POJ #1050 - To the Max

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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: http://blog.csdn.net/hitwhylz/article/details/11848439. 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:    http://blog.csdn.net/hitwhylz/article/details/11848439#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;}

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