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): 1571    Accepted Submission(s): 499

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

59 3 2 5 97 4 7 5 46 9 3 9 35 2 8 7 29 9 4 1 90

 

Sample Output

72 1

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup

 

题意：

一个矩阵，每层可以绕中点旋转求对角线的和的最大值。

分析：

代码：

#include <cstdio>#include <iostream>using namespace std;int main(){    int n, a[15][15];    while(~scanf("%d", &n), n)    {        for(int i = 0; i < n; i++)            for(int j = 0; j < n; j++)                scanf("%d", &a[i][j]);        int sum = 0, ICnt = 0;        int x1, y1, x2, y2, x3, y3, x4, y4; //四个点的坐标        for(x1 = 0; x1 < n / 2; x1++)       //控制环，n / 2 个        {            x2 = y1 = x1;       //定位四个点            y2 = n - y1 - 1;            y3 = y2;            x3 = n - x2 - 1;            x4 = x3;            y4 = y1;            int ma = a[x1][y1] + a[x2][y2] + a[x3][y3] + a[x4][y4]; //初始环的最大值            int cnt = 0, tc;            int ty1 = y1;            for(int i = y1 + 1; i < y2; i++)    //旋转            {                y1++;                x2++;                y3--;                x4--;                int tma = a[x1][y1] + a[x2][y2] + a[x3][y3] + a[x4][y4];                int ty = (ty1 + y2) / 2;        //中点纵坐标                if(tma > ma)    //更新最大值                {                    ma = tma;                    if(i > ty) cnt = y2 - i;    //在中点右边，次数为向右旋转                    else cnt = i - ty1;         //在中点左边，同理                }                if(tma == ma)               //两个点对应的值的和相等                {                    if(i > ty) tc = y2 - i;                    else tc = i - ty1;                    cnt = min(tc, cnt);     //取旋转次数小的                }            }            ICnt += cnt;            sum += ma;        }        sum += a[n / 2][n / 2];        printf("%d %d\n", sum, ICnt);    }    return 0;}