hdu 5877（树状数组+离散化）

题意：查找节点与它的父亲节点相乘小于k的种类数。

思路：用一个树状数组维护，查询的话就是它的父亲节点小于k/a[i]的个数。当跑完这个分支之后，要对他进行清除操作。

PS:数比较大，需要离散化。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<bitset>#include<cmath>#include<deque>#include<set>#include<vector>using namespace std;#define clr(x,y) memset(x,y,sizeof x)typedef long long ll;const int maxn = 1e5 + 10;ll a[maxn],b[maxn],c[maxn];struct Edge{int to,next;}edge[maxn*2];int edge_num;int in[maxn];int head[maxn];int tree[maxn];ll ans;int n;ll m;int len;inline int lowbit(int x){return x &(-x);}void update(int x,int val){for(int i = x;i < maxn;i += lowbit(i))tree[i] += val;}int get(int x){int ret = 0;for(int i = x; i > 0; i -= lowbit(i))ret += tree[i];return ret;}void dfs(int u){    int pos = lower_bound(b+1,b+len+1,m/a[u]) - b;if(pos == len +1 || b[pos] > m/a[u])pos --;    ans += get(pos);    if(head[u] == -1)return;    update(c[u],1);    for(int i = head[u];i != -1; i = edge[i].next)    {        int v = edge[i].to;dfs(v);    }    update(c[u],-1);}int main(){    int Tcase;scanf("%d",&Tcase);    while(Tcase --)    {        clr(in,0);edge_num = 0;clr(head,-1);        scanf("%d%I64d",&n,&m);        for(int i = 1; i <= n; i ++)scanf("%I64d",&a[i]),b[i] = a[i];        sort(b+1,b+n + 1);len = unique(b+1,b+n+1) - b - 1;        for(int i = 1; i <= n; i ++)c[i] = lower_bound(b+1,b+len+1,a[i])-b;        for(int i = 1; i < n; i ++)        {            int x,y;scanf("%d%d",&x,&y);in[y] ++;            edge[edge_num] = (Edge){y,head[x]};head[x] = edge_num ++;        }        int root;for(int i = 1; i <= n; i ++)if(!in[i])root = i;        clr(tree,0);        ans = 0;dfs(root);        printf("%I64d\n",ans);    }    return 0;}