2016大连网络赛 Weak Pair
Weak Pair
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
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
分析:dfs+树状数组+离散化;
代码:
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <climits> #include <cstring> #include <string> #include <set> #include <map> #include <queue> #include <stack> #include <vector> #include <list> #define rep(i,m,n) for(i=m;i<=n;i++) #define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++) #define mod 1000000007 #define inf 0x3f3f3f3f #define vi vector<int> #define pb push_back #define mp make_pair #define fi first #define se second #define ll long long #define pi acos(-1.0) #define pii pair<int,int> #define Lson L, mid, rt<<1 #define Rson mid+1, R, rt<<1|1 const int maxn=2e5+10; using namespace std; ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);} ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;} int n,m,k,t,h[maxn],tot,q[maxn],fa[maxn],num; ll ans,a[maxn],b[maxn],c[maxn]; struct node { int to,nxt; }e[maxn]; void add(int x,int y) { tot++; e[tot].to=y; e[tot].nxt=h[x]; h[x]=tot; } void gao(int x,int y) { for(int i=x;i<=num+5;i+=(i&(-i))) q[i]+=y; } int get(int x) { int ret=0; for(int i=x;i;i-=(i&(-i))) ret+=q[i]; return ret; } void dfs(int now) { ans+=get(num+5)-get(a[now]-1); gao(b[now],1); for(int i=h[now];i;i=e[i].nxt) { dfs(e[i].to); } gao(b[now],-1); } int main() { int i,j; scanf("%d",&t); while(t--) { ans=0; tot=0; j=0; ll p; memset(h,0,sizeof h); memset(fa,0,sizeof fa); memset(q,0,sizeof q); scanf("%d%lld",&n,&p); rep(i,1,n){ scanf("%lld",&a[i]); if(a[i]==0)b[i]=1e19; else b[i]=p/a[i]; c[j++]=a[i],c[j++]=b[i]; } sort(c,c+j); num=unique(c,c+j)-c; rep(i,1,n)a[i]=lower_bound(c,c+num,a[i])-c+2,b[i]=lower_bound(c,c+num,b[i])-c+2; rep(i,1,n-1) { int x,y; scanf("%d%d",&x,&y); add(x,y); fa[y]=x; } rep(i,1,n)if(!fa[i])dfs(i); printf("%lld\n",ans); } //system("Pause"); return 0; }