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C++ program to find out the maximum possible tally from given integers
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
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