UVa 108 - Maximum Sum

链接：

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=113&page=show_problem&problem=44

原题：

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 themaximal sub-rectangle. A sub-rectangle is any contiguous sub-array of sizeor 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 anarray of integers. The input begins with a single positive integerNon a line by itself indicating the size of the square two dimensional array. This is followed byintegers separated by white-space (newlines and spaces). Theseintegers 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.).Nmay 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

题目大意：

给一个N*N大小的矩阵，每个格子上有一个数字。 求这个矩阵中的子矩阵上面数字之和最小的数字是多少。

分析与总结：

经典的一道题。 也可以说是“最大连续和"那题（UVa 507 - Jill Rides Again）O(n)算法的升级版本。

这题的关键在于转化， 就是把矩阵上面的那些数转换成”最大连续和“。

看下图（画得实在有点对不起观众 = =）

把那些格子看成数字， 上面有红色斜线阴影的代表其中一个高度为2的子矩阵， 我们可以把这两层看做是一层， 上下层数字之和代表一个新的数字， 就可以用“最大连续和”的O（n）方法求出斜线阴影部分最大的一个2*x的矩阵之和。

要求整个大矩阵的最大子矩阵数字和， 就需要枚举所层数的情况，需要用两层for循环。 加上里面那个O(n)的线性时间扫描，最终复杂度是O(n^3).

代码：

/*
 *  UVa:  108 - Maximum Sum 
 *  Time: 0.012s
 *  Author: D_Double
 *
 */
#include<iostream>
#include<cstdio>
#include<cstring>
#define MAXN 102
using namespace std;

int arr[MAXN][MAXN],sum[MAXN][MAXN];

int mx,sm;
int main()
{
    int T,m,n,x,y;
    int i,j;
    
    while(~scanf("%d",&n)){
         memset(sum,0,sizeof(sum));
         memset(arr,0,sizeof(arr));

        for(i=1; i<=n; ++i){
            for(j=1; j<=n; ++j) {
                scanf("%d",&arr[i][j]);
				// sum[i][j] 表示对角顶点为(1,1)，(i,j)这个矩阵里面所有原数之和
                sum[i][j] = sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+arr[i][j]; 
            }
        }
        int maxSum=-2147483645;
        for(int i=1; i<=n; ++i){
            for(int j=0; j<i; ++j){
                int t, min;
                min=0;
                for(int k=1; k<=n; ++k){ // 线性时间扫描
                    t=sum[i][k]-sum[j][k]-min;
                    if(t>maxSum) maxSum=t;
                    if(sum[i][k]-sum[j][k] < min)
                        min=sum[i][k]-sum[j][k];
                }
            }
        }
        printf("%d\n", maxSum);
    }
    return 0;
}

附：这题的又一次升级版UVa 10827 - Maximum sum on a torus

—— 生命的意义，在于赋予它意义。

原创http://blog.csdn.net/shuangde800，By D_Double (转载请标明)