POJ #1050 - To the Max

Looks quite intuitive at the very first sight... I thought:
rect[x, y, w+1, h+1] = rect[x, y, w, h] + num[x + w + 1, y..y+h+1] + num[x..x+w+1, y+h+1] - num[x+w+1, y+h+1]

But that results in O(n^4) at least.

Reference: /article/1385567.html. Lesson learnt: usually 2D issue can be converted to 1D problem to solve. Cool thought. And what is more important: decoding problem thoroughly and identify familiar pattern in disguise from it.

Here is my AC code:

//    1050
//    Ref:    /article/1385567.html

#include <stdio.h>

#define MAX_N 100
#define Max(a, b) (a) > (b) ? (a) : (b)

int max_sum_1D(int rowSum[MAX_N + 1], int n)
{
//    dp[i] = max{a[i], dp[i-1] + a[i]}
int ret = -2147483648;
int dp[MAX_N + 1] = { 0 };
for (int k = 1; k <= n; k ++)
{
dp[k] = Max(rowSum[k], dp[k-1] + rowSum[k]);
ret = Max(ret, dp[k]);
}
return ret;
}

int calc(int in[MAX_N + 1][MAX_N + 1], unsigned n)
{
int maxSum = -2147483648;

for (int i = 1; i <= n; i++)
{
int rowSum[MAX_N + 1] = { 0 };    // for compact

for (int j = i; j <= n; j++)    // Row[i]..Row[j]
{
//    Compact to 1D array, from row[i] to row[j] for each column
for (int k = 1; k <= n; k++)
{
rowSum[k] += in[j][k];
}

int tmp_sum = max_sum_1D(rowSum, n);
maxSum = Max(tmp_sum, maxSum);
}
}

return maxSum;
}

int main()
{
int n; scanf("%d", &n);
int in[MAX_N + 1][MAX_N + 1];

//    Get input
for (int i = 1; i <= n; i ++)
for (int j = 1; j <= n; j++)
{
scanf("%d", &(in[i][j]));
}

//
int ret = calc(in, n);
printf("%d\n", ret);

return 0;
}

View Code