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
1 #include<bits/stdc++.h> 2 3 #define inf 0x3f3f3f3f 4 5 const int maxn=1000000; 6 7 using namespace std; 8 9 typedef long long ll; 10 11 int n; 12 13 ll cnt; 14 15 ll ans; 16 17 ll tot; 18 19 vector<ll> seg; 20 21 map<ll,int> m; 22 23 struct node{ 24 ll l,r; 25 }s[maxn+10]; 26 27 ll a[maxn+10]; 28 29 ll c[maxn+10]; 30 31 ll t[maxn+10]; 32 33 int lowbit(int x){ 34 return x&(-x); 35 } 36 37 void add(int x,int y){ 38 while(x<=cnt){ 39 c[x]+=y; 40 x+=lowbit(x); 41 } 42 } 43 44 ll sum(int x){ 45 ll res=0; 46 while(x){ 47 res+=c[x]; 48 x-=lowbit(x); 49 } 50 return res; 51 } 52 53 int main() 54 { 55 scanf("%d",&n); 56 for(int i=1;i<=n;i++){ 57 scanf("%I64d%I64d",&s[i].l,&s[i].r); 58 seg.push_back(s[i].l); 59 seg.push_back(s[i].r); 60 } 61 unique(seg.begin(),seg.end()); 62 sort(seg.begin(),seg.end()); 63 for(size_t i=0;i<seg.size();i++){ 64 if(i==0){ 65 a[++cnt]=1; 66 t[cnt]=cnt; 67 m[seg[i]]=cnt; 68 } else { 69 if(seg[i]-seg[i-1]==1){ 70 a[++cnt]=1; 71 m[seg[i]]=cnt; 72 t[cnt]=cnt; 73 } else { 74 a[++cnt]=seg[i]-seg[i-1]-1; 75 t[cnt]=cnt; 76 a[++cnt]=1; 77 m[seg[i]]=cnt; 78 t[cnt]=cnt; 79 } 80 } 81 } 82 for(int i=1;i<=n;i++){ 83 int x=m[s[i].l]; 84 int y=m[s[i].r]; 85 swap(t[x],t[y]); 86 } 87 for(int i=1;i<=cnt;i++){ 88 tot+=a[i]; 89 add(t[i],a[i]); 90 int temp=sum(t[i])-1; 91 ans+=(tot-temp-1)*a[i]; 92 } 93 printf("%I64d\n",ans); 94 return 0; 95 }
