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=1000000;

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+10];

ll a[maxn+10];

ll c[maxn+10];

ll t[maxn+10];

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=0;
    while(x){
        res+=c[x];
        x-=lowbit(x);
    }
    return res;
}

int main()
{
    scanf("%d",&n);
    for(int i=1;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=0;i<seg.size();i++){
           if(i==0){
                  a[++cnt]=1;
                  t[cnt]=cnt;
                  m[seg[i]]=cnt;
           } else {
           if(seg[i]-seg[i-1]==1){
                a[++cnt]=1;
                m[seg[i]]=cnt;
                t[cnt]=cnt;
           } else {
              a[++cnt]=seg[i]-seg[i-1]-1;
              t[cnt]=cnt;
              a[++cnt]=1;
              m[seg[i]]=cnt;
              t[cnt]=cnt;
           }
           }
    }
    for(int i=1;i<=n;i++){
        int x=m[s[i].l];
        int y=m[s[i].r];
        swap(t[x],t[y]);
    }
    for(int i=1;i<=cnt;i++){
        tot+=a[i];
        add(t[i],a[i]);
        int temp=sum(t[i])-1;
        ans+=(tot-temp-1)*a[i];
    }
    printf("%I64d\n",ans);
    return 0;
}