其他一些数状数组的题

APPLE TREE-POJ 3321

树上树状数组，lef记录次节点的编号，rig记录他的字节点中编号最大的，则sum(rig)-sum(lef-1)即为

所求值，此题会卡vector常数（我也不知道是什么东西）

#include<iostream>
#include<string.h> 
#include<cmath>
#include<vector>
#define maxn 100000+10
using namespace std;
int n,m,lef[maxn],rig[maxn],sum[maxn],tot,s[maxn];
vector<vector<int> > q(maxn);
void add(int x,int d)
{
    while(x<=n)
    {
        sum[x]+=d;
        x+=x&(-x);
    }
}
int susm(int x)
{
    int ans=0;
    while(x)
    {
        ans+=sum[x];
        x-=x&(-x);
    }
    return ans;
}
void dfs(int x)
{
    lef[x]=tot;
    for(int i=0;i<q[x].size();i++)
    {
        tot++;
        dfs(q[x][i]);
    }
    rig[x]=tot;
}                    
int main()
{
    while(cin>>n)
    {
    memset(sum,0,sizeof(sum));
    memset(lef,0,sizeof(lef));
    memset(rig,0,sizeof(rig));
    memset(s,0,sizeof(s));
    for(int i=0;i<maxn;i++) q[i].clear();
   for(int i=1;i<=n-1;i++)
   {
    int x,y;
    cin>>x>>y;
    q[x].push_back(y);
   }
   tot=1;
   dfs(1);
   for(int i=1;i<=n;i++)
   {
       s[i]=1;
       add(i,1);
   }
   cin>>m;
   while(m--)
   {
         char s1[5];
         int x;
      scanf("%s%d",s1,&x);
      if(s1[0]=='Q')
      {
          //int x;
          //cin>>x;
          cout<<susm(rig[x])-susm(lef[x]-1)<<endl;
       } 
       else
       {
             //int x;
             //cin>>x;
             if(s[x]) add(lef[x],-1);
             else add(lef[x],1);
             s[x]=!s[x];
             
      }
   }
    }
   return 0;
}

Disharmony Trees   HDU - 3015

树状数组加数据离散化，用三次排序解决即可

#include<iostream>
#include<algorithm>
using namespace std;
#define maxn (int)1e5+5
#include<cstring>
typedef long long ll;
int lowbit(int x)
{
    return x&(-x);
}
ll n,c[maxn],c1[maxn];  //c tong ji he zhi c1 tong ji ge shu
struct node
{
    ll x,h;
}tree[maxn];
bool cmp(node a,node b)
{
    return a.x<b.x;
}
bool cmp1(node a,node b)
{
    return a.h<b.h;
}
void add(ll x,ll num,ll *s)
{
    while(x<=n)
    {
        s[x]+=num;
        x+=lowbit(x);
    }
}
ll sum(ll x,ll *s)
{
    ll ans=0;
    while(x)
    {
        ans+=s[x];
        x-=lowbit(x);
    }
    return ans;
}
int main()
{
    while(cin>>n)
    {
        for(int i=1;i<=n;i++) cin>>tree[i].x>>tree[i].h;
        sort(tree+1,tree+1+n,cmp);
        int ji=tree[1].x;
        tree[1].x=1;
        for(int i=2;i<=n;i++)
        {
            if(tree[i].x==ji) tree[i].x=tree[i-1].x;
            else
            {
                ji=tree[i].x;
                tree[i].x=i;
            }
        }
        int jilu=tree[n].x;
        sort(tree+1,tree+1+n,cmp1);
        ji=tree[1].h;
        tree[1].h=1;
        for(int i=2;i<=n;i++)
        {
           if(tree[i].h==ji) tree[i].h=tree[i-1].h;
           else
           {
               ji=tree[i].h;
               tree[i].h=i;
           }    
        }
        sort(tree+1,tree+1+n,cmp1);
        memset(c,0,sizeof(c));
        memset(c1,0,sizeof(c1));
        for(int i=1;i<=n;i++)
        {
          add(tree[i].x,tree[i].x,c);
          add(tree[i].x,1,c1);    
        }
        int xiao,da;
        ll ans=0;
        for(int i=1;i<=n;i++)
        {
            xiao=sum(tree[i].x-1,c1)*tree[i].x-sum(tree[i].x-1,c);
            da=(sum(jilu,c)-sum(tree[i].x,c))-(sum(jilu,c1)-sum(tree[i].x,c1))*tree[i].x;
            ans+=tree[i].h*(xiao+da);
            add(tree[i].x,-tree[i].x,c);
            add(tree[i].x,-1,c1);
            
        }
        cout<<ans<<endl;    
    }    
    return 0;
}