ACM/ICPC 之 递归(POJ2663-完全覆盖+POJ1057(百练2775)-旧式文件结构图)
POJ2663-完全覆盖
题解见首注释
//简单递推-三个米诺牌(3*2)为一个单位打草稿得出规律
//题意-3*n块方格能被1*2的米诺牌以多少种情况完全覆盖
//Memory 132K Time: 0 Ms #include<iostream>
#include<cstring>
#include<cstdio>
using namespace std; int ans; //开始平铺
int Tiling(int n)
{
int sum = ;
if (n == )
return ;
else if (n >= )
{
sum += * Tiling(n - );
while (n -= , n - >= )
sum += *Tiling(n-);
}
return sum;
} int main()
{
int n;
while (scanf("%d",&n),n != -)
{
ans = Tiling(n);
printf("%d\n", ans);
}
return ;
}
POJ1057(百练2775)-旧式文件结构图
非递归版-注意记录状态
//旧式shell文件结构图模拟-分析文件(可递归)
//Memory: 148K Time: 0 Ms #include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std; #define MAX 62 char buff[MAX][MAX]; /* File name */
struct File{
char f[MAX];
}; /* Folder property */
struct Folder{
int catalog;
int num;
File f[MAX];
}dir[MAX]; /* sort_cmp */
bool cmp(File a,File b)
{
return strcmp(a.f, b.f) < ;
} void open_Folder(int cur,int num) //cur-当前文件目录,num-文件数
{
int i, j;
sort(dir[cur].f, dir[cur].f + num, cmp);
for (i = ; i < num; i++)
{
for (j = ; j < dir[cur].catalog; j++)
printf("| ");
printf("%s\n", dir[cur].f[i].f);
}
} int main()
{
int T, n;
int i, j, k;
T = ;
while (scanf("%s", buff[]), buff[][] != '#')
{
/*Reading all*/
n = ;
k = ;
while (scanf("%s", buff[++n]), buff[n][] != '*');
printf("DATA SET %d:\n",++T);
printf("ROOT\n");
dir[].num = ; //Init for (i = ; i <= n; i++)
{
if (buff[i][] == 'd') //Folder
{
k++;
dir[k].catalog = dir[k-].catalog+;
dir[k].num = ;
for (j = ; j < dir[k].catalog; j++)
printf("| ");
printf("%s\n", buff[i]);
}
else if (buff[i][] == 'f') //File
strcpy(dir[k].f[dir[k].num++].f, buff[i]);
else if (buff[i][] == ']') //end_Folder
{
open_Folder(k, dir[k].num);
k--;
}
else if (buff[i][] == '*') //end_root
open_Folder(k, dir[k].num);
}
printf("\n");
}
return ;
}
递归版-用形参记录状态
-PS:这个版本是本文发表当天写的,所以和上面版本(前一年做过的)的习惯和极值有些差异。
//递归模拟
//Memory:184K Time:0Ms
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std; #define MAX 31 int T; //当前为第T个样例
char cur[MAX]; //当前输入字符串 //FILE *out = fopen("out.txt", "w"); struct File {
char name[MAX];
friend bool operator < (File a, File b)
{
if (strcmp(a.name, b.name) < ) return true;
else return false;
}
}; struct Catalog {
char name[MAX];
int num; //该目录下的文件数
File f[MAX];
Catalog(char *name) { strcpy(this->name, name); num = ; }
}; void open(char *name,int level)
{
if (name[] == 'R') //ROOT前输入
printf("DATA SET %d:\n", ++T);
//fprintf(out, "\nDATA SET %d:\n", ++T);
while (level--)
printf("| ");
//fprintf(out,"| ");
printf("%s\n", name);
//fprintf(out, "%s\n", name);
} void root(Catalog dir,int level)
{
scanf("%s", cur); if (cur[] == '#') return; //END open(dir.name, level); do{
if (cur[] == 'd')
{
Catalog newDir(cur); //创建新文件夹,命名为cur
root(newDir, level + );
}
else if(cur[] == 'f')
strcpy(dir.f[dir.num++].name, cur);
else { //cur[0] == ']' or cur[0] == '*'
sort(dir.f, dir.f + dir.num);
for (int i = ; i < dir.num; i++)
open(dir.f[i].name, level);
return;
}
}while (scanf("%s", cur));
} int main()
{
while (cur[] != '#')
{
Catalog r("ROOT"); //创建根目录
root(r,);
printf("\n");
} return ;
}
ACM/ICPC 之 递归(POJ2663-完全覆盖+POJ1057(百练2775)-旧式文件结构图)的更多相关文章
- 【转】lonekight@xmu·ACM/ICPC 回忆录
转自:http://hi.baidu.com/ordeder/item/2a342a7fe7cb9e336dc37c89 2009年09月06日 星期日 21:55 初识ACM最早听说ACM/ICPC ...
- 2017 ACM ICPC Asia Regional - Daejeon
2017 ACM ICPC Asia Regional - Daejeon Problem A Broadcast Stations 题目描述:给出一棵树,每一个点有一个辐射距离\(p_i\)(待确定 ...
- HDU 4041 Eliminate Witches! (模拟题 ACM ICPC 2011亚洲北京赛区网络赛)
HDU 4041 Eliminate Witches! (模拟题 ACM ICPC 2011 亚洲北京赛区网络赛题目) Eliminate Witches! Time Limit: 2000/1000 ...
- ACM ICPC 2017 Warmup Contest 9 I
I. Older Brother Your older brother is an amateur mathematician with lots of experience. However, hi ...
- 训练报告 (2014-2015) 2014, Samara SAU ACM ICPC Quarterfinal Qualification Contest
Solved A Gym 100488A Yet Another Goat in the Garden B Gym 100488B Impossible to Guess Solved C Gym ...
- 2014嘉杰信息杯ACM/ICPC湖南程序设计邀请赛暨第六届湘潭市程序设计竞赛
比赛链接: http://202.197.224.59/OnlineJudge2/index.php/Contest/problems/contest_id/36 题目来源: 2014嘉杰信息杯ACM ...
- ACM/ICPC 之 BFS(离线)+康拓展开(TSH OJ-玩具(Toy))
祝大家新年快乐,相信在新的一年里一定有我们自己的梦! 这是一个简化的魔板问题,只需输出步骤即可. 玩具(Toy) 描述 ZC神最擅长逻辑推理,一日,他给大家讲述起自己儿时的数字玩具. 该玩具酷似魔方, ...
- ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 G. Garden Gathering
Problem G. Garden Gathering Input file: standard input Output file: standard output Time limit: 3 se ...
- ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 D. Delay Time
Problem D. Delay Time Input file: standard input Output file: standard output Time limit: 1 second M ...
随机推荐
- Vim 的 tab 设置
文章转自:http://blog.csdn.net/shell_picker/article/details/6033023 摘自 Vim 手册: 选项:1. tabstop:表示一个 tab 显示出 ...
- 降维PCA技术
降维技术使得数据变得更易使用,并且它们往往能够去除数据中的噪声,使得机器学习任务往往更加精确. 降维往往作为预处理步骤,在数据应用到其它算法之前清洗数据.有很多技术可以用于数据降维,在这些技术中,独立 ...
- django 文件上传
模板文件: <form method='post' action='/script/upload/' enctype="multipart/form-data" accept ...
- Linux网络参数设置
1.ifconfig 查询.设定网络卡与ip 设置桥接网络 # vi /etc/sysconfig/network-script/ifcfg-br0 DEVICE=br0 ...
- 跨域攻击xss
要完全防止跨域攻击是很难的,对于有些站点是直接 拦截跨域的访问,在你从本站点跳转到其他站点时提醒,这算是一种手段吧. 而跨域攻击的发起就是在代码(包括html,css,js或是后台代码,数据库数据)里 ...
- 微信电脑版真的要来了 微信Windows版客户端1.0 Alpha推出
微信电脑版的搜索量一直很大,但只有网页版,之前也写了微信网页版APP - 网页微信客户端电脑版体验,在键盘上打字的感觉就是快.现在微信Windows版客户端1.0 Alpha推出了,来一睹芳容吧(20 ...
- ICloud没有密码怎么注销?
忘了密码的用这方法就可以啦1.现在我说你做 . 先进入 iCloud 点击 某某账户(要删的账户)2.第二栏密码那里删除原来的再输入ios(任意密码)3.然后点完成 之后会出现错误.请点 好.然后左上 ...
- fileUpload1.HasFile的返回值永远都是false的问题处理
在aspnet项目中,如果有页面使用了fileupload,不巧你也在此页面使用了updatepanel局部刷新控件,那马就会出现一个很奇怪的问题:就是不管你选择文件了没有,fileUpload1.H ...
- N-Queens leetcode
The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens ...
- leetcode 102. Binary Tree Level Order Traversal
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, ...