【33.18%】【hdu 5877】Weak Pair (3种解法)

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1637    Accepted Submission(s): 531

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?

 

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

 

Output

For each test case, print a single integer on a single line denoting the number of weak pairs in the tree.

 

Sample Input

12 31 21 2

 

Sample Output

1

 

Source

2016 ACM/ICPC Asia Regional Dalian Online

 

【题解】

这道题的意思是，这两个条件同时满足才是一对weak pair，而不是说其中之一。

解法是，从根节点往下遍历树。

可以看到我们的遍历顺序是1->2->3.

然后我们把之前走过的点记在栈中(实际上dfs本身就是利用栈),然后走到3，之前的点就都是3的祖先了。

但是有个问题就是，当我们走到4之后。3也在栈中。而3不是4的祖先。于是我们要把3给退栈,实现方法如下。

void dfs(int x)

{
查找某种数据结构中小于等于k/a[x]的元素个数

把当前点加入到某种数据结构中

dfs(儿子);

把当前点从某种数据结构中退出来。

}

可以选用的数据结构

1.平衡树

平衡树的话应该就很好理解了吧。

就是把某个节点加入平衡树。查找小于m的值。然后把某个节点删掉。

这三个操作不会难。

(这里用的是SBT)

2.树状数组

因为ai过大。所以需要用到离散化的思想。

具体的实现方法:

seq[1..n]存a[1..n]

在seq[n+1..2*n]存k/a[i-n](i∈[n+1..2*n])，特殊的,如果a[i-n]==0,则seq[i] = INF.

然后把seq数组升序排。

然后当我们要找小于等于k/a[x]加入的时候。就在seq数组中索引k/a[x]的下标key。

因为k/a[x]也加进去了。所以如果要找小于等于k/a[x]的数则只要获取前缀和就可以了。

然后把a[x]加入，就索引seq中a[x]的下标key2.然后递增其值，然后修改与其相关的点。

把a[x]删去，就仍索引seq中a[x]的下标(其实已经索引过了就是key2),然后递减其值，再修改相关点。

3.线段树

仍然按照树状数组的方法离散化。然后就是把树状数组改成线段树了而已。单点递增。很容易的。直接递增答案query(1,key)即可;

【代码1】平衡树

#include <cstdio>#include <cstring>#include <vector>const int MAXN = 100009;using namespace std;int T, n, father[MAXN], root_tree, root;int si_ze[MAXN * 2], l[MAXN * 2], r[MAXN * 2], totn = 0;__int64 key[MAXN * 2], k, seq[MAXN], ans = 0;vector <int> son[MAXN];void init(){memset(father, 0, sizeof(father));for (int i = 1; i <= 100000; i++)son[i].clear();root = 0;totn = 0;ans = 0;}void input_data(){scanf("%d%I64d", &n, &k);for (int i = 1; i <= n; i++)scanf("%I64d", &seq[i]);int u, v;for (int i = 1; i <= n - 1; i++){scanf("%d%d", &u, &v);father[v] = u;son[u].push_back(v);}for (int i = 1; i <= n; i++)//找根节点if (father[i] == 0){root_tree = i;break;}}void right_rotation(int &t)//右旋{int k = l[t];l[t] = r[k];r[k] = t;si_ze[k] = si_ze[t];si_ze[t] = si_ze[l[t]] + si_ze[r[t]] + 1;t = k;}void left_rotation(int &t) //左旋{int k = r[t];r[t] = l[k];l[k] = t;si_ze[k] = si_ze[t];si_ze[t] = si_ze[l[t]] + si_ze[r[t]] + 1;t = k;}void maintain(int &t, bool flag){if (flag){if (si_ze[l[l[t]]] > si_ze[r[t]]) // 这是/型right_rotation(t);elseif (si_ze[r[l[t]]] > si_ze[r[t]])//....{left_rotation(l[t]);right_rotation(t);}elsereturn;//是平衡的就结束}else{if (si_ze[r[r[t]]] > si_ze[l[t]])left_rotation(t);elseif (si_ze[l[r[t]]] > si_ze[l[t]]){right_rotation(r[t]);left_rotation(t);}elsereturn;}maintain(l[t], true);maintain(r[t], false);maintain(t, true);maintain(t, false);}void insert(int &t, __int64 data2){if (t == 0){t = ++totn;l[t] = r[t] = 0;si_ze[t] = 1;key[t] = data2;}else{si_ze[t]++;if (data2 < key[t])insert(l[t], data2);elseinsert(r[t], data2);maintain(t, data2 < key[t]);}}void de_lete(__int64 data2, int &t) {si_ze[t]--;if (data2 == key[t]){bool flag1 = true, flag2 = true;//用于判断有无左右子树。if (l[t] == 0)flag1 = false;if (r[t] == 0)flag2 = false;if (!flag1 && !flag2)t = 0;elseif (!flag1 && flag2)t = r[t];elseif (flag1 && !flag2)t = l[t];elseif (flag1 && flag2){int temp = r[t];while (l[temp])temp = l[temp];key[t] = key[temp];de_lete(key[temp], r[t]);}}elseif (data2 < key[t])de_lete(data2, l[t]);elsede_lete(data2, r[t]);}__int64 find(__int64 what, int &t){if (t == 0)return 0;__int64 temp = 0;if (what >= key[t]) //左子树加上这个节点都小于等于what满足要求{temp += si_ze[l[t]] + 1;temp += find(what, r[t]);}elsereturn find(what, l[t]);return temp;}void dfs(int rt){__int64 temp = k / seq[rt];if (seq[rt] == 0)temp = 2100000000;//这里temp本应改成1e19的，但是貌似也对了。。ans += find(temp, root);insert(root, seq[rt]);int len = son[rt].size();for (int i = 0; i <= len - 1; i++)dfs(son[rt][i]);de_lete(seq[rt], root);}void output_ans(){printf("%I64d\n", ans);}int main(){//freopen("F:\\rush.txt", "r", stdin);//freopen("F:\\rush_out.txt", "w", stdout);scanf("%d", &T);while (T--){init();input_data();dfs(root_tree);output_ans();}return 0;}

【代码2】树状数组

#include <cstdio>#include <algorithm>#include <vector>using namespace std;const int MAXN = 101000;const long long INF = 10000000000000000000;int c[MAXN * 2], a[MAXN], father[MAXN], n, lena;long long seq[MAXN * 2], ans = 0, k;vector <int> son[MAXN];void init(){ans = 0;memset(c, 0, sizeof(c));memset(father, 0, sizeof(father));for (int i = 1; i <= 100010; i++)son[i].clear();}void input_data(){scanf("%d%I64d", &n, &k);for (int i = 1; i <= n; i++){scanf("%d", &a[i]);seq[i] = a[i];}for (int i = n + 1; i <= 2 * n; i++){long long temp = a[i - n];if (a[i - n] == 0) //除0就改为无限大seq[i] = INF;elseseq[i] = k / temp;}sort(seq + 1, seq + 1 + n * 2);//可能有重复的，但没关系,lower_bound返回的是一个特定值(seq是不会发生变化的)lena = 2 * n;}int lowbit(int x){return x & -x;}void dfs(int x){int key;if (a[x] == 0)key = lower_bound(seq + 1, seq + 1 + 2 * n, INF) - seq;elsekey = lower_bound(seq + 1, seq + 1 + 2 * n, k / a[x]) - seq;long long temp2 = 0;while (key){temp2 += c[key];key = key - lowbit(key);}ans += temp2;key = lower_bound(seq + 1, seq + 1 + 2 * n, a[x]) - seq;int temp = key;while (temp <= lena){c[temp]++;temp += lowbit(temp);}int len = son[x].size();for (int i = 0; i <= len - 1; i++)dfs(son[x][i]);while (key <= lena){c[key]--;key += lowbit(key);}}void get_ans(){int root = 1;for (int i = 1; i <= n - 1; i++){int u, v;scanf("%d %d", &u, &v);father[v] = u;son[u].push_back(v);}for (int i = 1; i <= n; i++)if (father[i] == 0){root = i;}dfs(root);}void output_ans(){printf("%I64d\n", ans);}int main(){//freopen("F:\\rush.txt", "r", stdin);//freopen("F:\\rush_out.txt", "w", stdout);int t;scanf("%d", &t);while (t--){init();input_data();get_ans();output_ans();}return 0;}

【代码3】线段树，基础的单点递增

#include <cstdio>#include <algorithm>#include <vector>#define lson begin, m, rt << 1#define rson m+1,end,rt<<1|1using namespace std;const int MAXN = 101000;const long long INF = 10000000000000000000;int  a[MAXN], father[MAXN], n, lena, sum[MAXN * 2 * 4]; //记住要开4倍long long seq[MAXN * 2], ans = 0, k;vector <int> son[MAXN];void build(int begin, int end, int rt){sum[rt] = 0;if (begin == end)return;int m = (begin + end) >> 1;build(lson);build(rson);}void init(){ans = 0;memset(father, 0, sizeof(father));for (int i = 1; i <= 100010; i++)son[i].clear();}void input_data(){scanf("%d%I64d", &n, &k);for (int i = 1; i <= n; i++){scanf("%d", &a[i]);seq[i] = a[i];}for (int i = n + 1; i <= 2 * n; i++){long long temp = a[i - n];if (a[i - n] == 0)seq[i] = INF;elseseq[i] = k / temp;}sort(seq + 1, seq + 1 + n * 2);lena = 2 * n;build(1, lena, 1);}long long query(int l, int r, int begin, int end, int rt)//求区间和。{if (l <= begin && end <= r)return sum[rt];int m = (begin + end) >> 1;long long temp = 0;if (l <= m)temp += query(l, r, lson);if (m < r)temp += query(l, r, rson);return temp;}void push_up(int rt){sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];}void updata(int pos, int num, int begin, int end, int rt){if (begin == end){sum[rt] += num;return;}int m = (begin + end) >> 1;if (pos <= m)updata(pos, num, lson);elseupdata(pos, num, rson);push_up(rt);}void dfs(int x){int key;if (a[x] == 0)key = lower_bound(seq + 1, seq + 1 + 2 * n, INF) - seq;elsekey = lower_bound(seq + 1, seq + 1 + 2 * n, k / a[x]) - seq;long long temp2 = query(1, key, 1, lena, 1);ans += temp2;key = lower_bound(seq + 1, seq + 1 + 2 * n, a[x]) - seq;updata(key, 1, 1, lena, 1);int len = son[x].size();for (int i = 0; i <= len - 1; i++)dfs(son[x][i]);updata(key, -1, 1, lena, 1);}void get_ans(){int root = 1;for (int i = 1; i <= n - 1; i++){int u, v;scanf("%d %d", &u, &v);father[v] = u;son[u].push_back(v);}for (int i = 1; i <= n; i++)if (father[i] == 0){root = i;}dfs(root);}void output_ans(){printf("%I64d\n", ans);}int main(){//freopen("F:\\rush.txt", "r", stdin);//freopen("F:\\rush_out.txt", "w", stdout);int t;scanf("%d", &t);while (t--){init();input_data();get_ans();output_ans();}return 0;}