【HDU5877】Weak Pair(线段树+dfs+离散化)

记录一个菜逼的成长。。

题目大意：
给你一个有根树，每个节点有权值，如果u是v的祖先，并且a[u] * a[v] <= k则（u,v）是一个虚弱对，求有多少组这样的对。

dfs序，dfs到某一节点时，路径上走过的点都是此节点的祖先。

///#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <iostream>#include <cstring>#include <string>#include <algorithm>#include <cstdlib>#include <vector>#include <set>#include <map>#include <queue>#include <stack>#include <list>#include <deque>#include <cctype>#include <bitset>#include <cmath>using namespace std;#define ALL(v) (v).begin(),(v).end()#define cl(a,b) memset(a,b,sizeof(a))#define bp __builtin_popcount#define pb push_back#define fin freopen("D://in.txt","r",stdin)#define fout freopen("D://out.txt","w",stdout)#define lson t<<1,l,mid#define rson t<<1|1,mid+1,r#define seglen (node[t].r-node[t].l+1)#define pi 3.1415926#define e  2.718281828459typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> PII;typedef pair<LL,LL> PLL;typedef vector<PII> VPII;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;template <typename T>inline void read(T &x){    T ans=0;    char last=' ',ch=getchar();    while(ch<'0' || ch>'9')last=ch,ch=getchar();    while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();    if(last=='-')ans=-ans;    x = ans;}inline bool DBread(double &num){    char in;double Dec=0.1;    bool IsN=false,IsD=false;    in=getchar();    if(in==EOF) return false;    while(in!='-'&&in!='.'&&(in<'0'||in>'9'))        in=getchar();    if(in=='-'){IsN=true;num=0;}    else if(in=='.'){IsD=true;num=0;}    else num=in-'0';    if(!IsD){        while(in=getchar(),in>='0'&&in<='9'){            num*=10;num+=in-'0';}    }    if(in!='.'){        if(IsN) num=-num;            return true;    }else{        while(in=getchar(),in>='0'&&in<='9'){                num+=Dec*(in-'0');Dec*=0.1;        }    }    if(IsN) num=-num;    return true;}template <typename T>inline void write(T a) {    if(a < 0) { putchar('-'); a = -a; }    if(a >= 10) write(a / 10);    putchar(a % 10 + '0');}/******************head***********************/const int maxn = 100000 + 10;int sum[maxn<<3];//数组开小了竟然判TLE  orz..找了好几个小时。。LL a[maxn<<1],b[maxn<<1],k,ans;int deep[maxn<<1];int n,m;int cnt,head[maxn];//边数较大，前向星建边struct Edge{    int to,next;}edge[maxn<<1];void add(int u,int v){    edge[cnt].to = v;    edge[cnt].next = head[u];    head[u] = cnt++;}//线段树基本操作。void pushup(int t){    sum[t] = sum[t<<1] + sum[t<<1+1];}void build(int t,int l,int r){    if(l == r){        sum[t] = 0;        return ;    }    int mid = (r + l) >> 1;    build(lson);    build(rson);    pushup(t);}int query(int t,int l,int r,int ql,int qr){    if(l >= ql && r <= qr){        return sum[t];    }    int mid = (l + r) >> 1;    if(ql > mid)return query(t<<1|1,mid + 1,r,ql,qr);    else if(qr <= mid)return query(t<<1,l,mid,ql,qr);    else {        return query(t<<1,l,mid,ql,mid) + query(t<<1|1,mid+1,r,mid+1,qr);    }}void update(int t,int l,int r ,int pos,LL v){    if(l== r){        sum[t] += v;        return ;    }    int mid = (l + r) >> 1;    if(pos > mid)update(t<<1|1,mid+1,r,pos,v);    else if(pos <= mid)update(t<<1,l,mid,pos,v);    pushup(t);}void dfs(int u){    int l = lower_bound(b+1,b+1+m,k/a[u]) - b;    ans += 1LL * query(1,1,m,1,l);//查询此节点之前有几个满足条件    int pos = lower_bound(b+1,b+1+m,a[u]) - b;    update(1,1,m,pos,1);//标记此节点，说明经过此节点    for( int i = head[u]; ~i; i = edge[i].next)dfs(edge[i].to);    update(1,1,m,pos,-1);//清除标记}int main(){    int T;scanf("%d",&T);    while(T--){        cl(deep,0);cl(head,-1);        cl(sum,0);        ans = cnt = 0;        scanf("%d%lld",&n,&k);        for( int i = 1; i <= n; i++ ){            scanf("%lld",a+i);            b[i] = a[i];        }        m = n;        //将k / a[i]离散化        for( int i = 1; i <= n; i++ ){            b[++m] = k / a[i];        }        sort(b+1,b+1+m);        m = unique(b+1,b+1+m) - (b+1);        build(1,1,m);        for( int u,v,i = 0; i < n - 1; i++ ){            scanf("%d%d",&u,&v);            add(u,v);            deep[v]++;        }        for( int i = 1; i <= n; i++ ){            if(deep[i] == 0)dfs(i);//从根节点开始        }        printf("%lld\n",ans);    }    return 0;}