Solution to a problem P3157 [[CQOI2011] dynamic reverse order]

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Solution [CQOI2011] dynamic reverse order

Title effect: Given a \ (\ n-) number of arranged sequentially delete \ (m \) elements, each element of the interrogation before the delete number reverse

Analysis: For this turn remove elements of the problem, our usual solution is to turn back the time delete becomes reverse insertion order , insert a number for each question then converted after (before the corresponding deletion) asks for the number of reverse.

We set elements \ (i \) timestamp is \ (T_i \) (For those elements are not removed, \ (T_i = 0 \) ), the location is \ (P_i \) , the right value of \ (V_i \ )

So, if the insert element \ (J \) , the element \ (I \) and elements \ (J \) constitute reverse order, they need to meet

\ [\ Begin {cases} T_i \ leq T_j \ quad \ text {i j can only be inserted into the first configuration for reverse} \\ P_i <P_j \\ V_i> V_j \ quad \ text {Since the arrangement is, we may not consider equals} \ end {cases} \]

或者
\[\begin{cases}T_i \leq T_j \\ P_i > P_j \\ V_i < V_j \end{cases}\]

Then the problem becomes a three-dimensional partial order problem, because they do not by force, we consider \ (cdq \) partition, we have an example to the first case (the second case Yihuhuhuapiao analogy can)

We first triplet () \ (T, P, V) \ According to the first dimension from small to large, and then follow the second dimension from small to large order of merging, because you have to count is the second dimension than this number small numbers in the third dimension is larger than the number the number of this number (probably a bit difficult to understand, forgive me bad language), and then find the third dimension can be larger than the current number in the number of data structures

The second case analogy, I wrote two lazy \ (cdq \) partition,Anyway, do not copy and paste tired (fog

Do you think this problem will be solved?naive

We actually calculated at the time \ (t \) insert elements \ (j \) after the new reverse number (you can see from the first dimension of the partial order out of this), so the actual answer to a request prefix and

Then no pit, and note that read elements instead of the subscript

Then remember to open \ (long \; long \) just fine, although I have not been seriously considered will not fry

On the code:

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 100;
int val[maxn],to[maxn],n,m;
namespace BIT{//树状数组,清空BIT我选择了打时间戳
    int f[maxn],t[maxn],n,tot;
    inline void init(int x){n = x;}
    inline int lowbit(int x){return x & (-x);}
    inline void clear(){tot++;}
    inline void add(int pos,int x){
        while(pos <= n){
            if(t[pos] < tot)f[pos] = 0,t[pos] = tot;
            f[pos] += x;
            pos += lowbit(pos);
        }
    }
    inline int query(int pos){
        int ret = 0;
        while(pos){
            if(t[pos] < tot)f[pos] = 0,t[pos] = tot;
            ret += f[pos];
            pos -= lowbit(pos);
        }
        return ret;
    }
    inline int query(int a,int b){return query(b) - query(a - 1);}
}
ll ans[maxn];
struct Node{//三元组,本来打算用tuple,但是担心常数还是选择手写
    int x,y,z;
    bool operator < (const Node &rhs)const{
        return x < rhs.x;
    }
};
namespace cdqa{//对应上文第一种情况
    Node val[maxn],tmp[maxn];
    void cdq(int a,int b){
        if(a == b)return;
        int mid = (a + b) >> 1;
        cdq(a,mid),cdq(mid + 1,b);
        int p1 = a,p2 = mid + 1,p = a;
        while(p1 <= mid && p2 <= b){
            if(val[p1].y < val[p2].y)BIT::add(val[p1].z,1),tmp[p++] = val[p1++];//按第二维从小到大的顺序归并
            else ans[val[p2].x] += BIT::query(val[p2].z + 1,n),tmp[p++] = val[p2++];
        }
        while(p1 <= mid)tmp[p++] = val[p1++];
        while(p2 <= b)ans[val[p2].x] += BIT::query(val[p2].z + 1,n),tmp[p++] = val[p2++];
        for(int i = a;i <= b;i++)val[i] = tmp[i];
        BIT::clear();
    }
}
namespace cdqb{//第二种情况
    Node val[maxn],tmp[maxn];
    void cdq(int a,int b){
        if(a == b)return;
        int mid = (a + b) >> 1;
        cdq(a,mid),cdq(mid + 1,b);
        int p1 = a,p2 = mid + 1,p = a;
        while(p1 <= mid && p2 <= b){
            if(val[p1].y > val[p2].y)BIT::add(val[p1].z,1),tmp[p++] = val[p1++];
            else ans[val[p2].x] += BIT::query(1,val[p2].z - 1),tmp[p++] = val[p2++];
        }
        while(p1 <= mid)tmp[p++] = val[p1++];
        while(p2 <= b)ans[val[p2].x] += BIT::query(1,val[p2].z - 1),tmp[p++] = val[p2++];
        for(int i = a;i <= b;i++)val[i] = tmp[i];
        BIT::clear();
    }
}
int main(){
#ifdef LOCAL
    freopen("fafa.in","r",stdin);
#endif
    scanf("%d %d",&n,&m);
    BIT::init(n);
    for(int i = 1;i <= n;i++)scanf("%d",&val[i]),to[val[i]] = i;//由于读入的是元素,所以我们需要做一个映射得到下标
    for(int i = 1;i <= n;i++){//读入
        cdqa::val[i].x = 0,cdqa::val[i].y = i,cdqa::val[i].z = val[i];
        cdqb::val[i].x = 0,cdqb::val[i].y = i,cdqb::val[i].z = val[i];
    }
    for(int pos,i = 1;i <= m;i++){//得到时间戳
        scanf("%d",&pos);
        cdqa::val[to[pos]].x = m - i + 1;
        cdqb::val[to[pos]].x = m - i + 1;
    }
    sort(cdqa::val + 1,cdqa::val + 1 + n);
    sort(cdqb::val + 1,cdqb::val + 1 + n);
    cdqa::cdq(1,n);
    cdqb::cdq(1,n);//计算
    for(int i = 1;i <= m;i++)ans[i] += ans[i - 1];//前缀和
    for(int i = m;i >= 1;i--)printf("%lld\n",ans[i]);//询问的是删除之后的逆序对数量,因此需要倒序输出.没有询问一开始的数量因而不需要输出ans[0]
    return 0;
}