Uva 507 最大子矩阵和

A - Maximum Sum
Time Limit:3000MS    
Memory Limit:0KB     64bit IO Format:%lld & %llu
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Status

Description

Background

A problem that is simple to solve in one dimension is often much more difficult to solve in more than one dimension. Consider satisfying a boolean expression in conjunctive normal form in which each conjunct
consists of exactly 3 disjuncts. This problem (3-SAT) is NP-complete. The problem 2-SAT is solved quite efficiently, however. In contrast, some problems belong to the same complexity class regardless of the dimensionality of the problem.

The Problem

Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle
with the largest sum is referred to as the maximal sub-rectangle. A sub-rectangle is any contiguous sub-array of size 

 or
greater located within the whole array. As an example, the maximal sub-rectangle of the array:



is in the lower-left-hand corner:



and has the sum of 15.

Input and Output

The input consists of an 

 array
of integers. The input begins with a single positive integer N on a line by itself indicating the size of the square two dimensional array. This is followed by 

 integers
separated by white-space (newlines and spaces). These 

 integers
make up the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on the second row, left-to-right, etc.). N may be as large as 100. The numbers in the array will be in the range [-127, 127].

The output is the sum of the maximal sub-rectangle.

Sample Input

4
0 -2 -7  0 9  2 -6  2
-4  1 -4  1 -1
8  0 -2

Sample Output

15

来源： <http://acm.hust.edu.cn/vjudge/contest/view.action?cid=99978>
 题意：

最大子矩阵和的问题，期间注意sum的初始化和b的变化，当b>0的时候

b+=a[j];b<=0时 b=a[j];

方法：把i行到j行的上下行的数据相加。应用一维的最大字段和方法。

代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <queue>

#include <stack>

#include <string>

#include <set>

#include <map>

#include <cstdlib>

#include <cmath>

#include <algorithm>

#define MAXX 100+10

#define inf 0x3f3f3f3f

using namespace std;

int maxsum(int *a, int n)

{

int b=0, sum=-2147483645;

for (int i=0; i<n; i++)

{

if (b<0)

b = a[i];

else

b += a[i];

if (b>sum)

sum = b;

}

return sum;

}

int main()

{

#ifdef LOCAL

freopen("in.txt", "r", stdin);

//freopen("out.txt", "w", stdout);

#endif // LOCAL

int n;

while (scanf("%d", &n)!=EOF)

{

int a[MAXX][MAXX], b[MAXX];

    for (int i=0; i<n; i++)

for (int j=0; j<n; j++)

scanf("%d", &a[i][j]);

int ans = -2147483645;

    for (int i=0; i<n; i++)

{

memset(b, 0, sizeof(b));

for (int j=i; j<n; j++)

    {

for (int k=0; k<n; k++)

b[k] += a[j][k];

int maxx = maxsum(b, n);

if (maxx > ans)

ans = maxx;

}

}

printf("%d\n", ans);

}

return 0;

}

相关的知识： http://blog.csdn.net/liufeng_king/article/details/8632430