Robot


Problem Statement :


You have two arrays of integers, V = {V1,V2,...Vn} and P = {P1,P2,...,Pn}, where both have N number of elements. Consider the following function:

score = 0

int Go(step, energy) {
    if (step == N) {
        score += V[step];
        return (score);
    }
    else {
        int way = random(1, 2);
        if (way == 1) {
            score += V[step];
        }
        else {
            energy = P[step];
        }
        if (energy > 0) {
            Go(step + 1, energy - 1);
        }
        else {
            KillTheWorld();
        }
    }
}
What is the maximum possible value of score that we can get in the end, if we call Go(1,0)?.
Note that the function should never invoke KillTheWorld function. And random(1,2) generates a random integer from set [1, 2].
It is guaranteed there will be a solution that wont kill the world.

Input Format

The first line contains an integer N. Each of the following N lines contains a pair of integers. The i-th line contains a pair of numbers, Vi,Pi, separated by space.

Constraints

1 <= N <= 5*10^5
0 <= Vi <= 10^9
0 <= Pi <- 10^5

Output Format

Derive the maximum score given by return (score);.



Solution :



title-img


                            Solution in C :

In C++ :






#include <vector> 
#include <list> 
#include <map> 
#include <set> 
#include <queue>
#include <stack> 
#include <bitset> 
#include <algorithm> 
#include <numeric> 
#include <utility> 
#include <sstream> 
#include <iostream> 
#include <iomanip> 
#include <cstdio> 
#include <cmath> 
#include <cstdlib> 
#include <ctime> 
#include <cstring> 

using namespace std; 

typedef long long ll; 
typedef pair<int, long long> pii;

#define INF 1000000000
#define pb push_back 
#define itr iterator 
#define sz size() 
#define mp make_pair

multiset<long long> values;
vector< long long> adj[6000000];
int v, p;
int n;

int main() {
	scanf("%d", &n);

	adj[1].push_back(0);
	values.insert(0);
	long long tot = 0;

	for (int i = 0; i < n; i++) {
		scanf("%d %d", &v, &p);
		tot += v;

		for (int k = 0; k < adj[i].size(); k++) {
			values.erase(values.find(adj[i][k]));
		}

		long long cur = -v + *values.rbegin();
		//printf("cur = %lld %d %lld\n", tot, -v, *values.rbegin());
		values.insert(cur);
		int will_rem = min(n, i + p + 1);
		adj[will_rem].push_back(cur);

		if (i == n-1) {
			printf("%lld\n", tot + cur + v);
			return 0;
		}
	}
}









In Java :





import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.PrintWriter;

public class Solution {
  private static Reader in;
  private static PrintWriter out;

  public static void main(String[] args) throws IOException {
    in = new Reader();
    out = new PrintWriter(System.out, true);
    int N = in.nextInt();
    long[] V = new long[N + 1];
    int[] P = new int[N + 1];
    for (int i = 1; i <= N; i++) {
      V[i] = in.nextInt();
      P[i] = in.nextInt();
    }
    long[] dp = new long[N + 1];
    long[] psum = new long[N + 1];
    for (int i = 1; i <= N; i++)
      psum[i] = psum[i - 1] + V[i];
    dp[N] = V[N];
    SegmentTree root = new SegmentTree(1, N);
    root.update(N, dp[N] + psum[N - 1]);
    for (int i = N - 1; i >= 1; i--) {
      int zero = Math.min(N, i + P[i]);
      dp[i] = zero > i ? (root.query(i + 1, zero) - psum[i]) : -(1l << 60);
      root.update(i, dp[i] + psum[i - 1]);
    }
    out.println (dp[1]);
    out.close();
    System.exit(0);
  }
  
  static class SegmentTree {
    public SegmentTree left, right;
    public int start, end;
    public long max;
    
    public SegmentTree(int start, int end) {
      this.start = start;
      this.end = end;
      if (start != end) {
        int mid = (start + end) >> 1;
        left = new SegmentTree(start, mid);
        right = new SegmentTree(mid + 1, end);
      }
    }
    
    public void update(int pos, long v) {
      if (start == pos && end == pos) {
        this.max = v;
        return;
      }
      int mid = (start + end) >> 1;
      if (mid >= pos) left.update(pos, v);
      else right.update(pos, v);
      max =  Math.max(max, v);
    }
    
    public long query (int s, int e) {
      if (start == s && end == e) {
        return max;
      }
      int mid = (start + end) >> 1;
      if (mid >= e) return left.query(s, e);
      else if (mid < s) return right.query(s, e);
      else return Math.max(left.query(s, mid), right.query(mid + 1, e));
    }
  }
  
  static class Reader {
    final private int BUFFER_SIZE = 1 << 16;
    private DataInputStream din;
    private byte[] buffer;
    private int bufferPointer, bytesRead;

    public Reader() {
      din = new DataInputStream(System.in);
      buffer = new byte[BUFFER_SIZE];
      bufferPointer = bytesRead = 0;
    }

    public Reader(String file_name) throws IOException {
      din = new DataInputStream(new FileInputStream(file_name));
      buffer = new byte[BUFFER_SIZE];
      bufferPointer = bytesRead = 0;
    }

    public String readLine() throws IOException {
      byte[] buf = new byte[1024];
      int cnt = 0, c;
      while ((c = read()) != -1) {
        if (c == '\n')
          break;
        buf[cnt++] = (byte) c;
      }
      return new String(buf, 0, cnt);
    }

    public int nextInt() throws IOException {
      int ret = 0;
      byte c = read();
      while (c <= ' ')
        c = read();
      boolean neg = (c == '-');
      if (neg)
        c = read();
      do {
        ret = ret * 10 + c - '0';
      } while ((c = read()) >= '0' && c <= '9');
      if (neg)
        return -ret;
      return ret;
    }

    public long nextLong() throws IOException {
      long ret = 0;
      byte c = read();
      while (c <= ' ')
        c = read();
      boolean neg = (c == '-');
      if (neg)
        c = read();
      do {
        ret = ret * 10 + c - '0';
      } while ((c = read()) >= '0' && c <= '9');
      if (neg)
        return -ret;
      return ret;
    }

    public double nextDouble() throws IOException {
      double ret = 0, div = 1;
      byte c = read();
      while (c <= ' ')
        c = read();
      boolean neg = (c == '-');
      if (neg)
        c = read();
      do {
        ret = ret * 10 + c - '0';
      } while ((c = read()) >= '0' && c <= '9');
      if (c == '.')
        while ((c = read()) >= '0' && c <= '9')
          ret += (c - '0') / (div *= 10);
      if (neg)
        return -ret;
      return ret;
    }

    private void fillBuffer() throws IOException {
      bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
      if (bytesRead == -1)
        buffer[0] = -1;
    }

    private byte read() throws IOException {
      if (bufferPointer == bytesRead)
        fillBuffer();
      return buffer[bufferPointer++];
    }

    public void close() throws IOException {
      if (din == null)
        return;
      din.close();
    }
  }

}









In C :






/*
4
4 2
0 2
4 0
3 4

7
*/

#include <stdio.h>
#include <stdlib.h>
typedef struct _node{
  int w;
} node;
void heap_insert(node *heap,node *elem,int *size,int idx);
void heap_read(node *heap,int *size,int idx);
long long get_val(int w,int idx,int *flag);
int *V,*P;
long long *sum,*dp;

int main(){
  int N,i,heap_size=0,flag;
  long long ans,max,t;
  node *heap,elem;
  scanf("%d",&N);
  V=(int*)malloc(N*sizeof(int));
  P=(int*)malloc(N*sizeof(int));
  sum=(long long*)malloc(N*sizeof(long long));
  dp=(long long*)malloc(N*sizeof(long long));
  heap=(node*)malloc(2*N*sizeof(heap));
  for(i=0;i<N;i++)
    scanf("%d%d",V+i,P+i);
  for(i=0;i<N;i++)
    if(i==0)
      sum[i]=V[i];
    else
      sum[i]=sum[i-1]+V[i];
  for(i=0;i<N-1;i++){
    max=0;
    if(heap_size)
      do{
        t=get_val(heap[1].w,i,&flag);
        if(flag>0 && t>max)
          max=t;
        else if(flag==0 && P[i] && t-V[i]>max){
          max=t-V[i];
          heap_read(heap,&heap_size,i);
        }
        else if(flag<=0)
          heap_read(heap,&heap_size,i);
      }while(flag<=0 && heap_size);
    dp[i]=max;
    elem.w=i;
    heap_insert(heap,&elem,&heap_size,i);
  }
  max=0;
  if(heap_size)
    do{
      t=get_val(heap[1].w,i,&flag);
      if(flag>=0 && t>max)
        max=t;
      else if(flag<0)
        heap_read(heap,&heap_size,i);
    }while(flag<0 && heap_size);
  printf("%lld",max);
  return 0;
}
void heap_insert(node *heap,node *elem,int *size,int idx){
  (*size)++;
  int index=(*size);
  while(index>1 && get_val(elem->w,idx,0)>get_val(heap[index/2].w,idx,0)){
    heap[index]=heap[index/2];
    index/=2;
  }
  heap[index]=(*elem);
  return;
}
void heap_read(node *heap,int *size,int idx){
  int index=1;
  while(index*2<*size && get_val(heap[*size].w,idx,0)<get_val(heap[index*2].w,idx,0) || index*2+1<*size && get_val(heap[*size].w,idx,0)<get_val(heap[index*2+1].w,idx,0)){
    index=(get_val(heap[index*2].w,idx,0)<get_val(heap[index*2+1].w,idx,0))?index*2+1:index*2;
    heap[index/2]=heap[index];
  }
  heap[index]=heap[*size];
  (*size)--;
  return;
}
long long get_val(int w,int idx,int *flag){
  if(flag){
    if(idx-w>P[w])
      *flag=-1;
    else if(idx-w==P[w])
      *flag=0;
    else
      *flag=1;
  }
  long long ans;
  if(!w)
    ans=0;
  else
    ans=dp[w-1];
  return ans+sum[idx]-sum[w];
}









In Python3 :





import sys
from operator import itemgetter

N = int(input())

V, P = [None] * N, [None] * N 
for i in range(N):
   v_item, p_item = input().split()
   V[i] = int(v_item)
   P[i] = int(p_item)
   
games = []
for i in range(N):
   maxVal = -1
   games.sort(key=itemgetter(1))
   for j in range(len(games) - 1, -1, -1):
      game = games[j]
      if game[1] == 0:
         del games[0:j+1]
         break

      if maxVal < game[0]:
         maxVal = game[0]
      else:
         del games[j]

         
      game[0] += V[i]
      game[1] += -1
   if maxVal == -1:
      maxVal = 0
   games.append([maxVal, P[i]])
print(max(games, key=itemgetter(0))[0])
                        








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