poj1050 dp动态规划
Description
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*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.
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 * 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
题目翻译过来，就是在一个大矩阵里面，找一个（矩阵和最大）的子矩阵。
如题意：
0 -2 -7 0
9 2 -6 2 9 2
-4 1 -4 1 -4 1
-1 8 0 -2 -1 8 9+2+（-4）+1+（-1）+8 = = 15 所以输出15
做这题之前，先来了解一下 一维数组子串，找连续数组的最大和 例如 2 3 -7 9 2 -6 9 最大和为（9,2，-6,9） -> 14
如何用算法实现，当然用dp 顺推， 假设数组a 只要a的前项与后项的和大于0，保留，继续比较，在这个过程中要存储max的值
正如 上面例子
2 3 -7 9 2 -6 9
遍历一遍 (前项加本项等于本项的值，但形成负数的时候，要归零)
形成 2 5（2+3） 0 （5+-7小于0，归零） 9 11（9+2） 5 （11-6） 14（5+9）
同样本题是这个思想的延伸，扩展到了二维
第一步：
0 -2 -7 0 -> max=0
9 2 -6 2 -> max=11
-4 1 -4 1 -> max=1 四个合起来 ，得max=11
-1 8 0 -2 -> max=8
第二步（二维压缩成一维 ）：
第一行与第二行相加保留在第二行，然后继续压缩，最后压缩成了一维。中间过程要与max比较，取较大值.........
AC代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#define Max(a,b) ((a)>(b)?:(a):(b))
#define Min(a,b) ((a)<(b)?:(a):(b))
#define Mem0(x) memset(x,0,sizeof(x))
#define Mem1(x) memset(x,-1,sizeof(x))
#define MemX(x) memset(x,0x3f,sizeof(x))
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
int n,a[1010][1010];
/*
int dp(int b[])
{
int i,ans=b[0];
for (i=0;i<n;i++)
printf("%d\t",b[i]);
printf("******");
for (i=0;i<n;i++){
if (b[i]>0)
b[i+1]+=b[i];
if (ans<b[i])
ans=b[i];
printf("%d\t",b[i]);
}
printf("%d\t",b[i]);
return ans;
}*/
int main()
{
int n;
while (cin>>n){
int i,j,k,ans=-inf;
Mem0(a);
for (i=0;i<n;i++){
int tmp=0;
for (j=0;j<n;j++){
cin>>a[i][j];
if (tmp>0)
tmp+=a[i][j];
else tmp=a[i][j];
if (tmp>ans)
ans=tmp;
}
}
for (i=0;i<n-1;i++){
for (j=i+1;j<n;j++){
int tmp=0;
for (k=0;k<n;k++){
a[i][k]+=a[j][k];
if (tmp>0)
tmp+=a[i][k];
else
tmp=a[i][k];
if (tmp>ans)
ans=tmp;
}
}
}
printf("%d\n",ans);
}
// system("pause");