[ACM] HDU 5024 Wang Xifeng's Little Plot （构造，枚举）

Wang Xifeng's Little Plot

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

解题思路：

N * N的矩阵。 走的方向为8个方向，当中' . '表示可走，以下用点表示，' #' 表示不可走，找出一条最长的路径包括几个点。输出点的个数。当中路径的要求是 最多包括一个直角（拐一个弯）。

int dir[8][2]={{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1}}   定义方向。 依照 上。左上，右，右下，下，左下，左，左上 的顺序定义，编号为0,1,2,3,4,5,6,7

那么构成直角的两个方向有7对，各自是  0 2，  1 3 。 2  4， 3 5 ，4 6， 5 7 。6 0 。7 1

那么枚举每一个可走的点。求出该点向8个方向分别走的最远距离，然后依照上面的配对，找出一条最长的合法路径。

枚举全然部可走的点，也就得出答案了。

代码：

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
using namespace std;
 
char mp[102][102];
int n;
int dir[8][2]={{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1}};
int ds[8];//一个点每一个方向最长能够走多远
int all[8];//每一个点为直角的拐点两边最长走多远
 
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
            break;
        getchar();
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
                scanf("%c",&mp[i][j]);
            getchar();
        }
        int ans=-1;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
        {
            if(mp[i][j]=='.')
            {
                memset(all,0,sizeof(all));
                memset(ds,0,sizeof(ds));
                for(int k=0;k<8;k++)
                {
                    int x=i,y=j;
                    while(x>=0&&x<n&&y>=0&&y<n&&mp[x][y]=='.')
                    {
                        ds[k]++;//每一个方向上最远走多少
                        x+=dir[k][0];
                        y+=dir[k][1];
                    }
                }
                for(int i=0;i<8;i++)
                {
                    all[i]=ds[i]+ds[(i+2)%8];//构成直角的两个方向
                    ans=max(ans,all[i]);
                    //cout<<"ans"<<ans<<endl;
                }
            }
        }
        printf("%d\n",ans-1);
    }
    return 0;
}