2016 大连网赛---Weak Pair(dfs+树状数组)

Posted 2020-08-08 茶飘香~

题目链接

http://acm.split.hdu.edu.cn/showproblem.php?pid=5877

 

Problem Description

You are given a rooted tree of N nodes, labeled from 1 to N. To the ith node a non-negative value ai is assigned.An ordered pair of nodes (u,v) is said to be weak if
  (1) u is an ancestor of v (Note: In this problem a node u is not considered an ancestor of itself);
  (2) au×av≤k.

Can you find the number of weak pairs in the tree?

 

Input

There are multiple cases in the data set.
  The first line of input contains an integer T denoting number of test cases.
  For each case, the first line contains two space-separated integers, N and k, respectively.
  The second line contains N space-separated integers, denoting a1 to aN.
  Each of the subsequent lines contains two space-separated integers defining an edge connecting nodes u and v , where node u is the parent of node v.

  Constrains: 
  1≤N≤105 
  0≤ai≤109 
  0≤k≤1018

 

Output

For each test case, print a single integer on a single line denoting the number of weak pairs in the tree.

 

Sample Input

1

2 3

1 2

1 2

 

Sample Output

1

 

题意：输入n，k  表示有n个节点的一棵树，然后输入n个节点的权值和n-1条边，求点对（u，v）的对数，满足：

      1、u是v的祖先节点。

      2、a[u]*a[v]<=k，a[]是存储权值的数组。

 

思路：从根节点开始向下深搜，每到一个点时计算sum+=Sum（a[i]）,Sum(x)表示大于等于x的个数，然后向树状数组中加入k/a[i]，继续递归深搜，退栈时从树状数组中减去k/a[i] ,这样可以保证树状数组中存的一直是一条到根节点的路径值。大题思路如上，这里要做一个离散化的处理，输入的权值<=1e9  k<=1e18  而只有1e5个点，所以可以离散到2*1e5 后处理；

 

题解中提示用treap计算大于等于x的个数，这样可以不需要进行离散化；

第一次自己做出深搜的题，挺高兴的^_^  看样子我对深搜有了一点认识了

 

代码如下：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <map>
using namespace std;
const long long maxn=200003;
long long root;
long long sum,k;
long long in[100005];
vector<long long>g[100005];
long long a[100005];
long long b[200005];

long long c[200005];
map<long long,long long>q;

long long Lowbit(long long t)
{
    return t&(t^(t-1));
}
void add(long long x,long long t)
{
    while(x > 0)
    {
        c[x]+=t;
        x -= Lowbit(x);
    }
}
long long Sum(long long li)
{
    long long s=0;
    while(li<200005)
    {
        s+=c[li];
        li=li+Lowbit(li);
    }
    return s;
}

void dfs(long long t)
{
    long long n=g[t].size();
    for(long long i=0;i<n;i++)
    {
        long long v=g[t][i];
        sum+=(long long)Sum(q[a[v]]);
        if(a[v]==0) add(maxn,1);
        else   add(q[k/a[v]],1);
        dfs(v);
        if(a[v]==0) add(maxn,-1);
        else   add(q[k/a[v]],-1);
    }
}

int main()
{
    long long T,N;
    scanf("%lld",&T);
    while(T--)
    {
        q.clear();
        memset(in,0,sizeof(in));
        memset(c,0,sizeof(c));
        memset(b,0,sizeof(b));
        scanf("%lld%lld",&N,&k);
        for(long long i=1;i<=N;i++)
        {
            scanf("%lld",&a[i]);
            b[2*i-2]=a[i];
            if(a[i]!=0)
            b[2*i-1]=k/a[i];
            g[i].clear();
        }
        sort(b,b+2*N);
        long long tot=0,pre=-1;
        for(long long i=0;i<2*N;i++)
        {
            if(b[i]!=pre)
            {
                pre=b[i];
                q[pre]=++tot;
            }
        }
        for(long long i=0;i<N-1;i++)
        {
            long long aa,bb;
            scanf("%lld%lld",&aa,&bb);
            g[aa].push_back(bb);
            in[bb]++;
        }
        for(long long i=1;i<=N;i++)
        if(in[i]==0) { root=i; break; }

        sum=0;
        if(a[root]==0) add(maxn,1);
        else   add(q[k/a[root]],1);
        dfs(root);
        printf("%lld\n",sum);
    }
    return 0;
}