HDU 5877(Problem 1010) (DFS+树状数组+离散化）

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

题意：给定一颗树，然后对于每一个节点，找到它的任何一个祖先u，如果num[u] * num[v] <= k。则贡献加1

题解：dfs搜索一遍树结构，树状数组记录路径上的数信息，由于数太大，所以需要离散化一下

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <list>
#include <queue>
#include <map>
#include <bitset>
using namespace std;
#define L(i) i<<1
#define R(i) i<<1|1
#define INF  0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-3
#define maxn 300100
#define MOD 1000000007
struct Edge
{
int to,next;
}edge[maxn];
int tot,head[maxn];

int n,m;
int pre[maxn],num1[maxn],num2[maxn];
long long k;
long long a[maxn];
long long b[maxn],ans;
int c[maxn];
void update(int x,int num) //离散化函数
{
while(x <= maxn)
{
c[x] += num;
x += x&(-x);
}
}
int getsum(int x)
{
int s = 0;
while(x)
{
s += c[x];
x -= x&(-x);
}
return s;
}
void add_edge(int u,int v)
{
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
}
void dfs(int u)
{
//printf("%d\n",u);
update(num2[u],1);
for(int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
dfs(v);
//printf("%d\n",u);
}
//printf("%d\n",u);
update(num2[u],-1);
ans += getsum(num1[u]);
//printf("%d %lld\n",u,ans);

}
int main()
{
int t,C = 1;
scanf("%d",&t);
while(t--)
{
tot = 0;
memset(head,-1,sizeof(head));
memset(pre,-1,sizeof(pre));
memset(c,0,sizeof(c));
scanf("%d%lld",&n,&k);
for(int i = 1; i <= n; i++)
{
scanf("%lld",&a[i]);
b[2*i-2] = a[i];
b[2*i-1] = k/a[i];
}
sort(b,b+2*n);
m = unique(b,b+2*n) - b;
for(int i = 1; i <= n; i++)
{
num1[i] = lower_bound(b,b+m,k/a[i]) - b+1;
num2[i] = lower_bound(b,b+m,a[i]) - b+1;
}
for(int i = 0; i < n-1; i++)
{
int x,y;
scanf("%d%d",&x,&y);
pre[y] = x;
add_edge(x,y);
}
int s;
for(int i = 1; i <= n; i++)
if(pre[i] == -1)
s = i;
ans = 0;
dfs(s);
printf("%lld\n",ans);
}
return 0;
}