uva 10827 - Maximum sum on a torus(最大子矩阵升级版)

1、http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1768

2、题目大意：

给定一个数字矩阵，求一个最大子矩阵，注意最左边可以和最右边可以作为一个子矩阵，所以需要将原来的矩阵复制成四个矩阵，例如原矩阵为

123

456

789

复制完的矩阵为

123 123

456 456

789 789

123 123

456 456

789 789

然后按照最大子矩阵求，注意控制子矩阵的长度不大于n

2、题目：

Problem H
Maximum sum on a torus
Input: Standard Input

Output: Standard Output

 

A grid that wraps both horizontally and vertically is called a torus. Given a torus where each cell contains an integer, determine the sub-rectangle with the largest sum. The sum of a sub-rectangle is the sum of all the elements in that rectangle. The grid below shows a torus where the maximum sub-rectangle has been shaded.

 

1

-1

0

0

-4

2

3

-2

-3

2

4

1

-1

5

0

3

-2

1

-3

2

-3

2

4

1

-4

Input

The first line in the input contains the number of test cases (at most 18). Each case starts with an integer N (1≤N≤75) specifying the size of the torus (always square). Then follows N lines describing the torus, each line containing N integers between -100 and 100, inclusive.

 

Output

For each test case, output a line containing a single integer: the maximum sum of a sub-rectangle within the torus.

 

Sample Input                                  Output for Sample Input

2

5

1 -1 0 0 -4

2 3 -2 -3 2

4 1 -1 5 0

3 -2 1 -3 2

-3 2 4 1 -4

3

1 2 3

4 5 6

7 8 9

15

45

3、代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int a[160][160];int dp[160];int b[160];int maxsum(int *bb,int n){    int maxx=-9999999;    int max=-9999999;    int i,j;    for(j=1; j<=n; j++)    {        memset(b,0,sizeof(b));        b[j-1]=maxx;        for(i=j; i<j+n&&j+n<=2*n; i++)        {            if(b[i-1]>0)                b[i]=b[i-1]+bb[i];            if(b[i-1]<0)                b[i]=bb[i];            if(b[i]>max)                max=b[i];        }        //printf("&&&%d %d %d %d\n",j,i-1,max,b[i-1]);    }    //printf("\n");    return max;}int main(){    int n,t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        int max=-99999999;        memset(dp,0,sizeof(dp));        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        for(int i=1; i<=n; i++)        {            for(int j=1; j<=n; j++)            {                scanf("%d",&a[i][j]);                a[i][j+n]=a[i+n][j]=a[i+n][j+n]=a[i][j];            }        }//         for(int i=1;i<=n+n;i++)//            {//                for(int j=1;j<=n+n;j++)//                printf("%d ",a[i][j]);//                printf("\n");//            }        for(int i=1; i<=2*n; i++)        {            memset(dp,0,sizeof(dp));            for(int j=i; j<i+n&&i+n<=2*n; j++)            {                for(int k=1; k<=2*n; k++)                {                    dp[k]+=a[j][k];                    //printf("$$$%d %d  %d \n",i,j,dp[k]);                }                //printf("\n");                int sum=maxsum(dp,n);                if(sum>max)                    max=sum;            }            memset(dp,0,sizeof(dp));        }        printf("%d\n",max);    }    return 0;}/*251 -1 0 0 -42 3 -2 -3 24 1 -1 5 03 -2 1 -3 2-3 2 4 1 -431 2 34 5 67 8 9*/