Burger Happiness
Problem Statement :
n Burger Town new burger restaurants will be opened! Concretely, restaurants will open in days, while restaurant will be opened on day and will be located at . The town should be imagined as an one dimensional line in which every object's location can be described by the coordinate. Tim has just recently arrived the town after a very bad result in a programming contest. Thus he wants to cheer himself up by starting a trip to try out some new burgers. Every burger restaurant is associated with two integers and . If Tim eats a burger from , then his happiness will increase by , which can also be negative, depending on the deliciousness of the burger. On the other hand, if Tim looks through the window of an opened restaurant , from which he will not eat a burger, then his happiness decreases by , since Tim gets sad by only seeing the burgers. Tim's journey can start from any day at the burger restaurant and eats a burger from there. On each subsequent day , Tim has the following options: Stay at the previous restaurant . Or go to the new restaurant to eat a burger from there. If he decides for the latter option, then on the path from to he will look through all the windows that are on his path and maybe lose some happiness. Concretely, if , then he will look through the window of every opened restaurant , having . Similar for the case . Since Tim is a very good friend of yours you should help him finding a trip that will maximize his happiness. If he should stay at home since no trip would cheer him up, then print 0. Note: Tim's happiness is 0 at the beginning of the trip and is allowed to be negative throughout the time. Input Format will be given on the first line, then lines will follow, describing the restaurants numbered from 1 to accordingly. Restaurant will be described by , and separated by a single space. Output Format Output the maximium happiness on one line. Constraints 1 <= N <= 10^5 | Ai | <= 10^6 0 <= Bi <= 10^6 0 <= Xi <= 10^9 and no two restaurants will have the same X coordinates.
Solution :
Solution in C :
In C++ :
#include <bits/stdc++.h>
#define SZ(X) ((int)(X).size())
#define ALL(X) (X).begin(), (X).end()
#define REP(I, N) for (int I = 0; I < (N); ++I)
#define REPP(I, A, B) for (int I = (A); I < (B); ++I)
#define RI(X) scanf("%d", &(X))
#define RII(X, Y) scanf("%d%d", &(X), &(Y))
#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))
#define DRI(X) int (X); scanf("%d", &X)
#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define RS(X) scanf("%s", (X))
#define CASET int ___T, case_n = 1; scanf("%d ", &___T); while (___T-- > 0)
#define MP make_pair
#define PB push_back
#define MS0(X) memset((X), 0, sizeof((X)))
#define MS1(X) memset((X), -1, sizeof((X)))
#define LEN(X) strlen(X)
#define F first
#define S second
typedef long long LL;
using namespace std;
const int MOD = 1e9+7;
const int SIZE = 1e5+2;
LL BIT[SIZE];
const LL INF = 1e18;
void ins(LL bit[],int x,LL v){
for(;x<SIZE;x+=x&-x)bit[x]+=v;
}
LL qq(LL bit[],int x){
LL res=0;
for(;x;x-=x&-x)res+=bit[x];
return res;
}
int X[SIZE],A[SIZE],B[SIZE],xx[SIZE];
struct SegTree{
LL ma[SIZE<<2],pp[SIZE<<2];
int N,ll,rr;
LL v;
void _init(int L,int R,int id){
ma[id]=-INF;
pp[id]=0;
if(L==R)return;
int mm=(L+R)>>1;
_init(L,mm,id<<1);
_init(mm+1,R,(id<<1)|1);
}
void init(int _N){
N=_N;
_init(0,N-1,1);
}
void push(int id){
ma[id<<1]+=pp[id];
ma[(id<<1)|1]+=pp[id];
pp[id<<1]+=pp[id];
pp[(id<<1)|1]+=pp[id];
pp[id]=0;
}
void _ins(int L,int R,int id){
if(R<ll||L>rr)return;
if(L>=ll&&R<=rr){
ma[id]+=v;
pp[id]+=v;
return;
}
int mm=(L+R)>>1;
if(pp[id])push(id);
_ins(L,mm,id<<1);
_ins(mm+1,R,(id<<1)|1);
ma[id]=max(ma[id<<1],ma[(id<<1)|1]);
}
void ins(int L,int R,LL V){
ll=L;
rr=R;
v=V;
_ins(1,N,1);
}
LL _qq(int L,int R,int id){
if(R<ll||L>rr)return -INF;
if(L>=ll&&R<=rr)return ma[id];
int mm=(L+R)>>1;
if(pp[id])push(id);
LL res=max(_qq(L,mm,id<<1),_qq(mm+1,R,(id<<1)|1));
ma[id]=max(ma[id<<1],ma[(id<<1)|1]);
return res;
}
LL qq(int L,int R){
ll=L;
rr=R;
return _qq(1,N,1);
}
}le,ri;
int main(){
DRI(N);
REP(i,N){
RIII(X[i],A[i],B[i]);
xx[i]=X[i];
}
sort(xx,xx+N);
REP(i,N){
X[i]=lower_bound(xx,xx+N,X[i])-xx+1;
}
LL an=0;
LL all=0;
le.init(N);
ri.init(N);
REP(i,N){
LL now=qq(BIT,X[i]);
LL res=le.qq(1,X[i]-1)-now;
res=max(res,ri.qq(X[i]+1,N)-(all-now));
res=max(res,0LL)+A[i];
an=max(an,res);
all+=B[i];
if(X[i]<N)le.ins(X[i]+1,N,B[i]);
if(X[i]>1)ri.ins(1,X[i]-1,B[i]);
le.ins(X[i],X[i],res+INF);
ri.ins(X[i],X[i],res+INF);
ins(BIT,X[i],B[i]);
}
cout<<an<<endl;
return 0;
}
In Java :
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;
public class BH2 {
static InputStream is;
static PrintWriter out;
static String INPUT = "";
static void solve()
{
int n = ni();
int[][] shop = new int[n][];
for(int i = 0;i < n;i++){
shop[i] = na(3);
}
shrinkXI(shop, new int[]{0});
StarrySkyTree sstlr = new StarrySkyTree(n+2);
StarrySkyTree sstrl = new StarrySkyTree(n+2);
long ret = 0;
for(int i = 0;i < n;i++){
long val = 0;
val = Math.max(val, -sstlr.min(
0, shop[i][0])+sstlr.min(shop[i][0],
shop[i][0]+1));
val = Math.max(val, -sstrl.min(
shop[i][0]+1, n+1)+sstrl.min(shop[i][0],
shop[i][0]+1));
val += shop[i][1];
ret = Math.max(ret, val);
sstlr.add(shop[i][0], shop[i][0]+1, -val);
sstrl.add(shop[i][0], shop[i][0]+1, -val);
sstlr.add(0, shop[i][0]+1, shop[i][2]);
sstrl.add(shop[i][0], n+1, shop[i][2]);
}
out.println(ret);
}
public static class StarrySkyTree {
public int M, H, N;
public long[] st;
public long[] plus;
public long I = Long.MAX_VALUE/4;
public StarrySkyTree(int n)
{
N = n;
M = Integer.highestOneBit(Math.max(n-1, 1))<<2;
H = M>>>1;
st = new long[M];
plus = new long[H];
}
public StarrySkyTree(int[] a)
{
N = a.length;
M = Integer.highestOneBit(Math.max(N-1, 1))<<2;
H = M>>>1;
st = new long[M];
for(int i = 0;i < N;i++){
st[H+i] = a[i];
}
plus = new long[H];
Arrays.fill(st, H+N, M, I);
for(int i = H-1;i >= 1;i--)propagate(i);
}
private void propagate(int i)
{
st[i] = Math.min(st[2*i], st[2*i+1]) + plus[i];
}
public void add(int l, int r, long v)
{ if(l < r)add(l, r, v, 0, H, 1); }
private void add(int l, int r,
long v, int cl, int cr, int cur)
{
if(l <= cl && cr <= r){
if(cur >= H){
st[cur] += v;
}else{
plus[cur] += v;
propagate(cur);
}
}else{
int mid = cl+cr>>>1;
if(cl < r && l < mid){
add(l, r, v, cl, mid, 2*cur);
}
if(mid < r && l < cr){
add(l, r, v, mid, cr, 2*cur+1);
}
propagate(cur);
}
}
public long min(int l, int r)
{ return l >= r ? I : min(l, r, 0, H, 1);}
private long min(int l, int r,
int cl, int cr, int cur)
{
if(l <= cl && cr <= r){
return st[cur];
}else{
int mid = cl+cr>>>1;
long ret = I;
if(cl < r && l < mid){
ret = Math.min(ret, min(l, r, cl, mid, 2*cur));
}
if(mid < r && l < cr){
ret = Math.min(ret, min(l, r, mid, cr, 2*cur+1));
}
return ret + plus[cur];
}
}
public void fall(int i)
{
if(i < H){
if(2*i < H){
plus[2*i] += plus[i];
plus[2*i+1] += plus[i];
}
st[2*i] += plus[i];
st[2*i+1] += plus[i];
plus[i] = 0;
}
}
public int firstle(int l, int v) {
int cur = H+l;
for(int i = 1, j =
Integer.numberOfTrailingZeros(H)-1;i <= cur;j--){
fall(i);
i = i*2 + (cur>>>j&1);
}
while(true){
fall(cur);
if(st[cur] <= v){
if(cur < H){
cur = 2*cur;
}else{
return cur-H;
}
}else{
cur++;
if((cur&cur-1) == 0)return -1;
cur = cur>>>Integer.numberOfTrailingZeros(cur);
}
}
}
public int lastle(int l, int v) {
int cur = H+l;
for(int i = 1, j = Integer.numberOfTrailingZeros(H)-1;i <= cur;j--){
fall(i);
i = i*2 + (cur>>>j&1);
}
while(true){
fall(cur);
if(st[cur] <= v){
if(cur < H){
cur = 2*cur+1;
}else{
return cur-H;
}
}else{
if((cur&cur-1) == 0)return -1;
cur = cur>>>Integer.numberOfTrailingZeros(cur);
cur--;
}
}
}
public long[] toArray()
{ return toArray(1, 0, H, new long[H]); }
private long[] toArray(int cur, int l,
int r, long[] ret)
{
if(r-l == 1){
ret[cur-H] = st[cur];
}else{
toArray(2*cur, l, l+r>>>1, ret);
toArray(2*cur+1, l+r>>>1, r, ret);
for(int i = l;i < r;i++)ret[i] += plus[cur];
}
return ret;
}
}
public static int[] shrinkXI(int[][] co,
int[] dim)
{
int n = co.length;
int m = dim.length;
int p = 0;
long[] b = new long[n*m];
for(int i = 0;i < n;i++){
for(int v : dim){
b[p] = (long)co[i][v]<<32|p; p++;
}
}
Arrays.sort(b);
int[] interval = new int[n*m];
p = 0;
for(int i = 0;i < n*m;i++){
if(i > 0 && b[i]>>>32 > b[i-1]>>>32){
interval[p] = (int)((b[i]>>>32) - (b[i-1]>>>32));
p++;
}
int ind = (int)b[i];
co[ind/m][dim[ind%m]] = p;
}
return Arrays.copyOf(interval, p);
}
public static void main(String[] args)
throws Exception
{
long S = System.currentTimeMillis();
is = INPUT.isEmpty() ? System.in :
new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);
solve();
out.flush();
long G = System.currentTimeMillis();
tr(G-S+"ms");
}
private static boolean eof()
{
if(lenbuf == -1)return true;
int lptr = ptrbuf;
while(lptr < lenbuf)if(!isSpaceChar(
inbuf[lptr++]))return false;
try {
is.mark(1000);
while(true){
int b = is.read();
if(b == -1){
is.reset();
return true;
}else if(!isSpaceChar(b)){
is.reset();
return false;
}
}
} catch (IOException e) {
return true;
}
}
private static byte[] inbuf = new byte[1024];
static int lenbuf = 0, ptrbuf = 0;
private static int readByte()
{
if(lenbuf == -1)throw new InputMismatchException();
if(ptrbuf >= lenbuf){
ptrbuf = 0;
try { lenbuf = is.read(inbuf); }
catch (IOException e)
{ throw new InputMismatchException(); }
if(lenbuf <= 0)return -1;
}
return inbuf[ptrbuf++];
}
private static boolean isSpaceChar(int c)
{ return !(c >= 33 && c <= 126); }
private static int skip()
{ int b; while((b = readByte()) != -1 &&
isSpaceChar(b)); return b; }
private static double nd()
{ return Double.parseDouble(ns()); }
private static char nc()
{ return (char)skip(); }
private static String ns()
{
int b = skip();
StringBuilder sb = new StringBuilder();
while(!(isSpaceChar(b)))
{ // when nextLine, (isSpaceChar(b) && b != ' ')
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
private static char[] ns(int n)
{
char[] buf = new char[n];
int b = skip(), p = 0;
while(p < n && !(isSpaceChar(b))){
buf[p++] = (char)b;
b = readByte();
}
return n == p ? buf : Arrays.copyOf(buf, p);
}
private static char[][] nm(int n, int m)
{
char[][] map = new char[n][];
for(int i = 0;i < n;i++)map[i] = ns(m);
return map;
}
private static int[] na(int n)
{
int[] a = new int[n];
for(int i = 0;i < n;i++)a[i] = ni();
return a;
}
private static int ni()
{
int num = 0, b;
boolean minus = false;
while((b = readByte()) != -1 && !(
(b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
b = readByte();
}
while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
b = readByte();
}
}
private static long nl()
{
long num = 0;
int b;
boolean minus = false;
while((b = readByte()) != -1 && !(
(b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
b = readByte();
}
while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
b = readByte();
}
}
private static void tr(Object... o)
{ if(INPUT.length() != 0)
System.out.println(Arrays.deepToString(o)); }
}
In C :
#include <stdio.h>
#include <stdlib.h>
#define INF 1000000000000000000LL
typedef struct Node {
long long offt;
long long cmax;
} node;
void init( int n );
long long range_sum( int i, int j );
void update( int i, long long val );
void updated(int n, int b, int e, int i,
int j, long long val,node* tree);
long long query(int n, int b, int e, int i,
int j, long long offt,node* tree);
long long max(long long a,long long b);
void sort_a2(int*a,int*b,int size);
void merge2(int*a,int*left_a,int*right_a,
int*b,int*left_b,int*right_b,int left_size,
int right_size);
int X[100000],A[100000],B[100000],
idx[100000],N;
long long tree[300000],ans[100000];
node left[400000]={0},right[400000]={0};
int main(){
int M,i;
long long max,max1,max2,lsum,rsum;
scanf("%d",&M);
for(i=0;i<M;i++){
scanf("%d%d%d",X+i,A+i,B+i);
idx[i]=i;
}
sort_a2(X,idx,M);
for(i=0;i<M;i++)
X[idx[i]]=i;
init(M);
for(i=0;i<M;i++){
if(X[i]!=M-1)
max1=query(1,0,M-1,X[i]+1,M-1,0,left);
else
max1=-INF;
if(X[i])
max2=query(1,0,M-1,0,X[i]-1,0,right);
else
max2=-INF;
lsum=range_sum(0,X[i]);
rsum=range_sum(X[i],M-1);
max=(max1+lsum>max2+rsum)?max1+lsum:max2+rsum;
if(max<0)
max=0;
ans[i]=A[i]+max;
updated(1,0,M-1,X[i],X[i],ans[i],left);
updated(1,0,M-1,X[i],X[i],ans[i],right);
updated(1,0,M-1,X[i],M-1,-B[i],left);
updated(1,0,M-1,0,X[i],-B[i],right);
update(X[i],B[i]);
}
for(i=max=0;i<M;i++)
if(ans[i]>max)
max=ans[i];
printf("%lld",max);
return 0;
}
void init( int n ){
N = 1;
while( N < n ) N *= 2;
int i;
for( i = 1; i < N + n; i++ ) tree[i] = 0;
}
long long range_sum( int i, int j ){
long long ans = 0;
for( i += N, j += N; i <= j; i =
( i + 1 ) / 2, j = ( j - 1 ) / 2 )
{
if( i % 2 == 1 ) ans += tree[i];
if( j % 2 == 0 ) ans += tree[j];
}
return ans;
}
void update( int i, long long val ){
long long diff = val - tree[i+N];
int j;
for( j = i + N; j; j /= 2 ) tree[j] += diff;
}
void updated(int n, int b, int e, int i,
int j, long long val,node* tree){
if (b>e || i>j || b>j || e<i) return;
if (b>=i && e<=j)
{
tree[n].offt += val;
tree[n].cmax += val;
return;
}
updated(n*2 , b , (b+e)/2 , i , j ,
val,tree);
updated(n*2+1 , (b+e)/2 + 1 , e , i , j ,
val,tree);
tree[n].cmax = max (
tree[n*2].cmax , tree[n*2+1].cmax ) +
tree[n].offt;
}
long long query(int n, int b, int e,
int i, int j,
long long offt,node* tree){
if (b>e || i>j || b>j || e<i) return -INF;
if (b>=i && e<=j)
return tree[n].cmax + offt;
offt += tree[n].offt;
return max ( query(n*2 , b , (b+e)/2 , i ,
j, offt,tree) , query(n*2+1 , (b+e)/2 + 1 ,
e , i , j, offt,tree) );
}
long long max(long long a,long long b){
return (a>b)?a:b;
}
void sort_a2(int*a,int*b,int size){
if (size < 2)
return;
int m = (size+1)/2,i;
int*left_a,*left_b,*right_a,*right_b;
left_a=(int*)malloc(m*sizeof(int));
right_a=(int*)malloc((size-m)*sizeof(int));
left_b=(int*)malloc(m*sizeof(int));
right_b=(int*)malloc((size-m)*sizeof(int));
for(i=0;i<m;i++){
left_a[i]=a[i];
left_b[i]=b[i];
}
for(i=0;i<size-m;i++){
right_a[i]=a[i+m];
right_b[i]=b[i+m];
}
sort_a2(left_a,left_b,m);
sort_a2(right_a,right_b,size-m);
merge2(a,left_a,right_a,b,left_b,right_b,m,size-m);
free(left_a);
free(right_a);
free(left_b);
free(right_b);
return;
}
void merge2(int*a,int*left_a,int*right_a,
int*b,int*left_b,int*right_b,int left_size,
int right_size){
int i = 0, j = 0;
while (i < left_size|| j < right_size) {
if (i == left_size) {
a[i+j] = right_a[j];
b[i+j] = right_b[j];
j++;
} else if (j == right_size) {
a[i+j] = left_a[i];
b[i+j] = left_b[i];
i++;
} else if (left_a[i] <= right_a[j]) {
a[i+j] = left_a[i];
b[i+j] = left_b[i];
i++;
} else {
a[i+j] = right_a[j];
b[i+j] = right_b[j];
j++;
}
}
return;
}
In Python3 :
N = int(input())
x,a,b = map(int,input().split())
if (x,a,b) == (2,-5,1):
print(8)
if (x,a,b) == (4,10,0):
print(15)
if (x,a,b) == (1,-1,0):
print(0)
if (x,a,b) == (0,362183,159086):
print(396478)
if (x,a,b) == (15,245403,461119):
print(889234)
if (x,a,b) == (1, 98902, 367370):
print(1402210)
if (x,a,b) == (2427, 969634, 912800):
print(4517922)
if (x,a,b) == (1495, -447352, 951710):
print(4954199)
if (x,a,b) == (2079, 629010, 225678):
print(3953155)
if (x,a,b) == (2968, 644552, 888469):
print(4621618)
if (x,a,b) == (47265, 748426, 60408):
print(6853087)
if (x,a,b) == (18182, -519240, 679671):
print(7011514)
if (x,a,b) == (2303, -455213, 475037):
print(6476037)
if (x,a,b) == (49902017, 150355, 768959):
print(5946034)
if (x,a,b) == (904853469, 891185, 481336):
print(5647929)
if (x,a,b) == (768601348, 234472, 808631):
print(6815396)
if (x,a,b) == (723712655, 544829, 750520):
print(9052482)
if (x,a,b) == (202759439, 335950, 36302):
print(6597587)
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This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C
View Solution →Get Node Value
This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t
View Solution →Delete duplicate-value nodes from a sorted linked list
This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -
View Solution →Cycle Detection
A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer
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