Maximum Sum 最大子矩阵和+dp+（最大连续子序列的变形）

Maximum Sum

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

解决方案：可枚举行区间，求出每列的和，再用d[i]=max(d[i-1]+sum[i],sum[i])动规公式，即可求出最大子矩阵和。

代码：
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int Matrix[103][103];
int Sum[103];
int dp[103];
int n;
int main(){
while(~scanf("%d",&n)){

for(int i=0;i<n;i++)
for(int j=0;j<n;j++){
scanf("%d",&Matrix[i][j]);
}
int Max=0;
for(int i=0;i<n;i++){

for(int j=0;j<n;j++){
memset(Sum,0,sizeof(Sum));
memset(dp,0,sizeof(dp));
if(i<=j){
for(int k=0;k<n;k++)
for(int s=i;s<=j;s++){
Sum[k]+=Matrix[s][k];

}
dp[0]=Sum[0];
for(int h=1;h<n;h++){
if(dp[h-1]+Sum[h]<Sum[h]){
dp[h]=Sum[h];
}
else dp[h]=dp[h-1]+Sum[h];
if(dp[h]>Max)
Max=dp[h];
}
}
}
}
printf("%d\n",Max);
}

return 0;}