SPOJ COT - Count on a tree树链剖分加主席树

B - Count on a tree

Time Limit:129MS    Memory Limit:1572864KB    64bit IO Format:%lld & %llu

SubmitStatusPracticeSPOJ COT

Description

You are given a tree with N nodes.The tree nodes are numbered from1 to N.Each node has an integer weight.

We will ask you to perform the following operation:

• u v k : ask for the kth minimum weight on the path from nodeu to node v

 

Input

In the first line there are two integers N and M.(N,M<=100000)

In the second line there are N integers.The ith integer denotes the weight of the ith node.

In the next N-1 lines,each line contains two integers u v,which describes an edge (u,v).

In the next M lines,each line contains three integers u v k,which means an operation asking for the kth minimum weight on the path from nodeu to node v.

Output

For each operation,print its result.

Example

Input:      
        8 5      
8 5105 2 9 3 8 5 7 71 2        1 31 43 53 63 74 82 5 12 5 22 5 32 5 47 8 2 

Output:2891057 

题目大意：给你一颗树，m次询问，问节点u到v第k大的数是多少。

思路：若是求一个数组的L到R的第k大的话，直接一个主席树就可以搞定了，现在变成了树上，然后其实是直接树剖，把u到v分成最多logn段的连续序列，对每个序列求出T[r] - T[l-1]然后相加起来，就是这logn段的左区间和，做法跟求数组的第k大一样。还有就是建主席树的时候放到树剖节点映射线段树那一部分，因为要保证重链上的节点是连续的。

#include <cstdio>#include <cstring>#include <queue>#include <cmath>#include <algorithm>#include <iostream>#include <cstdlib>using namespace std;//typedef __int64 LL;const int maxn = 100005;int w[maxn],fa[maxn],dep[maxn],num[maxn],son[maxn],top[maxn],vis[maxn],head[maxn],cnt,cou;int T[maxn],lson[maxn*50],rson[maxn*50],c[maxn*50],tot,len,b[maxn],tt,user[maxn],usel[maxn];int n,m;struct Edge{    int v,next;}e[maxn*2];void add(int a,int b){    e[cnt].v = b;    e[cnt].next = head[a];    head[a] = cnt++;}void init(){    cou = cnt = tot = len = 0;    tt = 1;    memset(head,-1,sizeof(head));    memset(son,-1,sizeof(son));}void hase(int k){    sort(b,b+k);    len = unique(b,b+k) - b;}int get_hase(int now){    return (lower_bound(b,b+len,now) - b);}int build(int l,int r){    int root = tot++;    c[root] = 0;    if(l != r)    {        int mid = (l + r) >> 1;        lson[root] = build(l,mid);        rson[root] = build(mid+1,r);    }    return root;}void dfs1(int u,int v,int deep){    fa[v] = u;    num[v] = 1;    dep[v] = deep;    int i;    for(i = head[v]; i != -1; i = e[i].next)    {        int end = e[i].v;        if(end == u)            continue;        dfs1(v,end,deep+1);        num[v] += num[end];        if(son[v] == -1 || num[son[v]] < num[end])        {            son[v] = end;        }    }}int insert1(int root,int pos,int val){    int newroot = tot++;    c[newroot] = c[root] + val;    int tmp = newroot;    int l = 0,r = len - 1;    while(l < r)    {        int mid = (l + r) >> 1;        if(pos <= mid)        {            lson[newroot] = tot++;            rson[newroot] = rson[root];            root = lson[root];            newroot = lson[newroot];            r = mid;        }        else        {            lson[newroot] = lson[root];            rson[newroot] = tot++;            root = rson[root];            newroot = rson[newroot];            l = mid + 1;        }        c[newroot] = c[root] + val;    }    return tmp;}void dfs2(int u,int v){    top[v] = u;    T[tt] = insert1(T[tt-1],get_hase(w[v]),1);    tt++;    vis[v] = ++cou;    if(son[v] != -1)    {        dfs2(u,son[v]);    }    else        return;    int i;    for(i = head[v]; i != -1; i = e[i].next)    {        int end = e[i].v;        if(end == son[v] || end == fa[v])            continue;        dfs2(end,end);    }}void solve(int a,int b1,int pos){    int q = top[a],w = top[b1];    int t = 0;    while(q != w)    {        if(dep[q] < dep[w])        {            swap(q,w);            swap(a,b1);        }        user[t] = T[vis[a]],usel[t] = T[vis[q]-1];        t++;        a = fa[q];        q = top[a];    }    if(dep[a] > dep[b1])        swap(a,b1);    user[t] = T[vis[b1]],usel[t] = T[vis[a]-1];    t++;    int l = 0,r = len - 1;    int rootr,rootl,i;    while(l < r)    {        int sum = 0;        int mid = (l + r) >> 1;        for(i = 0; i < t; i++)        {            sum += c[lson[user[i]]] - c[lson[usel[i]]];        }        if(sum < pos)        {            pos -= sum;            l = mid + 1;            for(i = 0; i < t; i++)            {                user[i] = rson[user[i]];                usel[i] = rson[usel[i]];            }        }        else        {            r = mid;            for(i = 0; i < t; i++)            {                user[i] = lson[user[i]];                usel[i] = lson[usel[i]];            }        }    }    printf("%d\n",b[l]);}int main(){    while(scanf("%d %d",&n,&m) != EOF)    {        int i;        init();        for(i = 1; i <= n; i++)        {            scanf("%d",&w[i]);            b[len++] = w[i];        }        hase(len);        int q,w;        for(i = 1; i < n; i++)        {            scanf("%d %d",&q,&w);            add(q,w);            add(w,q);        }        T[0] = build(0,len-1);        dfs1(1,1,1);        dfs2(1,1);        while(m--)        {            int A,B,C;            scanf("%d %d %d",&A,&B,&C);            solve(A,B,C);        }    }    return 0;}