To The Max
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8528 Accepted Submission(s): 4142
Problem Description
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 1 or greater located within the whole array. 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.
As an example, the maximal sub-rectangle of the array:
0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2
is in the lower left corner:
9 2
-4 1
-1 8
and has a sum of 15.
Input
The input consists of an N x N 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 N 2 integers separated by whitespace (spaces and newlines).
These are the N 2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in 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].
Output
Output 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
Source
Greater New York 2001
题目大意：给你一个N，接下来是N*N的矩阵。数有正有负，求最大的子矩阵
和。
思路：1003题是一维的求连续子序列最大和，dp[i] = max(dp[i-1]+a[i],a[i])
这道题是求二维的子矩阵最大和。考虑将二维转化为一维的。二维数组的每一
行都可以看做一个一维数组。map[i][j]数组上存的是第i行前j列上的和
第k行上，dp[k][i] = max(dp[k][i-1]+map[k][i],map[k][i])，但是这样只能知道
第k行前i个数的最大和是多少，而不知道是第k行上从第几列到第几列上得来的最
大和。状态转移方程换为，以第k行为终点，从第i列到第j列上的最大和
dp[i][j] = max(map[k][j]-map[k][i-1]+dp[i][j],map[k][j]-map[k][i-1]);
然后用一个Max，求出最大的dp[i][j]。因为不涉及求具体的子矩阵的情况，只需
要得到最优解，那么状态转移方程也可以写成
ans = max(map[k][j]-map[k][i-1]+ans,map[k][j]-map[k][i-1]);
ans求的就是是前k行，以第k行为终点，第i到第j列的最大和
第一种代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int map[110][110],dp[110][110];
int main()
{
int N,a;
while(~scanf("%d",&N) && N)
{
memset(map,0,sizeof(map));
memset(dp,0,sizeof(dp));
for(int i = 1; i <= N; i++)
{
for(int j = 1; j <= N; j++)
{
scanf("%d",&a);
map[i][j] = map[i][j-1] + a;
//map[i][j]表示第i行前j列的和
}
}
int Max = -0xffffff0;
for(int j = 1; j <= N; j++)
{
for(int i = 1; i <= j; i++)
{
dp[i][j] = 0;
for(int k = 1; k <= N; k++)
{
//ans求的是前k行，第i到第j列的最大和
dp[i][j]= max(dp[i][j]+map[k][j]-map[k][i-1],map[k][j]-map[k][i-1]);
if(dp[i][j] > Max)
Max = dp[i][j];
}
}
}
printf("%d\n",Max);
}
return 0;
}
第二种代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int map[110][110];
int main()
{
int N,a;
while(~scanf("%d",&N) && N)
{
memset(map,0,sizeof(map));
for(int i = 1; i <= N; i++)
{
for(int j = 1; j <= N; j++)
{
scanf("%d",&a);
map[i][j] = map[i][j-1] + a;
}
}
int Max = -0xffffff0;
for(int j = 1; j <= N; j++)
{
for(int i = 1; i <= j; i++)
{
int ans = 0;
for(int k = 1; k <= N; k++)
{
ans = max(ans+map[k][j]-map[k][i-1],map[k][j]-map[k][i-1]);
if(ans > Max)
Max = ans;
}
}
}
printf("%d\n",Max);
}
return 0;
}