Problem link: SPOJ – YODANESS – Yodaness Level
Solution : This problem is of relatively easy logic as a BIT problem. About like the logic of the problem SPOJ – Inversion Count. But do not compare with that. Because the logic is not same but the concept is about similar. So you have to catch the concept by yourself in pen n paper, you will get your simple logic 🙂
#include <bits/stdc++.h>
#define pf printf
#define sf(a) scanf("%d",&a)
#define sfl(a) scanf("%lld",&a)
#define sff(a,b) scanf("%d %d",&a,&b)
#define sffl(a,b) scanf("%lld %lld",&a,&b)
#define sfff(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define sffff(a,b,c,d) scanf("%d %d %d %d",&a,&b,&c,&d)
#define sffffl(a,b,c,d) scanf("%lld %lld %lld %lld",&a,&b,&c,&d)
#define sfffff(a,b,c,d,e) scanf("%d %d %d %d %d",&a,&b,&c,&d,&e)
#define sfffffl(a,b,c,d,e) scanf("%lld %lld %lld %lld %lld",&a,&b,&c,&d,&e)
#define sfc(a) scanf("%c",&a)
#define ms(a,b) memset(a,b,sizeof(a))
#define pb(a) push_back(a)
#define db double
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define ff first
#define ss second
#define sz(x) (int)x.size()
#define all(x) x.begin(),x.end()
#define CIN ios_base::sync_with_stdio(0); cin.tie(0)
#define max3(a, b, c) max(a, b) > max(b, c) ? max(a, b) : max(b, c)
#define min3(a, b, c) min(a, b) < min(b, c) ? min(a, b) : min(b, c)
#define for0(i,n) for(int i=0;i<n;i++)
#define for1(i,n) for(int i=1;i<=n;i++)
#define forrev(i,n) for(int i=n-1; i>=0; i--)
#define forab(i,a,b) for(int i=a;i<=b;i++)
#define forba(i,b,a) for(int i=b;i>=a;i--)
#define stlloop(x) for(__typeof(x.begin()) it=x.begin();it!=x.end();it++)
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) ((a)*((b)/gcd(a,b)))
#define case1(z) cout<<"Case "<<z<<": "
#define case2(z) printf("Case %d: ",z)
#define PI acos(-1) //3.14159265358979323846264338328
#define valid(tx,ty) tx>=0 && tx<row && ty>=0 && ty<col
#define intlim 2147483648
#define MAX 100005
#define inf 100000008
ll power(int n,int p)
{
ll ans=1;
for1(i,p) ans*=n;
return ans;
}
ll m=1000001;
ll bigmod(ll n,ll p)
{
if(p==0) return 1;
else if(p%2==0)
{
ll x=(bigmod(n,p/2))%m;
return (x*x)%m;
}
else return ((n%m)*(bigmod(n,p-1)%m))%m;
}
/*------------------------------Graph Moves----------------------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1}; // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1}; // Kings Move
//const int fx[]={-2, -2, -1, -1, 1, 1, 2, 2}; // Knights Move
//const int fy[]={-1, 1, -2, 2, -2, 2, -1, 1}; // Knights Move
/*---------------------------------------------------------------------*/
using namespace std;
ll tree[30004],input[30004];
map<string,int> mp;
vector<string> v;
ll query(ll idx)
{
ll sum=0;
while(idx>0)
{
sum+=tree[idx];
idx-= idx & (-idx);
}
return sum;
}
void update(ll idx,ll val, int n)
{
while(idx<=n)
{
tree[idx]+=val;
idx+=idx&(-idx);
}
}
int main()
{
CIN;
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int t;
cin>>t;
for1(z,t)
{
int n,c=0;
cin>>n;
string s,ss,str;
for1(i,n)
{
cin>>s;
v.pb(s);
}
for1(i,n)
{
cin>>s;
c++;
mp[s]=c;
}
for0(i,sz(v)) {input[i+1]=mp[v[i]];}
ll ans=0;
for1(i,n)
{
update(input[i],1,n);
ll x=query(input[i]-1);
ans+=(input[i]-x-1);
}
cout<<ans<<endl;
mp.clear();
v.clear();
ms(tree,0);
}
return 0;
}
Fall out, Codes! 