Weak Pair
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1445 Accepted Submission(s): 468
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?
(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
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】给你一棵有根树,一个定值k,以及树上每个结点的值a[i]，对于有序对(u,v),如果(1)u是v的祖先,且(2)a[u]*a[v]<=k,则称该有序对(u,v)是弱的，问树中有多少对有序对(u,v)是弱的。
【分析&&推荐】可以去看这篇博客，写得非常好http://blog.csdn.net/queuelovestack/article/details/52505856
【解题方法】和上面这篇文章介绍的方法一样，直接给出我的代码了。
【AC 代码】
//
//Created by just_sort 2016/9/12 16:50
//Copyright (c) 2016 just_sort.All Rights Reserved
//
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 1e5+6;
const LL inf = 1e18;
int n,q;LL k;
LL a[maxn];
LL b[maxn*2];
LL ans;
int c[maxn*2];
bool vis[maxn];
int head[maxn],edgecnt;
struct edge{
int v,next;
}E[maxn*2];
void init()
{
edgecnt=0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v)
{
E[edgecnt].v=v,E[edgecnt].next=head[u],head[u]=edgecnt++;
}
int lowbit(int x){
return x&(-x);
}
void update(int i,int v)
{
while(i<=q){
c[i]+=v;
i+=lowbit(i);
}
}
int query(int x)
{
int ret=0;
while(x){
ret += c[x];
x -= lowbit(x);
}
return ret;
}
void dfs(int u)
{
ans += query(lower_bound(b,b+q,a[u]?k/a[u]:inf)-b+1);
update(lower_bound(b,b+q,a[u])-b+1,1);
for(int i=head[u]; ~i; i=E[i].next){
int v = E[i].v;
dfs(v);
}
update(lower_bound(b,b+q,a[u])-b+1,-1);
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
q = 0;
init();
memset(c,0,sizeof(c));
memset(vis,false,sizeof(vis));
memset(b,0,sizeof(b));
scanf("%d%lld",&n,&k);
for(int i=1; i<=n; i++){
scanf("%I64d",&a[i]);
b[q++] = a[i];
if(!a[i]) b[q++]=inf;
else b[q++]=k/a[i];
}
sort(b,b+q);
q=unique(b,b+q)-b;
for(int i=1; i<n; i++){
int u,v;
scanf("%d%d",&u,&v);
addedge(u,v);
//addedge(v,u);
vis[v] = 1;
}
int ss=-1;
for(int i=1; i<=n; i++){
if(!vis[i]) ss=i;
}
ans = 0;
dfs(ss);
printf("%I64d\n",ans);
}
return 0;
}
【解法2】 利用Treap，复杂度是nlogn级别的，加个输入挂可以跑得很快。
【AC 代码】
//
//Created by just_sort 2016/9/12 16:50
//Copyright (c) 2016 just_sort.All Rights Reserved
//
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 1e5+6;
const LL inf = 1e18;
int n;
LL k,ans;
LL a[maxn];
struct FastIO
{
static const int S = 1310720;
int wpos;
char wbuf[S];
FastIO() : wpos(0) {}
inline int xchar()
{
static char buf[S];
static int len = 0, pos = 0;
if (pos == len)
pos = 0, len = fread(buf, 1, S, stdin);
if (pos == len) return -1;
return buf[pos ++];
}
inline int xuint()
{
int c = xchar(), x = 0;
while (c <= 32) c = xchar();
for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
return x;
}
inline int xint()
{
int s = 1, c = xchar(), x = 0;
while (c <= 32) c = xchar();
if (c == '-') s = -1, c = xchar();
for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
return x * s;
}
inline LL xlongint()
{
LL s = 1, c = xchar(), x = 0;
while (c <= 32) c = xchar();
if (c == '-') s = -1, c = xchar();
for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
return x * s;
}
inline void xstring(char *s)
{
int c = xchar();
while (c <= 32) c = xchar();
for (; c > 32; c = xchar()) * s++ = c;
*s = 0;
}
inline void wchar(int x)
{
if (wpos == S) fwrite(wbuf, 1, S, stdout), wpos = 0;
wbuf[wpos ++] = x;
}
inline void wint(LL x)
{
if (x < 0) wchar('-'), x = -x;
char s[24];
int n = 0;
while (x || !n) s[n ++] = '0' + x % 10, x /= 10;
while (n--) wchar(s[n]);
}
inline void wstring(const char *s)
{
while (*s) wchar(*s++);
}
~FastIO()
{
if (wpos) fwrite(wbuf, 1, wpos, stdout), wpos = 0;
}
} io;
/************************************Treap***************************************/
struct node
{
int l,r,w,rnd,size;
LL v;
};
node tr[100011];
int num=0,root=0,size=0;
void update(int k)
{
tr[k].size=tr[tr[k].l].size+tr[tr[k].r].size+tr[k].w;
return;
}
void rturn(int &k)
{
int t;t=tr[k].l;
tr[k].l=tr[t].r;tr[t].r=k;
tr[t].size=tr[k].size;
update(k);k=t;
return;
}
void lturn(int &k)
{
int t;t=tr[k].r;
tr[k].r=tr[t].l;tr[t].l=k;
tr[t].size=tr[k].size;
update(k);k=t;
return;
}
void insert(int &k,LL x)
{
if (k==0)
{
++num;k=num;tr[k].size=tr[k].w=1;
tr[k].v=x;tr[k].rnd=rand();
return;
}
tr[k].size++;
if (x==tr[k].v) tr[k].w++;
else
if (x>tr[k].v)
{
insert(tr[k].r,x);
if (tr[tr[k].r].rnd<tr[k].rnd) lturn(k);
}
else
{
insert(tr[k].l,x);
if (tr[tr[k].l].rnd<tr[k].rnd) rturn(k);
}
return;
}
void del(int &k,LL x)
{
if (k==0) return;
if (x==tr[k].v)
{
if (tr[k].w>1)
{
tr[k].w--;tr[k].size--;return;
}
else
{
if (tr[k].l==0||tr[k].r==0) k=tr[k].l+tr[k].r;
else
if (tr[tr[k].l].rnd<tr[tr[k].r].rnd)
{
rturn(k);del(k,x);
}
else
{
lturn(k);del(k,x);
}
}
}
else
if (x>tr[k].v) tr[k].size--,del(tr[k].r,x);
else tr[k].size--,del(tr[k].l,x);
}
int query_rank(int k,LL x)
{
if (k==0) return 0;
if (x==tr[k].v) return tr[tr[k].l].size+1;
if (x>tr[k].v) return tr[tr[k].l].size+tr[k].w+query_rank(tr[k].r,x);
else return query_rank(tr[k].l,x);
}
int query_num(int k,int x)
{
if (x<=tr[tr[k].l].size) return query_num(tr[k].l,x);
if (x>tr[tr[k].l].size+tr[k].w) return query_num(tr[k].r,x-tr[tr[k].l].size-tr[k].w);
return tr[k].v;
}
//int ans;
//void query_pro(int k,int x)//5
//{
// if (k==0) return;
// if (tr[k].v<x)
// {
// ans=k;query_pro(tr[k].r,x);
// }
// else query_pro(tr[k].l,x);
//}
//void query_sub(int k,int x)//6
//{
// if (k==0) return;
// if (tr[k].v>x)
// {
// ans=k;query_sub(tr[k].l,x);
// }
// else query_sub(tr[k].r,x);
//}
/******************************************************************************/
//
int head[maxn],edgecnt;
bool vis[maxn];
struct edge{
int v,next;
}E[maxn*2];
void init(){
edgecnt=0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v){
E[edgecnt].v=v,E[edgecnt].next=head[u],head[u]=edgecnt++;
}
void dfs(int u,int fa)
{
LL exe = inf;
if(a[u]!=0) exe = k/(a[u]);
ans += query_rank(root,exe);
insert(root,a[u]);
for(int i=head[u]; ~i; i=E[i].next){
int v=E[i].v;
if(v==fa) continue;
dfs(v,u);
}
del(root,a[u]);
}
int main()
{
int T;
T=io.xint();
while(T--)
{
init();
size=num=root=0;
for(int i=1; i<maxn; i++) tr[i].l=tr[i].r=0;
memset(vis,false,sizeof(vis));
//scanf("%d%lld",&n,&k);
n=io.xint();k=io.xlongint();
for(int i=1; i<=n; i++) a[i]=io.xlongint();
for(int i=1; i<n; i++){
int u,v;
//scanf("%d%d",&u,&v);
u=io.xint(),v=io.xint();
addedge(u,v);
addedge(v,u);
vis[v]=1;
}
int s=-1;
for(int i=1; i<=n; i++){
if(!vis[i]){
s=i;
break;
}
}
ans = 0;
dfs(s,0);
//io.wint(ans);
printf("%lld\n",ans);
}
return 0;
}
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