逆序数 cdq分治 归并排序

题目链接

思路
利用归并排序,每次merge()就能消除多个逆序对,把每次消除的逆序对数量加起来,就得到总共的逆序对总数

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
ll num[100020];
ll num2[100020];
ll ans;
ll n;
void read(ll &x){
    
    
    char c;
    ll flag = 1;
    while((c = getchar()) && (c < '0' || c > '9'))if(c == '-') flag = -1;
    x = c -'0';
    while((c = getchar()) && (c >= '0' && c <= '9')) x = x*10 + c - '0';
    x*=flag;
}
void merge(ll l,ll m,ll r){
    
    
    ll l1 = l,l2 = m;
    ll index = l1;
    while(l1 < m && l2 < r){
    
    
        if(num[l1] <= num[l2])
            num2[index++] = num[l1++];
         else{
    
    
             num2[index++] = num[l2++];
             ans+=(m - l1);//记录num[l2]对逆序对的贡献
        }
    }
    while(l1 < m)
        num2[index++] = num[l1++];
    while(l2 < r)
        num2[index++] = num[l2++];
        
    for(ll i = l;i < r;i++)
        num[i] = num2[i];
    return ;
}
void msort(ll l,ll r){
    
    
    if(l + 1 == r)//左闭右开 
        return;
    msort(l,(l + r) >> 1);
    msort((l + r) >> 1,r);
    merge(l,(l + r) >> 1,r);
    
    return ;
}

int main(){
    
    
    read(n);
   
    for(ll i = 1;i <= n;i++)
    	read(num[i]);  
    msort(1,n+1);
    cout << ans << endl;
    return 0;
}