Find Maximum Possible Tally from Given Integers in C++

Suppose, we are given two integers n and m and there are k tuples of integers that contain four integer numbers {ai, bi, ci, di}. Four arrays a, b, c, d are given, and a[i] signifies the i-th tuple's a value. Now, let us consider a sequence dp that has n positive integers and 1 <= dp[1] < dp[2] < ..... < dp[n] <= m. We define a metric 'tally'. The metric tally is the sum of d[i] over all indices i such that dp[b[i]] − dp[a[i]] = c[i]. If there is no such i, the tally is 0. We have to find out the maximum possible tally of dp.

So, if the input is like n = 4, m = 5, k = 4, a = {2, 2, 3, 5}, b = {4, 3, 4, 6}, c = {4, 3, 3, 4}, d = {110, 20, 20, 40}, then the output will be 130.

Steps

To solve this, we will follow these steps −

Define arrays A, B, C, D, and dp of sizes: 100, 100, 100, 100, 10 respectively.
Define a function depthSearch(), this will take c, l,
if c is same as n, then:
   total := 0
   for initialize i := 0, when i < k, update (increase i by 1), do:
      if dp[B[i]] - dp[A[i]] is same as C[i], then:
         total := total + D[i]
   res := maximum of res and total
   return
for initialize j := l, when j <= m, update (increase j by 1), do:
   dp[c] := j
   depthSearch(c + 1, j)
for initialize i := 0, when i < k, update (increase i by 1), do:
   A[i] := a[i], B[i] := b[i], C[i] := c[i], D[i] := d[i]
   decrease A[i] by 1
   decrease B[i] by 1
depthSearch(0, 1)
return res

Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;

int n, m, k, res = 0;
int A[100], B[100], C[100], D[100], dp[10]; 

void depthSearch(int c, int l){
   if(c == n){
      int total = 0;
      for(int i = 0; i < k; i++) {
         if(dp[B[i]] - dp[A[i]] == C[i]) total += D[i];
      }
      res = max(res, total);
      return;
   }
   for(int j = l; j <= m; j++){
      dp[c] = j;
      depthSearch(c + 1, j);
   }
}
int solve(int a[], int b[], int c[], int d[]){
   for(int i = 0; i < k; i++){
       A[i] = a[i], B[i] = b[i], C[i] = c[i], D[i] = d[i]; A[i]--, B[i]--;
   }
   depthSearch(0, 1);
   return res;
}
int main() {
   n = 4, m = 5, k = 4;
   int a[] = {2, 2, 3, 5}, b[] = {4, 3, 4, 6}, c[] = {4, 3, 3, 4}, d[] = {110, 20, 20, 40};
   cout<< solve(a, b, c, d);
   return 0;
}

Input

4, 5, 4, {2, 2, 3, 5}, {4, 3, 4, 6}, {4, 3, 3, 4}, {110, 20, 20, 40}

Output

130