Can anyone explain the solution of this memoization/dynamic programming prob/puzzle?

Question

This is the problem statement:

This is a two player game. Initially there are n integer numbers in an array and players A and B get chance to take them alternatively. Each player can take one or more numbers from the left or right end of the array but cannot take from both ends at a time. He can take as many consecutive numbers as he wants during his time. The game ends when all numbers are taken from the array by the players. The point of each player is calculated by the summation of the numbers, which he has taken. Each player tries to achieve more points from other. If both players play optimally and player A starts the game then how much more point can player A get than player B?

Input

The input consists of a number of cases. Each case starts with a line specifying the integer n (0 < n ≤100), the number of elements in the array. After that, n numbers are given for the game. Input is terminated by a line where n=0.

Output

For each test case, print a number, which represents the maximum difference that the first player obtained after playing this game optimally.

Sample Input                                Output for Sample Input

4

4 -10 -20 7                                 7

4

1 2 3 4                                     10

5

4 -10 -20 7 19                              12

0

This is the solution of this problem.

#include<stdio.h>
#include<stdlib.h>
#define maxn 103

//typedef long long bg;
typedef long bg;

bg Table[maxn][maxn];
bg Seq[maxn];
bool Flag[maxn][maxn];
bg N;

bg Sum(int l, int r) {
    int i, sum = 0;
    for (i = l; i <= r; i++)
        sum += Seq[i];
    return sum;
}

bg Recur(int L, int R) {
    bg max, i, d, k;
    if (L == R)
        return Seq[L];
    if (Flag[L][R])
        return Table[L][R];
    max = Sum(L, R);
    d = Seq[L];
    for (i = L + 1; i <= R; i++) {
        k = Recur(i, R);
        if ((d - k) > max)
            max = d - k;
        d += Seq[i];
    }
    d = Seq[R];
    for (i = R - 1; i >= L; i--) {
        k = Recur(L, i);
        if ((d - k) > max)
            max = d - k;
        d += Seq[i];
    }
    Flag[L][R] = true;
    Table[L][R] = max;
    return max;
}

void Cal() {
    bg max, i, d, k;
    max = Sum(1, N);
    d = Seq[1];
    for (i = 2; i <= N; i++) {
        k = Recur(i, N);
        if ((d - k) > max)
            max = d - k;
        d += Seq[i];
    }
    d = Seq[N];
    for (i = N - 1; i >= 1; i--) {
        k = Recur(1, i);
        if ((d - k) > max)
            max = d - k;
        d += Seq[i];
    }
    printf("%ld\n", max);
}

void Reset() {
    for (int i = 1; i <= 100; i++) {
        for (int j = 1; j <= 100; j++)
            Flag[i][j] = false;
    }
}
int main() {
    //freopen("in.txt", "r", stdin);
    int i;
    while (scanf("%ld", &N) && N) {
        for (i = 1; i <= N; i++)
            scanf("%ld", &Seq[i]);
        Cal();
        Reset();
    }
    return 0;
}

I did debug it but found it difficult to understand the solution.

Can anyone explain the code or the solution of this problem. Or can anyone post code to solve this problem with dynamic programming not recursion?

Answer 1

The state here is Table[left][right] which represents the best solution if you have a sequence that includes the elements from left to right inclusive. At each step every possible move is tried - take N elements from the left or N elements from the right, where N is between 1 and right - left.

Example:
4 -10 -20 7

Take from the left:

Table[1][4] = max(sum(1, 1) - Table[2][4], sum(1, 2) - Table[3][4], sum(1, 3) - Table[4][4], sum(1, 4)).

Take from the right:

Table[1][4] = max(sum(4, 4) - Table[1][3], sum(3, 4) - Table[1][2], sum(2, 4) - Table[1][1], sum(1, 4)).

sum(L, R) is the sum of array numbers between L and R. I am subtracting because of the next opponent turn.