SPOJ: BGSHOOT ( SegmentTree-DS)


SPOJ: BGSHOOT ( Segment Tree-DS )


///==================================================///
/// HELLO WORLD !! ///
/// IT'S ME ///
/// BISHAL GAUTAM ///
/// [ bsal.gautam16@gmail.com ] ///
///==================================================///
#include<bits/stdc++.h>
#define X first
#define Y second
#define mpp make_pair
#define nl printf("\n")
#define SZ(x) (int)(x.size())
#define pb(x) push_back(x)
#define pii pair<int,int>
#define pll pair<ll,ll>
///---------------------
#define S(a) scanf("%d",&a)
#define P(a) printf("%d",a)
#define SL(a) scanf("%lld",&a)
#define S2(a,b) scanf("%d%d",&a,&b)
#define SL2(a,b) scanf("%lld%lld",&a,&b)
///------------------------------------
#define all(v) v.begin(),v.end()
#define CLR(a) memset(a,0,sizeof(a))
#define SET(a) memset(a,-1,sizeof(a))
#define fr(i,a,n) for(int i=a;i<=n;i++)
#define UNIK(v) sort(all(v)),v.resize( unique(all(v)) -v.begin() );
using namespace std;
typedef long long ll;
///==========CONSTANTS=============///
/// Digit 0123456789012345678 ///
#define MX 200005
#define inf 1000000010
#define MD 1000000000LL
#define eps 1e-9
///===============================///

int xx[MX+2],qx[MX+2],yy[MX+2],qy[MX+2];
int lim,ar[2*MX+2],dp[2*MX+2][22];

void Rmq( ) {
for(int i=1; i<=lim; i++)dp[i][0]=ar[i];
for(int j=1; (1<<j)<=lim; j++) {
for(int i=1; i+(1<<j)-1<=lim; i++) {
dp[i][j]=(dp[i][j-1]>dp[i+(1<<(j-1))][j-1])?dp[i][j-1]:dp[i+(1<<(j-1))][j-1];
}
}
}

int Qry(int l,int r) {
int k=log2(r-l+1);
return (dp[l][k]>dp[r-(1<<k)+1][k])?dp[l][k]:dp[r-(1<<k)+1][k];
}

vector<int>v;
int main() {
int tc,cs=1,i,j,k,n,m,q,x,y;
S(n);
fr(i,1,n) {
S2(xx[i],yy[i]);
v.pb( xx[i] );
v.pb( yy[i] );
}
S(m);
fr(i,1,m) {
S2(qx[i],qy[i]);
v.pb( qx[i] );
v.pb( qy[i] );
}
UNIK(v);
lim=0;
fr(i,1,n) {
x=lower_bound( all(v),xx[i] )-v.begin()+1;
y=lower_bound( all(v),yy[i] )-v.begin()+1;
ar[x]++;
ar[y+1]--;
lim=max(lim,y+2);
}
fr(i,1,lim) {
ar[i]+=ar[i-1];
}
Rmq( );
ll sm=0LL;
fr(i,1,m) {
x=lower_bound( all(v),qx[i] )-v.begin()+1;
y=lower_bound( all(v),qy[i] )-v.begin()+1;
int mx=Qry(x,y);
printf("%d\n",mx);
}
return 0;
}

Comments

  1. ar eta naki segment+compression kore solve kora jay ! kibhabe compression?

    ReplyDelete
  2. You can replace RMQ part of this solution with segment tree ,which will cost LogN for each query ,wheres RMQ cost O(1). For compression, Here I pushed each input on a vector-v, then unify vector. After that, compressed each value to 1-based index of on sorted vector.

    ReplyDelete

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