ZOJ1463:Brackets Sequence(间隙DP)
Let us define a regular brackets sequence in the following way:
1. Empty sequence is a regular sequence.
2. If S is a regular sequence, then (S) and [S] are both regular sequences.
3. If A and B are regular sequences, then AB is a regular sequence.
For example, all of the following sequences of characters are regular brackets sequences:
(), [], (()), ([]), ()[], ()[()]
And all of the following character sequences are not:
(, [, ), )(, ([)], ([(]
Some sequence of characters '(', ')', '[', and ']' is given. You are to find the shortest possible regular brackets sequence, that contains the given character sequence as a subsequence. Here, a string a1 a2 ... an is called a subsequence of the string b1 b2
... bm, if there exist such indices 1 = i1 < i2 < ... < in = m, that aj = bij for all 1 = j = n.
Input
The input file contains at most 100 brackets (characters '(', ')', '[' and ']') that are situated on a single line without any other characters among them.
Output
Write to the output file a single line that contains some regular brackets sequence that has the minimal possible length and contains the given sequence as a subsequence.
This problem contains multiple test cases!
The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank
line between input blocks.
The output format consists of N output blocks. There is a blank line between output blocks.
Sample Input
1
([(]
Sample Output
()[()]
关键在于输入与输出格式。神坑。
区间dp,dp[i][j]表示
区间 i 到j之间的匹配数,区间两端的 字符能否够刚好匹配,若能够匹配 状态转移就多了一个 dp[i][j] = max(dp[i][k]+dp[k+1][j],dp[i+1][j-1]+1),若不能匹配就是dp[i][j]
= max(dp[i][j],dp[i][k]+dp[k+1][j]);
若是两端能够匹配的,并且两端匹配了导致的dp值最大那么就标记一下。mark[i][j]
= -1,否则 就mark[i][j] = k,这样把全部区间都dp一遍,回头再用DFS寻找。若是两端匹配导致值最大的 那么就直接输出这个字符标记一下,继续往更小的区间去搜索,否则 就分开两个区间搜索 [i,k] [k+1,j]
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define up(i,x,y) for(i=x;i<=y;i++)
#define down(i,x,y) for(i=x;i>=y;i--)
#define mem(a,b) memset(a,b,sizeof(a))
#define w(a) while(a)
char str[105];
int t,len,dp[105][105],mark[105][105],pos[105];
void dfs(int i,int j)
{
if(mark[i][j]==-1)
{
pos[i]=pos[j]=1;
dfs(i+1,j-1);
}
else if(mark[i][j]>=0)
{
dfs(i,mark[i][j]);
dfs(mark[i][j]+1,j);
}
return;
}
int main()
{
int l,i,j,k;
scanf("%d%*c%*c",&t);
while(t--)
{
gets(str);
len=strlen(str);
if(!len)
{
printf("\n");
if(t)
printf("\n");
continue;
}
up(i,0,len-1)
up(j,0,len-1)
{
mark[i][j]=-2;
dp[i][j]=0;
}
mem(pos,0);
i=j=l=0;
w(l<len)
{
if(i==j)
{
i++,j++;
if(j==len)
i=0,l++,j=l;
continue;
}
if((str[i]=='('&&str[j]==')')||(str[i]=='['&&str[j]==']'))
{
up(k,i,j-1)
{
if(dp[i][j]<dp[i][k]+dp[k+1][j])
{
mark[i][j]=k;
dp[i][j]=dp[i][k]+dp[k+1][j];
}
}
if(dp[i][j]<dp[i+1][j-1]+1)
{
mark[i][j]=-1;
dp[i][j]=dp[i+1][j-1]+1;
}
}
else
{
up(k,i,j-1)
{
if(dp[i][j]<dp[i][k]+dp[k+1][j])
{
mark[i][j]=k;
dp[i][j]=dp[i][k]+dp[k+1][j];
}
}
}
i++,j++;
if(j==len)
{
l++;
i=0;
j=l;
}
}
dfs(0,len-1);
up(i,0,len-1)
{
if(pos[i]==1)
printf("%c",str[i]);
else if(str[i]=='('||str[i]==')')
printf("()");
else
printf("[]");
}
printf("\n");
if(t)
{
printf("\n");
getchar();
}
} return 0;
}
版权声明:本文博主原创文章,博客,未经同意不得转载。
ZOJ1463:Brackets Sequence(间隙DP)的更多相关文章
- poj 1141 Brackets Sequence 区间dp,分块记录
Brackets Sequence Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 35049 Accepted: 101 ...
- POJ 1141 Brackets Sequence(区间DP, DP打印路径)
Description We give the following inductive definition of a “regular brackets” sequence: the empty s ...
- poj 1141 Brackets Sequence (区间dp)
题目链接:http://poj.org/problem?id=1141 题解:求已知子串最短的括号完备的全序列 代码: #include<iostream> #include<cst ...
- [原]POJ1141 Brackets Sequence (dp动态规划,递归)
本文出自:http://blog.csdn.net/svitter 原题:http://poj.org/problem?id=1141 题意:输出添加括号最少,并且使其匹配的串. 题解: dp [ i ...
- URAL 1183 Brackets Sequence(DP)
题目链接 题意 : 给你一串由括号组成的串,让你添加最少的括号使该串匹配. 思路 : 黑书上的DP.dp[i][j] = min{dp[i+1][j-1] (sh[i] == sh[j]),dp[i] ...
- Ural 1183 Brackets Sequence(区间DP+记忆化搜索)
题目地址:Ural 1183 最终把这题给A了.. .拖拉了好长时间,.. 自己想还是想不出来,正好紫书上有这题. d[i][j]为输入序列从下标i到下标j最少须要加多少括号才干成为合法序列.0< ...
- poj 1141 Brackets Sequence ( 区间dp+输出方案 )
http://blog.csdn.net/cc_again/article/details/10169643 http://blog.csdn.net/lijiecsu/article/details ...
- UVA 1626 Brackets sequence 区间DP
题意:给定一个括号序列,将它变成匹配的括号序列,可能多种答案任意输出一组即可.注意:输入可能是空串. 思路:D[i][j]表示区间[i, j]至少需要匹配的括号数,转移方程D[i][j] = min( ...
- POJ 题目1141 Brackets Sequence(区间DP记录路径)
Brackets Sequence Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 27793 Accepted: 788 ...
随机推荐
- IOS 与ANDROID框架及应用开发模式对照一
IOS 和ANDROID操作系统都是眼下流行的移动操作系统,被移动终端和智能设备大量採用,两者都採用了先进的软件技术进行设计,为了方便应用开发两者都採用了先进的设计模式. 两者在框架设计上都採用了什么 ...
- SWT的ListVierer的使用
package com.test; import java.util.ArrayList; import java.util.List; import model.People; import org ...
- Determining IP information for eth1... failed; no link present. Check cable? 解决办法
有时候会遇到这种问题 解决办法为 进入网卡配置,将 BOOTPROTO 改为 none 然后 ifconfig –a 查看可以得到 eth1 已经可以寻到 iP 地址,如下: Service netw ...
- C#获取设备的IP和Mac类
/// <summary> /// 此类用于获得设备的Ip和Mac /// </summary> public class Mac { [DllImport("Iph ...
- WPF弹性模拟动画
原文:WPF弹性模拟动画 我们此次将要制作模拟物理中的弹性现象的交互动画,我们让一个小球向鼠标点击位置移动,这个移动的轨迹不是简单的位移,而是根据胡克定律计算得出的. 胡克定律:F=-kd F代表弹性 ...
- 【OC加强】NSDate的使用方法——日期时间在实际开发中比較有用
(1)日期的最主要知识点就是日期转换成字符串格式化输出,相反就是依照某个格式把字符串转换成日期. (2)一般关于时区的设置非常少用到,仅仅要了解就可以. #import <Foundation/ ...
- 同时显示多个 Notification
主要出在PendingIntent.getActivity();的第二个参数,API文档里虽然说是未被使用的参数(给出的例子也直接写0的),实际上是通过该参数来区别不同的Intent的,如果id相同, ...
- atitit.报告最佳实践oae 和报告引擎的选择
atitit.报告最佳实践oae 与报表引擎选型 1. 报表的基本的功能and结构 2 1.1. 查询设计器(配置化,metadata in html) ,anno 2 1.2. 查询引擎 2 1.3 ...
- BZOJ 2783 JLOI 2012 树 乘+二分法
标题效果:鉴于一棵树和一个整数s,问中有树木几个这样的路径,点和担保路径==s,深度增量点. 这一数额的输出. 思维:用加倍的想法,我们可以O(logn)在时间找点他第一n.因为点权仅仅能是正的,满足 ...
- 图像特效——摩尔纹 moir
%%% Moir %%% 摩尔纹 clc; clear all; close all; addpath('E:\PhotoShop Algortihm\Image Processing\PS Algo ...