UVA11990 ``Dynamic'' Inversion (树状数组套平衡树)

``Dynamic'' Inversion

Time Limit: 3000ms

Memory Limit: 131072KB

This problem will be judged on UVA. Original ID: 11990
64-bit integer IO format: %lld      Java class name: Main

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Problem D

"Dynamic" Inversion

You are given a permutation {1,2,3,...,n}. Remove m of them one by one, and output the number of inversion pairs before each removal. The number of inversion pairs of an array A is the number of ordered pairs (i,j) such that i < j and A[i] > A[j].

Input

The input contains several test cases. The first line of each case contains two integers n and m (1<=n<=200,000, 1<=m<=100,000). After that, n lines follow, representing the initial permutation. Then m lines follow, representing the removed integers, in the order of the removals. No integer will be removed twice. The input is terminated by end-of-file (EOF). The size of input file does not exceed 5MB.

Output

For each removal, output the number of inversion pairs before it.

Sample Input

5 4153425142

Output for the Sample Input

5221

Explanation

(1,5,3,4,2)->(1,3,4,2)->(3,4,2)->(3,2)->(3)

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Rujia Liu's Present 3: A Data Structure Contest Celebrating the 100th Anniversary of Tsinghua University
Special Thanks: Yi Chen, Dun Liang
Note: Please make sure to test your program with the gift I/O files before submitting!

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给你一个１到ｎ的排列，进行ｍ次删除操作,输出每次删除之前的逆序对数，静态逆序对可以用归并排序或者树状数组来求，动态的可以用树状数组套平衡树来求，具体来说，就是树状数组的每个节点是一颗平衡树，第x个节点保存x,x-1,x-2...,x-lowbit(x)+1的所有值（建成平衡树），假设要删除的数是x[i]在原排列中的位置是pos[i]，我们只需要知道１到pos[i-1]中有多少个数比x[i]大和pos[i+1]到结尾有多少个数比x[i]小，就能知道逆序对减少了多少(已经删除的点直接忽略掉)．那么我们可以通过在pos[i],pos[i]-lowbit(pos[i]),pos[i]-lowbit(pos[i]) - lowbit(pos[i]-lowbit(pos[i]))...中查找有多少个数比小于等于ｘ，记为t2，则１到pos[i]中节点总数减t2就得到前缀比x[i]大的，小于等于ｘ的总数（可以通过树状数组查询　ask(c,val)　）－t2就是后缀里比ｘ小的，然后更新pos[i],pos[i]+lowbit(pos[i]).....这些平衡树，给被删除的节点打上标记(--siz即可）,同时在原树状数组里x[i]位置上加上－１．

#include <bits/stdc++.h>int lowbit(int x) {return x & -x;}using namespace std;typedef long long ll;const int N = 2e5 + 10;const int M = 4e6 + 10;int tot;struct Splay {    int fa,key,siz;    int ch[2];    Splay() = default;    Splay(int fa,int key):fa(fa),key(key),siz(1){        ch[0] = ch[1] = 0;    }}Node[M];int newNode(int fa,int key){    int rt = ++tot;    Node[rt] = Splay(fa,key);    return rt;}int built(vector<int>&v,int fa,int l,int r){    int mid = (l + r) >> 1;    int rt = newNode(fa,v[mid-1]);    Splay & t = Node[rt];    if(mid > l) t.ch[0] = built(v,rt,l,mid-1);    if(mid < r) t.ch[1] = built(v,rt,mid+1,r);    t.siz += Node[t.ch[0]].siz + Node[t.ch[1]].siz;    return rt;}int ask(int rt,int x)///询问rt为根的子树里有多少不大于ｘ的节点{    if(!rt)return 0;    Splay & t = Node[rt];    if(t.key == x) {        return Node[t.ch[0]].siz+1;    }    bool flag = (t.siz - (Node[t.ch[0]].siz + Node[t.ch[1]].siz));///是否没被删除    if(t.key < x) return Node[t.ch[0]].siz + flag + ask(t.ch[1],x);    return ask(t.ch[0],x);}void remove(int rt,int x){    Splay & t = Node[rt];    --t.siz;    if(t.key==x)return;    remove(t.ch[t.key < x],x);}int n,m;int root[N],pos[N];vector<int>vec[N];int c[N];void add(int *T,int x,int d){    for(;x <= n; x += (x&-x)) {        T[x] += d;    }}int ask(int *T,int x){    int ans = 0;    for(;x > 0; x -= (x&-x)) {        ans += T[x];    }    return ans;}ll cnt;int init(){    if(scanf("%d%d",&n,&m)!=2) return 0;    memset(c,0,sizeof(int)*(n+1));    for(int i = 0; i <= n; ++i) vec[i].clear();    tot = cnt = 0;    Node[0].siz = 0;    for(int i = 1; i <= n; ++i) {        int x;        scanf("%d",&x);        cnt += (i-1)-ask(c,x);        add(c,x,1);        pos[x] = i;        for(int j = i; j <= n; j += (j&-j)) {            vec[j].push_back(x);        }    }    for(int i = 1; i <= n; ++i) {///可以通过打标记将删除变成静态的，直接按照完全二叉树来建树即可        sort(vec[i].begin(),vec[i].end());        root[i] = built(vec[i],0,1,vec[i].size());    }    return 1;}void change(int x,int val){    int t1 = 0,t2 = 0;    for(; x > 0; x -= (x&-x)) {        int t = ask(root[x],val); //查询root[x]里有多少个数小于等于val        t2 += t;        t1 += Node[root[x]].siz - t;//大于val的个数    }    t2 = ask(c,val) - t2;//后缀里小于val的个数    add(c,val,-1);//在val位置加-1    cnt -= (t1+t2);//删除之后对逆序对的影响    x = pos[val];    for( ;x <= n; x += (x&-x)) remove(root[x],val);//在受影响的平衡树中删除节点}int main(){    while(init()) {        for(int i = 0; i < m; ++i) {            int x;scanf("%d",&x);            printf("%lld\n",cnt);            change(pos[x],x);        }    }    return 0;}