codeforces540E-树状数组求逆序对

1-1e9的数据范围 但有1e5个区间 所以可以考虑把没有涉及到的区间缩成一个点 然后树状数组求逆序对

 #include<bits/stdc++.h>

 #define inf 0x3f3f3f3f

 const int maxn=;

 using namespace std;

 typedef long long ll;

 int n;

 ll cnt;

 ll ans;

 ll tot;

 vector<ll> seg;

 map<ll,int> m;

 struct node{
  ll l,r;
 }s[maxn+];

 ll a[maxn+];

 ll c[maxn+];

 ll t[maxn+];

 int lowbit(int x){
    return x&(-x);
 }

 void add(int x,int y){
     while(x<=cnt){
         c[x]+=y;
         x+=lowbit(x);
     }
 }

 ll sum(int x){
     ll res=;
     while(x){
         res+=c[x];
         x-=lowbit(x);
     }
     return res;
 }

 int main()
 {
     scanf("%d",&n);
     for(int i=;i<=n;i++){
         scanf("%I64d%I64d",&s[i].l,&s[i].r);
         seg.push_back(s[i].l);
         seg.push_back(s[i].r);
     }
     unique(seg.begin(),seg.end());
     sort(seg.begin(),seg.end());
     for(size_t i=;i<seg.size();i++){
            if(i==){
                   a[++cnt]=;
                   t[cnt]=cnt;
                   m[seg[i]]=cnt;
            } else {
            if(seg[i]-seg[i-]==){
                 a[++cnt]=;
                 m[seg[i]]=cnt;
                 t[cnt]=cnt;
            } else {
               a[++cnt]=seg[i]-seg[i-]-;
               t[cnt]=cnt;
               a[++cnt]=;
               m[seg[i]]=cnt;
               t[cnt]=cnt;
            }
            }
     }
     for(int i=;i<=n;i++){
         int x=m[s[i].l];
         int y=m[s[i].r];
         swap(t[x],t[y]);
     }
     for(int i=;i<=cnt;i++){
         tot+=a[i];
         add(t[i],a[i]);
         int temp=sum(t[i])-;
         ans+=(tot-temp-)*a[i];
     }
     printf("%I64d\n",ans);
     return ;
 }