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To the Max
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. 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 Source |
题意：
给你一个数字矩阵。要你找出一个子矩阵。使得该矩阵的和比任意其它子矩阵的和都大。求出最大值。
思路：
对于一维的最大连续和肯定很简单。dp[i]=max(arr[i],dp[i-1]+arr[i])。dp[i]表示最大连续和包含arr[i]的最大值。
那么二维就推广下。枚举子矩阵的上下边。把两边间的数字压缩成一条线就可以按一维处理了。
详细见代码：
#include<algorithm>
#include<iostream>
#include<string.h>
#include<sstream>
#include<stdio.h>
#include<math.h>
#include<vector>
#include<string>
#include<queue>
#include<set>
#include<map>
//#pragma comment(linker,"/STACK:1024000000,1024000000")
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=1e-8;
const double PI=acos(-1.0);
const int maxn=100010;
//typedef __int64 ll;
int sum[150][150],dp[150];
int main()
{
int data,n,i,j,k,ans,tp;
memset(sum,0,sizeof sum);
dp[0]=0;
while(~scanf("%d",&n))
{
ans=-INF;
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
scanf("%d",&data);
sum[j][i]=sum[j][i-1]+data;
}
}
for(i=1;i<=n;i++)
{
for(j=1;j<=i;j++)
{
for(k=1;k<=n;k++)
{
tp=sum[k][i]-sum[k][i-j-1];
dp[k]=max(tp,dp[k-1]+tp);
ans=max(dp[k],ans);
}
}
}
printf("%d\n",ans);
}
return 0;
}