HDU 5024 Wang Xifeng's Little Plot（暴力枚举+瞎搞）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5024

Problem Description

《Dream of the Red Chamber》(also 《The Story of the Stone》) is one of the Four Great Classical Novels of Chinese literature, and it is commonly regarded as the best one. This novel was created in Qing Dynasty, by Cao Xueqin. But the last 40 chapters of the original version is missing, and that part of current version was written by Gao E. There is a heart breaking story saying that after Cao Xueqin died, Cao's wife burned the last 40 chapter manuscript for heating because she was desperately poor. This story was proved a rumor a couple of days ago because someone found several pages of the original last 40 chapters written by Cao. 

In the novel, Wang Xifeng was in charge of Da Guan Yuan, where people of Jia family lived. It was mentioned in the newly recovered pages that Wang Xifeng used to arrange rooms for Jia Baoyu, Lin Daiyu, Xue Baochai and other teenagers. Because Jia Baoyu was the most important inheritor of Jia family, and Xue Baochai was beautiful and very capable , Wang Xifeng didn't want Jia Baoyu to marry Xue Baochai, in case that Xue Baochai might take her place. So, Wang Xifeng wanted Baoyu's room and Baochai's room to be located at two ends of a road, and this road should be as long as possible. But Baoyu was very bad at directions, and he demanded that there could be at most one turn along the road from his room to Baochai's room, and if there was a turn, that turn must be ninety degree. There is a map of Da Guan Yuan in the novel, and redists (In China English, one whose job is studying 《Dream of the Red Chamber》is call a "redist") are always arguing about the location of Baoyu's room and Baochai's room. Now you can solve this big problem and then become a great redist.

 

Input

The map of Da Guan Yuan is represented by a matrix of characters '.' and '#'. A '.' stands for a part of road, and a '#' stands for other things which one cannot step onto. When standing on a '.', one can go to adjacent '.'s through 8 directions: north, north-west, west, south-west, south, south-east,east and north-east.

There are several test cases.

For each case, the first line is an integer N(0<N<=100) ,meaning the map is a N × N matrix.

Then the N × N matrix follows.

The input ends with N = 0.

 

Output

For each test case, print the maximum length of the road which Wang Xifeng could find to locate Baoyu and Baochai's rooms. A road's length is the number of '.'s it includes. It's guaranteed that for any test case, the maximum length is at least 2.

 

Sample Input

   
   
   
   

    
    
    
    3
#.#
##.
..#
3
...
##.
..#
3
...
###
..#
3
...
##.
...
0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    3
4
3
5
   
   
   
   

 

Source

2014 ACM/ICPC Asia Regional Guangzhou Online

题意:

在一个给定的图中，‘.’代表可行路，‘#’是不可行的，问在最多有一个90度的转弯的路径中最远的两点的距离！

思路：

暴力枚举每一点的八个方向的可行路的距离，由于只能是90度的转弯，所以我们把八个方向分为两次枚举！这样就可以保证是90度了！枚举每一个‘.’的可行路径后再把其中的最远的两条路径相加减一即可！（也就是枚举每一点作为转弯的那一点）。

代码如下：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
#define MAXN 117
using namespace std;
char mm[MAXN][MAXN];
int k;
int xx[8]= {0,1,0,-1,-1,1,1,-1};
int yy[8]= {1,0,-1,0,1,1,-1,-1};
int main()
{
    int n;
    int re1[8],re2[8];
    while(~scanf("%d",&n) && n)
    {
        memset(mm,0,sizeof(mm));
        int maxx = -1;
        for(int i = 0; i < n; i++)
        {
            scanf("%s",mm[i]);
        }
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < n; j++)
            {
                memset(re1,0,sizeof(re1));
                if(mm[i][j] == '.')
                {
                    for(int l = 0; l < 4; l++)
                    {
                        int t1 = i, t2 = j;
                        while(1)
                        {
                            if(mm[t1][t2]=='.')
                            {
                                re1[l]++;
                            }
                            else
                                break;
                            t1+=xx[l], t2+=yy[l];
                        }
                    }
                    sort(re1,re1+4);
                    if(re1[2]+re1[3]> maxx)
                        maxx = re1[2]+re1[3];
                }
            }
        }
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < n; j++)
            {
                memset(re2,0,sizeof(re2));
                if(mm[i][j] == '.')
                {
                    for(int l = 4; l < 8; l++)
                    {
                        int t1 = i, t2 = j;
                        while(1)
                        {
                            if(mm[t1][t2]=='.')
                            {
                                re2[l-4]++;
                            }
                            else
                                break;
                            t1+=xx[l], t2+=yy[l];
                        }
                    }
                    sort(re2,re2+4);
                    if(re2[2]+re2[3]> maxx)
                        maxx = re2[2]+re2[3];
                }
            }
        }
        printf("%d\n",maxx-1);
    }
    return 0;
}