hduoj 4708 Rotation Lock Puzzle 2013 ACM/ICPC Asia Regional Online —— Warmup

http://acm.hdu.edu.cn/showproblem.php?pid=4708

Rotation Lock Puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

分析：

题意是求将水平阴影或者竖直阴影旋转最小的次数和，使得主对角线与副对角线的代数和最大。

直接模拟即可。

AC代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
long long  a[15][15];
long long step,ans;
void solve1(int t,int x1,int y1,int x2,int y2,int x3,int y3,int x4,int y4)
{
    int i;
    long long w;
    t = t-1;
    for(i=0;i<t;i++)
    {
        w = a[x1][y1] + a[x2][y2]+a[x3][y3]+a[x4][y4];
        y1++;
        x2++;
        y3--;
        x4--;
            if(w>ans)
            {
             ans = w;
             step = i;
            }
     }
}
void solve2(int t,int x1,int y1,int x2,int y2,int x3,int y3,int x4,int y4)
{
    int i;
    t =t-1;
    long long w;
    //printf("ans = %d %d
",ans,step);
    for(i=0;i<t;i++)
    {
        w = a[x1][y1] + a[x2][y2]+a[x3][y3]+a[x4][y4];
        x1++;
        y2--;
        x3--;
        y4++;
            if(w>=ans)
            {
              ans = w;
              if(step>i)
                 step=i;
            }
     }
}
int main()
{
    int T;
    long long ss,aa;
    while(scanf("%d",&T) && T)
    {
        ss = 0;aa = 0;
        for(int i=1;i<=T;i++)
            for(int j=1;j<=T;j++)
               scanf("%lld",&a[i][j]);
        int mid1 = (T+1)/2;
        for(int i=3,k=1;i<=T;i=i+2,k++)
        {
            step = 0;ans = 0;
            solve1(i,mid1 - k,mid1-k,mid1-k,mid1+k,mid1+k,mid1+k,mid1+k,mid1-k);
            solve2(i,mid1 - k,mid1-k,mid1-k,mid1+k,mid1+k,mid1+k,mid1+k,mid1-k);
            ss= ss +step;
            aa = ans+aa;
        }
   printf("%lld %lld
",aa+a[mid1][mid1],ss);
    }
    return 0;
}