• hdu 1081 To The Max(dp+化二维为一维)


    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1081

    To The Max

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 8839    Accepted Submission(s): 4281


    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
     
    题目大意:在一个数的矩阵里面找一个和最大矩阵。
    题目思路:化二维为一维,转化成1003即可。先求一列一列的和,这样就转化成了一维了。举个例子,我先计算第一行得到dp[0]=0dp[1]=-2,dp[2]=-7,dp[3]=0;这些就是所谓的和,然后在用1003的方法做。那么第二行,就会更新,dp[0]=9,dp[1]=0,dp[2]=-13,dp[3]=2;这些依旧是和,这也就是转换成了一维数组,用1003的方法解决,以此类推~~~
     
    详见代码。
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 
     5 using namespace std;
     6 
     7 int dp[110];
     8 int num[110][110];
     9 
    10 int main ()
    11 {
    12     int n;
    13     while (~scanf("%d",&n))
    14     {
    15         for (int i=0; i<n; i++)
    16             for (int j=0; j<n; j++)
    17                 scanf("%d",&num[i][j]);
    18         int ans=-99999;
    19         for (int i=0; i<n; i++)
    20         {
    21             memset(dp,0,sizeof(dp));
    22             for (int j=i; j<n; j++)
    23             {
    24                 int Max=-1;
    25                 for (int k=0; k<n; k++)
    26                 {
    27                     dp[k]+=num[j][k];       //先计算出所有的dp和
    28                 }
    29                 for (int k=0; k<n; k++)     //1003的做法,代码类似
    30                 {
    31                     if (Max+dp[k]<dp[k])
    32                         Max=dp[k];
    33                     else
    34                         Max=Max+dp[k];
    35                     if (ans<Max)              //不断更新最大值
    36                     {
    37                         ans=Max;
    38                         //cout<<j<<" "<<k<<endl;
    39                     }
    40                 }
    41             }
    42         }
    43         printf ("%d
    ",ans);
    44     }
    45     return 0;
    46 }
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  • 原文地址:https://www.cnblogs.com/qq-star/p/4328338.html
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