ural 1146. Maximum Sum（动态规划）

1146. Maximum Sum

Time limit: 1.0 second Memory limit: 64 MB

Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. 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. A sub-rectangle is any contiguous sub-array of size 1 × 1 or greater located within the whole array.

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-hand corner and has the sum of 15.

Input

The input consists of an N × N array of integers. The input begins with a single positive integer Non a line by itself indicating the size of the square two dimensional array. This is followed by N ²integers separated by white-space (newlines and spaces). These N ² integers make up the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on 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

The output is the sum of the maximal sub-rectangle.

Sample

input	output
4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2	15

题意：输入部分第一行是一个正整数N,说明矩阵的大小。下一行后面跟着 个整数。留意这些整数的输入格式可能比较混乱。矩阵是以行优先存储的。N可能和100一样大，矩阵中每个数字的范围为 [-127, 127].

思路1:对矩阵进行穷举。找出所有的子矩阵计算矩阵和并找出最大值。

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>

using namespace std;
int s[105][105]= {0};
int dp[105][105]= {0};



int main()
{
//    freopen("1.txt","r",stdin);
    int i,j,ki,kj;
    int n;
    while(cin>>n)
    {
        for(i=1; i<=n; i++)
            for(j=1; j<=n; j++)
            {
                cin>>s[i][j];
                s[i][j]+=s[i-1][j];
                s[i-1][j]+=s[i-1][j-1];
            }//输入矩阵并且计算从（1,1）到（i-1，j）的矩阵的和值
        for(i=1; i<=n; i++)
            s[n][i]+=s[n][i-1];//计算从（1,1）到（n，i）的矩阵的和值
        int max=-99999999;//特别注意因为
        for(i=1; i<=n; i++)
            for(j=1; j<=n; j++)
            {
                max=-99999999;
                for(ki=0; ki<i; ki++)
                    for(kj=0; kj<j; kj++)
                        if(max<s[i][j]-s[i][kj]-s[ki][j]+s[ki][kj])//计算（ki，kj）到（i,j)的子矩阵的和并且和max比较；
                            max=s[i][j]-s[i][kj]-s[ki][j]+s[ki][kj];
                dp[i][j]=max;
            }
        max=-99999999;
        for(i=1; i<=n; i++)
            for(j=1; j<=n; j++)
                if(max<=dp[i][j])
                    max=dp[i][j];
        cout<<max<<endl;
    }
    return 0;
}

思路2求最大子矩阵的和，先按列进行枚举起始列和结束列，用数据够快速求出每行起始列到结束列之间各数据的和，然后就可以降成一维的，问题就变成了求最大连续子序列了

#include<cstdio>
#include<cstring>
#include<iostream>

using namespace std;

#define N 1100
#define INF 0x3f3f3f3f
int s[N][N]= {0},dp[N],n;

void get_sum(int x ,int y)
{
    for(int i=1; i<=n; i++)
    {
        dp[i]=s[y][i]-s[x-1][i];
    }
    return ;
}

int DP()
{
    int sum=0,max=-INF;
    for(int i=1; i<=n; i++)
    {
        sum+=dp[i];
        max=sum>max?sum:max;
        if(sum<0) sum=0;
    }
    return max;
}



int main()
{
    int i,j;
//    freopen("1.txt","r",stdin);
    while(cin>>n)
    {
        for(i=1; i<=n; i++)
            for(j=1; j<=n; j++)
            {
                cin>>s[i][j];
                s[i][j]+=s[i-1][j];
            }//输入矩阵并且计算从（1,j）到（i，j）的矩阵的和值
        int max=-INF,ans;
        for(i=1; i<=n; i++)
            for(j=i; j<=n; j++)
            {
                get_sum(i,j);
                ans=DP();
                max=ans>max?ans:max;
            }
        printf("%d
",max);
    }
    return 0;
}