Maximum points collected by two persons allowed to meet once in C++


In this tutorial, we will be discussing a program to find maximum points collected by two persons allowed to meet once

For this we will be provided with a matrix with cells containing points. Our task is to find the path when two people starting from two corners meet such that they are having maximum points collected.

Example

 Live Demo

#include<bits/stdc++.h>
#define M 3
#define N 3
using namespace std;
int findMaxPoints(int A[][M]) {
   //storing points
   int P1S[M+1][N+1], P1E[M+1][N+1];
   memset(P1S, 0, sizeof(P1S));
   memset(P1E, 0, sizeof(P1E));
   int P2S[M+1][N+1], P2E[M+1][N+1];
   memset(P2S, 0, sizeof(P2S));
   memset(P2E, 0, sizeof(P2E));
   for (int i=1; i<=N; i++)
      for (int j=1; j<=M; j++)
         P1S[i][j] = max(P1S[i-1][j], P1S[i][j-1]) + A[i-1][j-1];
   for (int i=N; i>=1; i--)
      for (int j=M; j>=1; j--)
         P1E[i][j] = max(P1E[i+1][j], P1E[i][j+1]) + A[i-1][j-1];
   for (int i=N; i>=1; i--)
      for(int j=1; j<=M; j++)
         P2S[i][j] = max(P2S[i+1][j], P2S[i][j-1]) + A[i-1][j-1];
   for (int i=1; i<=N; i++)
      for (int j=M; j>=1; j--)
         P2E[i][j] = max(P2E[i-1][j], P2E[i][j+1]) + A[i-1][j-1];
   int ans = 0;
   for (int i=2; i<N; i++) {
      for (int j=2; j<M; j++) {
         int op1 = P1S[i][j-1] + P1E[i][j+1] + P2S[i+1][j] + P2E[i-1][j];
         int op2 = P1S[i-1][j] + P1E[i+1][j] + P2S[i][j-1] + P2E[i][j+1];
         ans = max(ans, max(op1, op2));
      }
   }
   return ans;
}
int main() {
   int A[][M] = {
      {100, 100, 100},
      {100, 1, 100},
      {100, 100, 100}
   };
   cout << "Max Points : " << findMaxPoints(A);
   return 0;
}

Output

Max Points : 800

Updated on: 09-Sep-2020

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