Hdu-524 Wang Xifeng's Little Plot(記憶化検索)
Wang Xifeng's Little Plot
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 71 Accepted Submission(s): 49
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(0Then 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
Sample Output
1つのn*nの平面の上で'#'は障害物を表して、'.'は道路を表します.点から別の点への経路の最長距離nがあるかどうかを尋ねる.この経路は、各点が8方向に進むことができ、1回だけ曲がることができ、90度しか曲がることができません.解析:記憶化検索、各'.'ノードについて、8方向を検索し、その方向、最も遠い距離を記録し、最後に各'.'ノードを遍歴し、すべてのノードの90度の値の最大値を求める.最大値-1は最終結果です(2つのセグメントの始点が繰り返し計算されるため).
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 71 Accepted Submission(s): 49
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
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
1つのn*nの平面の上で'#'は障害物を表して、'.'は道路を表します.点から別の点への経路の最長距離nがあるかどうかを尋ねる.この経路は、各点が8方向に進むことができ、1回だけ曲がることができ、90度しか曲がることができません.解析:記憶化検索、各'.'ノードについて、8方向を検索し、その方向、最も遠い距離を記録し、最後に各'.'ノードを遍歴し、すべてのノードの90度の値の最大値を求める.最大値-1は最終結果です(2つのセグメントの始点が繰り返し計算されるため).
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 105;
const int INF = 0x3f3f3f3f;
const int dr[] = {-1,-1,-1, 0, 1, 1, 1, 0};
const int dc[] = {-1, 0, 1, 1, 1, 0,-1,-1};
struct Node {
int step[8]; // 8
}p[N][N];
char grid[N][N];
int n;
void solve(int r,int c) {
int x,y;
for(int d = 0; d < 8; d++) {
x = r;
y = c;
while(x >= 0 && x < n && y >= 0 && y < n && grid[x][y] != '#') {
p[r][c].step[d]++;
x += dr[d];
y += dc[d];
}
}
}
int main() {
while(scanf("%d",&n) != EOF && n) {
memset(p,0,sizeof(p));
for(int i = 0; i < n; i++) {
scanf("%s",grid[i]);
}
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
if(grid[i][j] == '#') {
continue;
}
solve(i,j);
}
}
int maxn = -INF;
int sum,d;
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
if(grid[i][j] == '#') {
continue;
}
for(int k = 0; k < 8; k++) {
d = (k+6) % 8; //
sum = p[i][j].step[k] + p[i][j].step[d];
maxn = max(maxn,sum);
d = (k+2) % 8; //
sum = p[i][j].step[k] + p[i][j].step[d];
maxn = max(maxn,sum);
}
}
}
printf("%d
",maxn-1);
}
return 0;
}