可以把连续不改变的一条线段看做一个点，重新赋值，求逆序对即可。
注意开 \(long long\) ！
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<unordered_map>
using namespace std;
const int N = 5e5 + 1 ;
#define int long long
struct node {
int c [ N ] ;
int lb ( int x ){ return x & ( - x ) ; }
int update ( int x , int y ) {
for ( ; x < N ; x += lb ( x ) )
c [ x ] += y ;
}
int query ( int x ) {
int rhs = 0 ;
for ( ; x ; x -= lb ( x ) )
rhs += c [ x ] ;
return rhs ;
}
}tree;
int n , ls [ N ] , m , now ;
pair < int , int > a [ N ] ;
pair < int , int > True [ N ] ;
unordered_map < int , int > map;
signed main ( ) {
cin >> n ;
for ( int i = 1 ; i <= n ; i ++ ) {
scanf ( "%d %d" , & a [ i ] . first , & a [ i ] . second ) ;
ls [ ++ m ] = a [ i ] . first;
ls [ ++ m ] = a [ i ] . second;
}
std :: sort ( ls + 1 , ls + m + 1 ) ;
m = unique ( ls + 1 , ls + m + 1 ) - ls - 1 ;
for ( int i = 1 ; i <= m ; i ++ ) {
if ( ls [ i ] == ls [ i - 1 ] + 1 ) ;
else {
now ++ ;
True [ now ] = { now , ls [ i ] - ls [ i - 1 ] - 1 } ;
}
now ++ ;
True [ now ] = { now , 1 } ;
map [ ls [ i ] ] = now;
}
for ( int i = 1 ; i <= n ; i ++ ) {
swap ( True [ map [ a [ i ] . first ] ] , True [ map [ a [ i ] . second ] ] ) ;
}
int ans = 0 ;
for (int i = now ; i >= 1 ; i -- ) {
ans += tree . query ( True [ i ] . first - 1 ) * True [ i ] . second ;
tree . update ( True [ i ] . first , True [ i ] . second ) ;
}
printf ( "%lld\n" , ans ) ;
return 0 ;
}