luogu P1966 火柴排队

Posted 2020-10-03 ZlycerQan

二次联通门 : luogu P1966 火柴排队

 

 

 

 

/*
    luogu P1966 火柴排队

    神TM逆序对。。。
    noip怎么这么坑啊。。
    暴力都没得打
    此题模拟考试时爆了0

    做法
    将A数组排序，由于B数组与A数组是一一对应的
    那么B数组的位置也会发生相应的变化
    此时B数组逆序数对数即为答案
*/
#include <cstdio>
#include <iostream>
#include <algorithm>

const int BUF = 123123123;
char Buf[BUF], *buf = Buf;

inline void read (int &now)
{
    for (now = 0; !isdigit (*buf); ++ buf);
    for (; isdigit (*buf); now = now * 10 + *buf - ‘0‘, ++ buf);
}
#define Mod 99999997
#define Max 1232334

struct Data
{
    int x, Id;
    bool operator < (const Data &now) const
    {
        return this->x < now.x;
    }
};

Data a[Max], b[Max];
int key[Max];

struct Bit_Tree
{
    protected : int data[Max], N;
    public :
        inline void Prepare (int x) { this->N = x; }
        void C (int pos)
        {
            for (; pos <= N; pos += pos & -pos)
                ++ data[pos];
        }

        int Q (int pos)
        {
            register int i; int res = 0;
            for (i = N; i; i -= i & -i) res += data[i];
            for (i = pos - 1; i; i -= i & -i) res -= data[i];
            return res;
        }
};
Bit_Tree Bit;

int Main ()
{
    freopen ("match.in", "r", stdin);
    freopen ("match.ans", "w", stdout);
    fread (buf, 1, BUF, stdin);
    int N; read (N); register int i; int Answer = 0;
    for (i = 1; i <= N; ++ i) read (a[i].x), a[i].Id = i;
    for (i = 1; i <= N; ++ i) read (b[i].x), b[i].Id = i;
    std :: sort (a + 1, a + 1 + N);
    std :: sort (b + 1, b + 1 + N);
    for (i = 1; i <= N; ++ i) key[a[i].Id] = b[i].Id;    
    for (i = 1, Bit.Prepare (N); i <= N; ++ i)
    {
        Bit.C (key[i]);
        Answer = (Answer + Bit.Q (key[i] + 1)) % Mod;
    }
    printf ("%d", Answer);
    return 0;
}    
int ZlycerQan = Main ();
int main (int argc, char *argv[]) {;}