hdu_5877_Weak Pair(离散+DFS+树状数组)

题目链接：hdu_5877_Weak Pair

题意：

给你一棵树，让你找有多少对满足那两个条件的weak pair

题解：

有人用Treap,我不会，然后我用树状数组+离散来替代Treap，用DFS搜到叶子，然后在树状数组中找小于k/a[u]的个数，注意a[u]可以为0,要处理一下

 #include<bits/stdc++.h>
 #define F(i,a,b) for(int i=a;i<=b;i++)
 using namespace std;
 typedef long long ll;

 const int N=1e5+;
 int t,n,v[N],g[N],nxt[N],ed,sum[N],in[N],edn;
 ll k,a[N],hsh[N],ans;

 inline int getid(ll x){return upper_bound(hsh+,hsh++edn,x)-hsh-;}
 inline void add(int x,int c){while(x<=n)sum[x]+=c,x+=x&-x;}
 inline int ask(int x){int an=;while(x)an+=sum[x],x-=x&-x;return an;}

 inline void adg(int x,int y){v[++ed]=y,nxt[ed]=g[x],g[x]=ed;}

 void dfs(int u)
 {
     add(getid(a[u]),);
     for(int i=g[u];i;i=nxt[i])dfs(v[i]);
     add(getid(a[u]),-);
     ll tp;
     if(a[u]!=)tp=k/a[u];else tp=hsh[edn]+;
     if(tp>hsh[edn])tp=hsh[edn]+;
     ans+=ask(getid(tp));
 }

 int main()
 {
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d%lld",&n,&k);
         F(i,,n)scanf("%lld",a+i),hsh[i]=a[i];
         sort(hsh+,hsh++n),memset(sum,,sizeof(sum));
         memset(in,,sizeof(in)),edn=;
         memset(g,,sizeof(g)),ed=,ans=;
         F(i,,n)if(hsh[i]!=hsh[edn])hsh[++edn]=hsh[i];
         F(i,,n-)
         {
             int x,y;
             scanf("%d%d",&x,&y);
             adg(x,y),in[y]++;
         }
         F(i,,n)if(!in[i]){dfs(i);break;}
         printf("%lld\n",ans);
     }
     return ;
 }