HDU 5024 Wang Xifeng's little plot (DFS + brute force)

Wang Xifeng's little plot Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/65536 K (Java/Others)
Total submission (s): 194 accepted submission (s): 131


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 shoshould be as long as possible. but Baoyu was very bad at directions ctions, and he demanded that there cocould 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.
Inputthe 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. when standing on '. ', one can go to adjacent '.'s through 8 directions ctions: 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.
Outputfor each test case, print the maximum length of the road which Wang Xifeng cocould find to locate Baoyu and Baochai's rooms. A road's length is the number '.'s it except des. it's guaranteed that for any test case, the maximum length is at least 2.
Sample Input

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

 
Sample output

3435

Idea: each vertex can be expanded in eight directions. A three-dimensional array is provided to record the Extensible length of each vertex in eight directions.

Then, enumerate the sum of eight vertical line segments and record the maximum value.

#include"iostream"#include"stdio.h"#include"string.h"#include"algorithm"#include"queue"#include"vector"using namespace std;#define N 105#define LL __int64const int inf=1000000;char g[N][N];int dx[8]={1,1,1,0,-1,-1,-1,0};int dy[8]={1,0,-1,-1,-1,0,1,1};int f[N][N][8],n;bool judge(int x,int y){    if(x<0||x>=n||y<0||y>=n)        return true;    return false;}int main(){    int i,j,k,x,y;    while(scanf("%d",&n),n)    {        memset(f,0,sizeof(f));        for(i=0;i<n;i++)            scanf("%s",g[i]);        for(i=0;i<n;i++)        {            for(j=0;j<n;j++)            {                if(g[i][j]=='.')                {                    for(k=0;k<8;k++)                    {                        x=i;y=j;                        int cnt=0;                        while(1)                        {                            if(judge(x,y)||g[x][y]=='#')                                break;                            cnt++;                            x+=dx[k];                            y+=dy[k];                        }                        f[i][j][k]=cnt;                    }                }            }        }        int ans=0;        for(i=0;i<n;i++)        {            for(j=0;j<n;j++)            {                if(g[i][j]=='#')                    continue;                for(k=0;k<8;k++)                {                    ans=max(ans,f[i][j][k]+f[i][j][(k+2)%8]);                }            }        }        printf("%d\n",ans==0?0:ans-1);    }    return 0;}

HDU 5024 Wang Xifeng's little plot (DFS + brute force)