题目大意:给一个变量列表和变量的大小关系,输出所有的满足约束的序列. 构建为有向图,然后就是拓扑排序,使用回溯输出所有的结果. #include <cstdio> #include <cstring> #include <cctype> #include <map> #include <algorithm> #include <vector> using namespace std; #define N 26 map<char,…

Following Orders Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4885   Accepted: 1973 Description Order is an important concept in mathematics and in computer science. For example, Zorn's Lemma states: ``a partially ordered set in which…

Following Orders Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4902   Accepted: 1982 Description Order is an important concept in mathematics and in computer science. For example, Zorn's Lemma states: ``a partially ordered set in which…

题目链接: 啊哈哈.点我点我 题意是: 第一列给出全部的字母数,第二列给出一些先后顺序. 然后按字典序最小的方式输出全部的可能性.. . 思路: 整体来说是拓扑排序.可是又非常多细节要考虑.首先要按字典序最小的方式输出,所以自然输入后要对这些字母进行排列.然后就是输入了.用scanf不能读空格.所以怎么建图呢??设置一个变量推断读入的先后顺序.那么建图完成后,就拓扑排序了.那么多种方式自然就是dfs回溯了..那么这个问题就得到了解决.. 题目: Following Orders Time Lim…

UVA - 1153 Keep the Customer Satisfied Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Description   Simon and Garfunkel Corporation (SG Corp.) is a large steel-making company with thousand of customers. Keeping the customer…

"Waiting for orders we held in the wood, word from the front never came By evening the sound of the gunfire was miles away Ah softly we moved through the shadows, slip away through the trees Crossing their lines in the mists in the fields on our hand…

uva 10192 Vacation The Problem You are planning to take some rest and to go out on vacation, but you really don't know which cities you should visit. So, you ask your parents for help. Your mother says "My son, you MUST visit Paris, Madrid, Lisboa an…

题意:给出几个圆的半径,贴着底下排放在一个长方形里面,求出如何摆放能使长方形底下长度最短. 由于球的个数不会超过8, 所以用全排列一个一个计算底下的长度,然后记录最短就行了. 全排列用next_permutation函数,计算长度时坐标模拟着摆放就行了. 摆放时折腾了不久,一开始一个一个把圆放到最左端,然后和前面摆好的圆比较检查是否会出现两个圆重叠,是的话就把当前圆向右移动到相切的位置.然后判断宽度. 结果发现过不了几个例子,检查了好久,终于发现问题出在初始位置上,如果出现一个特别小的圆放到最左…

10067 - Playing with Wheels 题目页:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1008 从一开始思路就不对,之后才焕然大悟……每次都是这样. 还有,感觉搜索和图遍历有点分不清呢. 在第63行加入 if (u == target) return; 可以提速很多,可以从300ms左右降低到100ms以内. ?…

状压dp,每个状态可以表示为一个n元组,且上限为8,可以用一个九进制来表示状态.但是这样做用数组开不下,用map离散会T. 而实际上很多九进制数很多都是用不上的.因此类似uva 1601 Morning after holloween的思想,先dfs预处理出所有状态,用map将状态离散, 预处理出算出状态的转移DAG,而不是转移的时候在解码编码判断是否可行,然后一层一层bfs就行了. 附上测试数据 一层一层的bfs转移类似滚动数组,注意初始化(具体问题具体分析,有些问题没有必要做这一步) 一层一…