hdu 5877 离散化+树状数组 （2016大连网赛）

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

题意：

n个节点的有根树，u是v的祖先，每个节点有个权值a[u].求 a[u] * a[v] <= k的有序对的数量。

分析：

这题一开始想到先求一下dfs序，然后把dfs序排列，二分找到每个结点的位置，然后在先前遍历找符合要求 的点，时间复杂度O(n^2)，显然会超时~~

然后就想dfs过程中维护一个数据结构，把访问过的祖先节点的a都存起来，然后找比当前节点a值小的数的个数，求一下所有结点的和即可~~

因为数据范围很大，显然要离散化，顺便把k/a[i]也放进去，然后用树状数组维护一下即可。时间复杂度O（nlogn）

#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <queue>
#include <set>

using namespace std;
typedef long long ll;
typedef pair<int, int>pii;
const double PI = acos (-1.0);
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 9;
ll a
;
int vis
, head
, p
, c
;
ll k, ans;
int n, cnt, tot;
struct Edge {
int v,  next;
} edge
;

void addedge (int u, int  v) {
edge[tot].v = v;
edge[tot].next = head[u];
head[u] = tot++;
}

int lowbit (int x) {
return x & (-x);
}
void update (int x, int val, int n) {
for (int i = x; i <= n; i += lowbit (i) )
c[i] += val;
}
int getsum (int x) {
int ans = 0;
for (int i = x; i > 0; i -= lowbit (i) )
ans += c[i];
return ans;
}
void dfs (int u) {
int pos = a[u + n];
ans += getsum (pos);
update (a[u], 1, n + n);
for (int i = head[u]; ~i; i = edge[i].next) {
int v = edge[i].v;
dfs (v);
}
update (a[u], -1, n + n);
}
bool cmp (int x, int y) {
if (a[x] == a[y]) return x < y;
return a[x] < a[y];
}

int main() {
//freopen ("f.txt", "r", stdin);
int u, v;
int T;
scanf ("%d", &T);
while (T--) {
scanf ("%d%lld", &n, &k);
memset (head, -1, sizeof (head) );
memset (vis, 0, sizeof (vis) );
ans = 0;
tot = 0;

for (int i = 1; i <= n; i++) {
scanf ("%lld", &a[i]);
if (a[i] == 0) a[n + i] = INF;
else a[n + i] = k / a[i];
p[i] = i;
p[n + i] = n + i;
}
sort (p + 1, p + 1 + n + n, cmp);

for (int i = 1; i <= n + n; i++) a[p[i]] = i;

for (int i = 0; i < n - 1; i++) {
scanf ("%d%d", &u, &v);
addedge (u, v);
vis[v]++;
}
for (int i = 1; i <= n; i++) if (vis[i] == 0) {
u = i;
break;
}
dfs (u);
printf ("%lld\n", ans);
}
return 0;
}