        dfs(1,-1);//遍历树
        //初始化
        for(int i=1; i<=n; ++i)
            update(i,1);
        memset(visit,false,sizeof(visit));
        scanf("%d",&m);
        for(int i=0; i<m; ++i)
        {
            scanf("%s%d",op,&a);
            if(op[0]=='C')//更新
            {
                if(visit[a])
                {
                    update(low[a],1);
                    visit[a]=false;
                }
                else
                {
                    update(low[a],-1);
                    visit[a]=true;
                }
            }
            else//查询
                printf("%d\n",gets_um(high[a])-get_sum(low[a]-1));
        }
    }
    return 0;
}*/

poj 的 judge error  还不知道哪里错了  明天继续看~~~~~