Sicily 6137. Removing Brackets
来源:互联网 发布:移动端js选择时间插件 编辑:程序博客网 时间:2024/05/04 18:42
6137. Removing Brackets
Constraints
Time Limit: 1 secs, Memory Limit: 256 MB
Description
Mirko was bored at his chemistry class, so he played Bomb Switcher on his cell phone. Unfortunately, he was spotted and was given a ridiculously heavy assignment for homework. For a given valid math expression with brackets, he must find all different expressions that can be obtained by removing valid pairs of brackets from the original expression. Two expressions are different if there is a character at which they differ.
For example, given (2+(2*2)+2), one can get (2+2*2+2), 2+(2*2)+2, and 2+2*2+2. (2+2*2)+2 and 2+(2*2+2) can?t be reached, since we would have to remove pairs of brackets that are not valid. More than one pairs of brackets can surround the same part of the expression.
Input
The first and only line of input contains one valid mathematical expression composed of nonnegative integers, basic arithmetic operations denoted with characters ?+?, ?*?, ?-? and ?/?, and brackets ?(? and ?)?.
Given expression won?t have more than 200 characters, and will have at least one, and no more than 10 pairs of brackets. Each expression is guaranteed to have at least one pair of brackets.
Output
Output all different expressions that can be obtained by removing valid pairs of brackets, sorted lexicographically.
Sample Input
样例1:(0/(0))样例2:(2+(2*2)+2)样例3:(1+(2*(3+4)))
Sample Output
样例1:(0/0) 0/(0) 0/0样例2:(2+2*2+2) 2+(2*2)+2 2+2*2+2样例3:(1+(2*3+4)) (1+2*(3+4)) (1+2*3+4) 1+(2*(3+4)) 1+(2*3+4) 1+2*(3+4) 1+2*3+4
Problem Source
COCI 2012.4 2012年每周一赛第十一场
DFS,枚举所有,注意有可能会有((1 + 2))的情况;
#include <iostream>#include <string>#include <string.h>#include <algorithm>using namespace std;string c[1050];//答案string数组char ori_in[205];//原来读入的字符串,用于比较char in[205];//不断在DFS中变换的字符串int b_pos[11];//左括号位置int to_b_pos[11];//与左括号位置对应的用括号位置int counter = 0;//总共不同的答案数int brackets = 0;//括号对数bool was_here(int pos) {//检查是否已经出现过 for (int i = 0; i < pos; i++) { if (c[i] == c[pos]) { return true; } } return false;}void push_in() {//存入答案数组 for (int i = 0; i < (int)strlen(in); i++) { if (in[i] != ' ') { c[counter].push_back(in[i]); } } if (!was_here(counter)) { counter++; } else { c[counter].clear(); }}void DFS(int pos) { if (pos == brackets) { if (strcmp(ori_in, in)) push_in(); return; } DFS(pos + 1);//当前位置括号不变 in[b_pos[pos]] = ' '; in[to_b_pos[pos]] = ' '; DFS(pos + 1);//当前位置括号去掉 in[b_pos[pos]] = '('; in[to_b_pos[pos]] = ')';}bool is_new(int pos, int count) {//计算括号数目的时候用于判断(是否已经出现过 for (int i = 0; i < count; i++) { if (b_pos[i] == pos) { return false; } } return true;}int main() { ios_base::sync_with_stdio(false); cin >> in; strcpy(ori_in, in); brackets = 0; for (int i = 0; i < (int)strlen(in); i++) { if (in[i] == '(') { brackets++; } } int k = 0; for (int i = 0; i < (int)strlen(in); i++) { if (in[i] == ')') { for (int j = i - 1; j >= 0; j--) { if (in[j] == '(' && is_new(j, k)) { b_pos[k] = j; to_b_pos[k++] = i; break; } } } } DFS(0); sort(c, c + counter);//排序得字典序 for (int i = 0; i < counter; i++) { cout << c[i] << endl; } return 0;}
- Sicily 6137. Removing Brackets
- Sicily 1543 Completing Brackets
- Sicily 1543. Completing Brackets
- Sicily 1543. Completing Brackets
- <Sicily>Brackets Matching
- removing
- Sicily 1543 Completing Brackets (SOJ 1543) 【括号匹配】
- Brackets
- Brackets
- Brackets
- Brackets
- Brackets
- Brackets
- brackets
- Brackets
- Brackets! Brackets!
- Removing Columns
- Removing Columns
- JVM详解之Java垃圾回收机制详解和调优
- Jersey 入门与Javabean
- TMDS,LVDS,RSDS
- Color.FromArgb()方法详解
- c++ primer(第五版)笔记 第十四章 重载运算与类型转换
- Sicily 6137. Removing Brackets
- Cordova3.x自学系列之一 环境搭建及常用命令
- 【AC大牛陈鸿的ACM总结贴】【ID AekdyCoin】人家当初也一样是菜鸟
- 关于tableview的一些bug
- 为什么explorer.exe会占有大量的CPU?
- Sicily 6136. Windows
- 从B 树、B+ 树、B* 树谈到R 树(转)
- Spring Transaction + MyBatis SqlSession事务管理机制研究学习
- SIFT算法详解