hdu 4915 Parenthese sequence(高效)
来源:互联网 发布:stewart mac 编辑:程序博客网 时间:2024/05/22 03:15
题目链接:hdu 4915 Parenthese sequence
题目大意:给定一个序列,由(,),?组成?可以表示(或者),问说有一种、多种或者不存在匹配。
解题思路:从左向右,从右向左,分别维护左括号和右括号可能的情况,区间上下界。如果过程中出现矛盾,则为None,否则要判断唯一解还是多解。枚举每个问号的位置,假设该问号可为左右括号,则有多解。
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1e6+5;char s[maxn];int n, l[maxn][2], r[maxn][2];bool check (int x) { int ldown = l[x-1][0] + 1; int lup = l[x-1][1] + 1; if (ldown > r[x+1][1] || lup < r[x+1][0]) return false; int rdown = r[x+1][0] + 1; int rup = r[x+1][1] + 1; if (rdown > l[x-1][1] || rup < l[x-1][0]) return false; return true;}int judge () { n = strlen(s+1); if (n&1) return 0; memset(l[0], 0, sizeof(l[0])); memset(r[n+1], 0, sizeof(r[n+1])); for (int i = 1; i <= n; i++) { if (s[i] == '(') { l[i][0] = l[i-1][0] + 1; l[i][1] = l[i-1][1] + 1; } else if (s[i] == ')') { if (l[i-1][1] == 0) return 0; l[i][0] = (l[i-1][0] == 0 ? l[i-1][0] + 2 : l[i-1][0]) - 1; l[i][1] = l[i-1][1] - 1; } else { l[i][0] = (l[i-1][0] == 0 ? l[i-1][0] + 2 : l[i-1][0]) - 1; l[i][1] = l[i-1][1] + 1; } } for (int i = n; i; i--) { if (s[i] == ')') { r[i][0] = r[i+1][0] + 1; r[i][1] = r[i+1][1] + 1; } else if (s[i] == '(') { if (r[i+1][1] == 0) return 0; r[i][0] = (r[i+1][0] == 0 ? r[i+1][0] + 2 : r[i+1][0]) - 1; r[i][1] = r[i+1][1] - 1; } else { r[i][0] = (r[i+1][0] == 0 ? r[i+1][0] + 2 : r[i+1][0]) - 1; r[i][1] = r[i+1][1] + 1; } } for (int i = 1; i <= n; i++) if (s[i] == '?' && check(i)) return 2; return 1;}int main () { while (scanf("%s", s+1) == 1) { int flag = judge(); if (flag == 2) printf("Many\n"); else if (flag == 1) printf("Unique\n"); else printf("None\n"); } return 0;}
1 0
- hdu 4915 Parenthese sequence(高效)
- HDU 4915 Parenthese sequence
- hdu 4915 Parenthese sequence
- hdu 4915 Parenthese sequence
- HDU 4915 Parenthese sequence
- HDU-4915-Parenthese sequence
- hdu 4915 Parenthese sequence
- HDU 4915 Parenthese sequence
- 【HDU】4915 Parenthese sequence 贪心
- HDU 4915 Parenthese sequence DP
- HDU 4915 Parenthese sequence(瞎搞题)
- hdu 4915 Parenthese sequence (贪心+模拟)
- hdu 4915 Parenthese sequence(贪心,模拟)
- hdu 4915 Parenthese sequence (贪心+模拟)
- 【HDOJ 4915】Parenthese sequence
- HDOJ 4915 Parenthese sequence
- HDU 4915 Parenthese sequence _(:зゝ∠)_ 呵呵
- hdu 4915 Parenthese sequence 多校第五场
- [leetcode 21] Remove Nth Node From End of List
- 协议,层次,接口
- 输入密码回显星号 -- C语言
- Hibernate_3_客户实例_数据库中自动生成主从表
- struts2自定义类型转换器
- hdu 4915 Parenthese sequence(高效)
- 重装Ubuntu 保留/home分区中的数据
- java中内部类的创建四种情况,三种方式,及内部数据访问权限
- 深拷贝 和 浅拷贝
- 【java】Integer == Integer?
- Codeforce 22B Bargaining Table
- log4cplus 日志应用
- c++学习——第一天
- 中断控制器及中断控制