Taxicab Driver's Problem


Problem Statement :


Burger Town is a city that consists of  special junctions and  pathways. There is exactly one shortest path between each pair of junctions. Junction  is located at  and the distance between two junctions  is defined by the Taxicab geometry.

Tim has recently afforded a taxicab to work as a taxicab driver. His vehicle was very cheap, but has a very big flaw. It can only drive  units horizontally and  units vertically before refueling.

If a customer wants to be brought from a junction  to another junction , then this car is only capable of driving the route, iff the sum of horizontal distances and the sum of vertical distances on this path are less than or equal to  and  respectively.

Also, there is a unique path between any two junctions.

Now he has thoughts about returning the vehicle back to the seller. But he first wants to know, if it's even worth it. That's why he wants to know the number of unordered pairs  such that it is not possible to drive a customer from junction  to junction .

Input Format

On the first line you will be given ,  and  separated by a single space.
Each of the next  lines contains two space separated integers , denoting the location of junction . Each of the next  lines contains two space separated integers describing a path existing between , i.e., there is a path between  and .

Output Format

Output the number of unordered pairs  such that it is not possible to drive from  to .

Constraints

2  <=   N  <=  10^5
0  <=  H , V  <=  10^14
0  <=  xi, yi  <=  10^9


Solution :



title-img 


                            Solution in C :

In   C++  :








#include <bits/stdc++.h>

#define FO(i,a,b) for (int i = (a); i < (b); i++)
#define sz(v) int(v.size())

using namespace std;

typedef long long ll;

typedef pair<ll,ll> E;
#define dx first
#define dy second
struct edge {
    int o;
    ll dx, dy;

    edge(int o=0, ll dx=0, ll dy=0) : o(o), dx(dx), dy(dy) {}
};

vector<edge> u[100005];
int n; ll DX, DY;
ll Y[100005], X[100005];
int sc[100005];

void delete_from(int x, int y) {
    FO(i,0,sz(u[x])) if (u[x][i].o == y) {
        swap(u[x][i],u[x].back());
        u[x].pop_back();
        return;
    }
}

void delete_node(int x) {
    FO(i,0,sz(u[x])) {
        delete_from(u[x][i].o,x);
    }
}

void resetsc(int x, int p) {
    sc[x] = 1;
    FO(i,0,sz(u[x])) if (u[x][i].o != p) {
        resetsc(u[x][i].o,x);
        sc[x] += sc[u[x][i].o];
    }
}

int find_centre(int x, int p, int tsz) {
    int lsz = tsz-sc[x];
    FO(i,0,sz(u[x])) if (u[x][i].o != p) {
        lsz = max(lsz, sc[u[x][i].o]);
        int y = find_centre(u[x][i].o,x,tsz);
        if (y != -1) return y;
    }
    if (2*lsz <= tsz+5) return x;
    return -1;
}

void getl(int x, int p, vector<E> &v, ll dx, ll dy) {
    if (dx != 0 || dy != 0) v.push_back(E(dx,dy));
    FO(i,0,sz(u[x])) if (u[x][i].o != p) {
        getl(u[x][i].o, x, v, dx+u[x][i].dx, dy+u[x][i].dy);
    }
}

bool cmp(E a, E b) {
    if (a.dx != b.dx) return a.dx < b.dx;
    else return a.dy < b.dy;
}

ll res;
vector<ll> bit;

void ub(int y, int dv) {
    for (;y<sz(bit);y+=y&-y) bit[y] += dv;
}

ll qb(int y) {
    ll r = 0;
    for (;y>0;y-=y&-y) r += bit[y];
    return r;
}

ll doitbf(vector<E> p) {
    ll RES = 0;
    FO(i,0,sz(p)) FO(j,0,i) if (p[i].dx+p[j].dx <= DX && p[i].dy+p[j].dy <= DY) RES++;
    return RES;
}

ll doit(vector<E> p) {
    //printf("DOIT\n");
    //FO(i,0,sz(p)) printf("%lld,%lld\n", p[i].dx, p[i].dy);
    ll RES = 0;
    vector<E> q;
    vector<ll> y;
    FO(i,0,sz(p)) {
        if (p[i].dx+p[i].dx <= DX && p[i].dy+p[i].dy <= DY) RES--;
        ll qdx = DX-p[i].dx;
        ll qdy = DY-p[i].dy;
        if (qdx >= 0 && qdy >= 0) q.push_back(E(qdx, qdy));
    }
    FO(i,0,sz(p)) y.push_back(p[i].dy);
    FO(i,0,sz(q)) y.push_back(q[i].dy);
    sort(y.begin(),y.end());
    y.resize(unique(y.begin(),y.end())-y.begin());
    bit.resize(sz(y)+5);
    FO(i,0,sz(bit)) bit[i] = 0;

    sort(p.begin(),p.end(),cmp);
    sort(q.begin(),q.end(),cmp);

    int pi = 0, qi = 0;
    while (qi < sz(q)) {
        if (pi < sz(p) && !cmp(q[qi],p[pi])) {
            int yv = lower_bound(y.begin(),y.end(),p[pi].dy)-y.begin()+1;
            ub(yv,1);
            pi++;
        } else {
            int yv = lower_bound(y.begin(),y.end(),q[qi].dy)-y.begin()+1;
            RES += qb(yv);
            qi++;
        }
    }
    RES /= 2;

    return RES;
}

void testdoit() {
    vector<E> v;
    DX = DY = 100;
    FO(i,0,1000) v.push_back(E(rand()%DX,rand()%DY));
    printf("%lld %lld\n", doit(v), doitbf(v));
}

void solve(int x) {
    if (sz(u[x]) == 0) return;
    resetsc(x,-1);
    x = find_centre(x,-1,sc[x]);
    vector<E> cv;
    cv.push_back(E(0,0));
    FO(i,0,sz(u[x])) {
        vector<E> v;
        getl(u[x][i].o,x,v,u[x][i].dx,u[x][i].dy);
        res -= doit(v);
        FO(j,0,sz(v)) cv.push_back(v[j]);
    }
    res += doit(cv);
    delete_node(x);
    FO(i,0,sz(u[x])) solve(u[x][i].o);
}

int main() {
    //testdoit();
    //return 0;

    scanf("%d %lld %lld", &n, &DX, &DY);
    FO(i,0,n) {
        scanf("%lld %lld", &X[i], &Y[i]);
    }
    FO(i,0,n-1) {
        int a,b; scanf("%d %d", &a, &b); a--; b--;
        u[a].push_back(edge(b,abs(X[a]-X[b]),abs(Y[a]-Y[b])));
        u[b].push_back(edge(a,abs(X[a]-X[b]),abs(Y[a]-Y[b])));
    }
    solve(0);
    ll T = (n * 1ll * (n-1)) / 2;
    printf("%lld\n", T - res);
}










In   Java :








import java.io.*;
import java.util.*;

public class Solution {

static class Pair implements Comparable<Pair> {
long fi;
long se;

public Pair(long fi, long se) {
this.fi = fi;
this.se = se;
}

@Override
public int compareTo(Pair o) {
if (fi != o.fi) {
return fi > o.fi ? 1 : -1;
}
if (se == o.se) {
return 0;
}
return se > o.se ? 1 : -1;
}
}

static boolean[] cut;
static int[] size;

static int getSize(int v, int p) {
size[v] = 1;
for (int u: e[v]) {
if (u != p && ! cut[u]) {
size[v] += getSize(u, v);
}
}
return size[v];
}

static Pair[] a;
static List<Integer>[] e;
static Pair[] b;

static int getDist(int v, int p, int i, long hh, long vv) {
hh += Math.abs(a[v].fi - a[p].fi);
vv += Math.abs(a[v].se - a[p].se);
b[i++] = new Pair(hh, vv);
for (int u: e[v]) {
if (u != p && ! cut[u]) {
i = getDist(u, v, i, hh, vv);
}
}
return i;
}

static public int lowerBound(long[] arr, int len, long key) {
if (key <= arr[0]) {
return 0;
}
if (key > arr[len - 1]) {
return 0;
}

int index = Arrays.binarySearch(arr, 0, len, key);
if (index < 0) {
index = - index - 1;
if (index < 0) {
return 0;
}
} 
while (index > 0 && arr[index-1] == key) {
index--;
}
return index;
}

static int upperBound(long[] arr, int len, long key) {
int index = Arrays.binarySearch(arr, 0, len, key);
if (index < 0) {
index = - index - 1;
if (index < 0) {
return 0;
}    
if (index >= len) {
return len;
}    
}
while (index < len && arr[index] == key) {
index++;
}
return index;
}

static int[] fenwick;
static long[] c;
static long h;
static long v;

static long calc(int l, int r) {
int n = r-l;
long ret = 0;
Arrays.sort(b, l, r);
for (int i = l; i < r; i++) {
c[i-l] = b[i].se;
}
Arrays.sort(c, 0, n);
Arrays.fill(fenwick, 0, n, 0);
for (int j = l, i = r; --i >= l; ) {
for (; j < r && b[j].fi+b[i].fi <= h; j++)
for (int x = lowerBound(c, n, b[j].se); x < n; x |= x+1) {
fenwick[x]++;
}
for (int x = upperBound(c, n, v-b[i].se); x > 0; x &= x-1) {
ret += fenwick[x-1];
}
}
return ret;
}


static Object[] divide(int l, int v) {
getSize(v, -1);
int nn = size[v];
int p = -1;
for(;;) {
int ch = -1;
for (int u: e[v]) {
if (u != p && ! cut[u] && 2*size[u] >= nn) {
ch = u;
break;
}
}
if (ch < 0) {
break;
}
p = v;
v = ch;
}
cut[v] = true;
b[l] = new Pair(0, 0);
int i = l+1;
long ret = 0;
for (int u: e[v]) {
if (! cut[u]) {
Object[] r = divide(i, u);
ret += (long)r[1];
getDist(u, v, i, 0, 0);
ret -= calc(i, (int)r[0]);
i = (int)r[0];
}
}
cut[v] = false;
ret += calc(l, i) - 1;
return new Object[]{i, ret};
}


public static void main(String[] args) 
throws IOException {
BufferedReader br = new BufferedReader(
    new InputStreamReader(System.in));
BufferedWriter bw = new BufferedWriter(
    new FileWriter(System.getenv("OUTPUT_PATH")));

StringTokenizer st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
h = Long.parseLong(st.nextToken());
v = Long.parseLong(st.nextToken());

a = new Pair[n];
e = new List[n];
for (int i = 0; i < n; i++) {
st = new StringTokenizer(br.readLine());
long fi = Long.parseLong(st.nextToken());
long se = Long.parseLong(st.nextToken());

a[i] = new Pair(fi, se);
e[i] = new LinkedList<>();
}

for (int i = 0; i < n - 1; i++) {
st = new StringTokenizer(br.readLine());
int u = Integer.parseInt(st.nextToken()) - 1;
int v = Integer.parseInt(st.nextToken()) - 1;

e[u].add(v);
e[v].add(u);
}

cut = new boolean[n];
size = new int[n];
b = new Pair[n];
fenwick = new int[n];
c = new long[n];
long result = (long)(n-1)*n - ((long)divide(0, 0)[1]) >> 1;
bw.write(String.valueOf(result));
bw.newLine();
bw.close();
br.close();
}
}








View More Similar Problems

Insert a Node at the head of a Linked List

Given a pointer to the head of a linked list, insert a new node before the head. The next value in the new node should point to head and the data value should be replaced with a given value. Return a reference to the new head of the list. The head pointer given may be null meaning that the initial list is empty. Function Description: Complete the function insertNodeAtHead in the editor below

View Solution →

Insert a node at a specific position in a linked list

Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e

View Solution →

Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

View Solution →

Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

View Solution →

Reverse a linked list

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio

View Solution →

Compare two linked lists

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis

View Solution →