Order is an important concept in mathematics and in computer science. For example, Zorn's Lemma states: ``a partially ordered set in which every chain has an upper bound contains a maximal element.'' Order is also important in reasoning about the fix-point semantics of programs.

This problem involves neither Zorn's Lemma nor fix-point semantics, but does involve order.
Given a list of variable constraints of the form x < y, you are to write a program that prints all orderings of the variables that are consistent with the constraints.

For example, given the constraints x < y and x < z there are two orderings of the variables x, y, and z that are consistent with these constraints: x y z and x z y.

Input

The input consists of a sequence of constraint specifications. A specification consists of two lines: a list of variables on one line followed by a list of contraints on the next line. A constraint is given by a pair of variables, where x y indicates that x < y.

All variables are single character, lower-case letters. There will be at least two variables, and no more than 20 variables in a specification. There will be at least one constraint, and no more than 50 constraints in a specification. There will be at least one, and no more than 300 orderings consistent with the contraints in a specification.

Input is terminated by end-of-file.

Output

For each constraint specification, all orderings consistent with the constraints should be printed. Orderings are printed in lexicographical (alphabetical) order, one per line.

Output for different constraint specifications is separated by a blank line.

Sample Input

a b f g
a b b f
v w x y z
v y x v z v w v

Sample Output

abfg
abgf
agbf
gabf wxzvy
wzxvy
xwzvy
xzwvy
zwxvy
zxwvy 思路:
简单的拓扑排序+dfs,通过入度判断,从1到26就有字典序,代码如下:
int in[], G[][], vis[], print[], num;
char ans[]; void init() {
num = ;
memset(in, , sizeof(in));
memset(G, , sizeof(G));
memset(vis, , sizeof(vis));
memset(print, , sizeof(print));
} void dfs(int u, int cnt) {
ans[cnt] = u - + 'a';
if(cnt == num) {
for(int i = ; i <= cnt; ++i)
cout << ans[i];
cout << "\n";
return;
}
// mark the point
print[u] = ;
for(int i = ; i <= ; ++i) {
if(vis[i] && G[u][i]) in[i]--;
}
for(int i = ; i <= ; ++i) {
if(vis[i] && !print[i] && !in[i]) {
dfs(i, cnt+);
}
// backtracing
}
for(int i = ; i <= ; ++i) {
if(vis[i] && G[u][i]) in[i]++;
}
print[u] = ;
} int main() {
ios::sync_with_stdio(false);
string t;
int t1, t2;
while(getline(cin, t)) {
init();
int siz = t.size();
for(int i = ; i < siz; i += ) {
vis[t[i] - 'a' + ] = ;
num++;
}
getline(cin, t);
siz = t.size();
for(int i = ; i + < siz; i += ) {
t1 = t[i] - 'a' + , t2 = t[i+] - 'a' + ;
G[t1][t2] = , in[t2]++;
}
for(int i = ; i <= ; ++i) {
if(vis[i] && !in[i]) {
dfs(i, );
}
}
cout << "\n";
} return ;
}

Day4 - H - Following Orders POJ - 1270的更多相关文章

  1. POJ 1270 Following Orders

    Following Orders Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4902   Accepted: 1982 ...

  2. H - Buy Tickets POJ - 2828 逆序遍历 树状数组+二分

    H - Buy Tickets POJ - 2828 这个题目还是比较简单的,其实有思路,不过中途又断了,最后写了一发别的想法的T了. 然后脑子就有点糊涂,不应该啊,这个题目应该会写才对,这个和之前的 ...

  3. POJ 1270 Following Orders 拓扑排序

    http://poj.org/problem?id=1270 题目大意: 给你一串序列,然后再给你他们部分的大小,要求你输出他们从小到大的所有排列. 如a b f g 然后 a<b ,b< ...

  4. POJ 1270 Following Orders (拓扑排序,dfs枚举)

    题意:每组数据给出两行,第一行给出变量,第二行给出约束关系,每个约束包含两个变量x,y,表示x<y.    要求:当x<y时,x排在y前面.让你输出所有满足该约束的有序集. 思路:用拓扑排 ...

  5. poj 1270 Following Orders (拓扑排序+回溯)

    Following Orders Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5473   Accepted: 2239 ...

  6. POJ 1270 Following Orders(拓扑排序)题解

    Description Order is an important concept in mathematics and in computer science. For example, Zorn' ...

  7. POJ 1270 Following Orders(拓扑排序)

    题意: 给两行字符串,第一行为一组变量,第二行时一组约束(每个约束包含两个变量,x y 表示 x <y).输出满足约束的所有字符串序列. 思路:拓扑排序 + 深度优先搜索(DFS算法) 课本代码 ...

  8. poj 1270(toposort)

    http://poj.org/problem?id=1270 题意:给一个字符串,然后再给你一些规则,要你把所有的情况都按照字典序进行输出. 思路:很明显这肯定要用到拓扑排序,当然看到discuss里 ...

  9. poj 1270(dfs+拓扑排序)

    题目链接:http://poj.org/problem?id=1270 思路:就是一简单的dfs+拓扑排序,然后就是按字典序输出所有的情况. http://paste.ubuntu.com/59872 ...

随机推荐

  1. 简单聊一聊Ansible自动化运维

    一.Ansible概述 Ansible是今年来越来越火的一款开源运维自动化工具,通过Ansible可以实现运维自动化,提高运维工程师的工作效率,减少人为失误.Ansible通过本身集成的非常丰富的模块 ...

  2. 让 el-dialog 居中,并且内容多的时候内部可以滚动

    .el-dialog { position: absolute; top: 50%; left: 50%; margin: 0 !important; transform: translate(-50 ...

  3. 腾讯云 docker 镜像 dotnet/core sdk aspnet

    ccr.ccs.tencentyun.com/mcr.microsoft.com/dotnetcoresdk  = mcr.microsoft.com/dotnet/core/sdk  => 3 ...

  4. nginx 书籍

    1.<实战nginx> 2.<深入理解nginx> 3.nginx开发从入门到精通 http://tengine.taobao.org/book/ 4.Nginx源码学习,配置 ...

  5. JS html页面

    js窗口置顶 if (window != top) top.location.href = location.href; js打开新窗口 js window.open()弹出窗口参数说明及居中设置 f ...

  6. Lightroom中几个重要名词术语的解释

    Lightroom是照片管理.处理.发布的综合性智能软件,里面有几个重要的专有技术名词,通过我的理解做一个总结: 一.目录(Catalog) 就是Lightroom的数据库,会把用户的照片的信息.照片 ...

  7. uniGUI之TUniHiddenPanel(14)

    TUniHiddenPanel是将不在界面上显示的  容器  控件.  只有uniDBGrid实际列才有对应的编辑控件,如果是外键列则无法设置 编辑控件. 里面的控件将不会 显示.将控件放入其中即可. ...

  8. 对Python中列表和数组的赋值,浅拷贝和深拷贝的实例讲解

    引用:https://www.jb51.net/article/142775.htm 列表赋值: 1 2 3 4 5 6 7 >>> a = [1, 2, 3] >>&g ...

  9. RadioButton 用法

    @Html.RadioButton("rdoNotice", "1ST", true, new { id = "rdoFirstNotice" ...

  10. cookie、sessionStorage和localStorage的区别

    cookie.sessionStorage.localStorage 都是用于本地存储的技术:其中 cookie 出现最早,但是存储容量较小,仅有4KB:sessionStorage.localSto ...