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

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;
}

[ACM] HDU 5024 Wang Xifeng&#39;s Little Plot (构造,枚举)的更多相关文章

  1. HDU 5024 Wang Xifeng&#39;s Little Plot 搜索

    pid=5024">点击打开链接 Wang Xifeng's Little Plot Time Limit: 2000/1000 MS (Java/Others)    Memory ...

  2. HDU 5024 Wang Xifeng's Little Plot (DP)

    题意:给定一个n*m的矩阵,#表示不能走,.表示能走,让你求出最长的一条路,并且最多拐弯一次且为90度. 析:DP,dp[i][j][k][d] 表示当前在(i, j)位置,第 k 个方向,转了 d ...

  3. HDU 5024 Wang Xifeng's Little Plot(枚举)

    题意:求一个图中只有一个90°拐点的路的最大长度. 分析:枚举每一个为'.'的点,求出以该点为拐点的八种路中的最大长度,再比较所有点,得出最大长度即可. 如上样例,这样是个90°的角... 注意:最多 ...

  4. 2014 网选 5024 Wang Xifeng's Little Plot

    题意:从任意一个任意一个可走的点开始找一个最长的路,这条路如果有转弯的话, 那么必须是 90度,或者没有转弯! 思路: 首先用dfs将所有可走点开始的 8 个方向上的线段的最长长度求出来 ! step ...

  5. hdu5024 Wang Xifeng's Little Plot (水

    http://acm.hdu.edu.cn/showproblem.php?pid=5024 网络赛 Wang Xifeng's Little Plot Time Limit: 2000/1000 M ...

  6. hdu 5024 最长的L型

    http://acm.hdu.edu.cn/showproblem.php?pid=5024 找到一个最长的L型,L可以是斜着的 简单的模拟 #include <cstdio> #incl ...

  7. HDU 4911 http://acm.hdu.edu.cn/showproblem.php?pid=4911(线段树求逆序对)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4911 解题报告: 给出一个长度为n的序列,然后给出一个k,要你求最多做k次相邻的数字交换后,逆序数最少 ...

  8. KMP(http://acm.hdu.edu.cn/showproblem.php?pid=1711)

    http://acm.hdu.edu.cn/showproblem.php?pid=1711 #include<stdio.h> #include<math.h> #inclu ...

  9. HDU-4632 http://acm.hdu.edu.cn/showproblem.php?pid=4632

    http://acm.hdu.edu.cn/showproblem.php?pid=4632 题意: 一个字符串,有多少个subsequence是回文串. 别人的题解: 用dp[i][j]表示这一段里 ...

随机推荐

  1. 【bzoj4999】This Problem Is Too Simple! 树链剖分+动态开点线段树

    题目描述 给您一颗树,每个节点有个初始值. 现在支持以下两种操作: 1. C i x(0<=x<2^31) 表示将i节点的值改为x. 2. Q i j x(0<=x<2^31) ...

  2. NetScaler通过DHCP服务器获取IP地址

    NetScaler通过DHCP服务器获取IP地址 DHCP 选项参考 https://www.iana.org/assignments/bootp-dhcp-parameters/bootp-dhcp ...

  3. POJ2152 Fire 【树形dp】

    题目链接 POJ2152 题解 经典老题,还真暴力 \(n \le 1000\),所以可以\(O(n^2)\)做 所以可以枚举每个点依附于哪一个点 设\(f[u]\)表示以\(u\)为根的子树的最小代 ...

  4. O(1)gcd学习笔记

    设最大权值为\(M\) \(T=\sqrt M\) 定理 任意一个\(\le M\)的数一定可以表示为abc三个数的乘积 满足这三个数要么\(\le T\),要么是一个质数 证明: 考虑反证 假设\( ...

  5. android 微信开发交流群

    有效期很短,可加个人微信 如果已过有效期,加我个人微信,我拉你进群

  6. Javascript&Html-延迟调用和间歇调用

    Javascript&Html-延迟调用和间歇调用 Javascript 是一种单线程语言,所有的javascript任务都会放到一个任务列表中,这些javascript任务会按照插入到列表中 ...

  7. 牛客挑战赛14-F细胞

    https://www.nowcoder.com/acm/contest/81/F 循环卷积的裸题,太久没做FFT了,这么裸的循环卷积都看不出来 注意一下本文的mod 都是指表示幂的模数,而不是NTT ...

  8. 常用 vim 命令总结

    学习 vim ,是需要经常使用的,而这些命令,是我这段时间最常用的命令,很有效率的提高的我的文本编辑效率.----------------------------------------------- ...

  9. usb 2.0 operation mode

    一般來說 USB 的通訊結構有如 Server/Client,以 PC 上的情形為例,位於主機上的 USB 裝置稱為『USB Host』,我們可以在上面外接上數個裝置(與 USB Host 相連的裝置 ...

  10. hdu 4970 树状数组 “改段求段”

    题意:塔防.给1--n,给出m个塔,每个塔有攻击力,给出k个怪兽的位子和血量,问有几只可以到达n点. 今天刚刚复习了树状数组,就碰到这个题,区间更新.区间求和类型.第三类树状数组可以斩. 注意一下大数 ...