poj 1050 To the Max(最大子矩阵权值)

To the Max

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 38220 Accepted: 20161

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

40 -2 -7 0 9 2 -6 2-4 1 -4  1 -18  0 -2

Sample Output

15

Source

Greater New York 2001

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;}