HDU 4708:Rotation Lock Puzzle

Rotation Lock Puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1836    Accepted Submission(s): 580

Problem Description

Alice was felling into a cave. She found a strange door with a number square matrix. These numbers can be rotated around the center clockwise or counterclockwise. A fairy came and told her how to solve this puzzle lock: “When the sum of main diagonal and anti-diagonal is maximum, the door is open.”.
Here, main diagonal is the diagonal runs from the top left corner to the bottom right corner, and anti-diagonal runs from the top right to the bottom left corner. The size of square matrix is always odd.



This sample is a square matrix with 5*5. The numbers with vertical shadow can be rotated around center ‘3’, the numbers with horizontal shadow is another queue. Alice found that if she rotated vertical shadow number with one step, the sum of two diagonals is maximum value of 72 (the center number is counted only once).

 

Input

Multi cases is included in the input file. The first line of each case is the size of matrix n, n is a odd number and 3<=n<=9.There are n lines followed, each line contain n integers. It is end of input when n is 0 .

 

Output

For each test case, output the maximum sum of two diagonals and minimum steps to reach this target in one line.

 

Sample Input

5
9 3 2 5 9
7 4 7 5 4
6 9 3 9 3
5 2 8 7 2
9 9 4 1 9
0

 

Sample Output

72 1

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup

 

迷失在幽谷中的鸟儿，独自飞翔在这偌大的天地间，却不知自己该飞往何方……

题意：给出一个n*n矩阵，n>=3&&n<=9，如果按照最大可能性，只有第二层和第四层有可能是垂直线段覆盖区域！垂直区域可以向逆时针或者顺时针旋转，求旋转几次可以使构成的新矩阵主对角线和副对角线的和最大！
一道简单可暴力解决的题目！因为n<=9！

#include<stdio.h>
#include<iostream>
using namespace std;
int ab[10][10];
int tl,tr,bl,br;
int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                scanf("%d",&ab[i][j]);
        int add=0,kk=0;
        for(int i=1; i<=(n-1)/2; i++)
        {
            int sum=-0xffffff,ji=0;
            for(int j=0; j<n-(i*2-1); j++)
            {
                tl=ab[i+j][i];
                tr=ab[i][n-i+1-j];
                bl=ab[n-i+1][i+j];
                br=ab[n-i+1-j][n-i+1];
                int oth=tl+tr+bl+br;
                if(oth>sum)sum=oth,ji=j;
                else if(oth==sum&&j<ji)ji=j;
            }
            for(int j=0; j<n-(i*2-1); j++)
            {
                tl=ab[i][i+j];
                tr=ab[i+j][n-i+1];
                bl=ab[n-i+1-j][i];
                br=ab[n-i+1][n-i+1-j];
                int oth=tl+tr+bl+br;
                if(oth>sum)sum=oth,ji=j;
                else if(oth==sum&&j<ji)ji=j;
            }
            add+=sum;
            kk+=ji;
        }
        printf("%d %d
",ab[(n+1)/2][(n+1)/2]+add,kk);
    }
    return 0;
}