Longest Increasing Subsequences(最长递增子序列)的两种DP实现
/**
@description: Longest Increasing Subsequence
@author: seiyagoo
@create: 2013.10.25
@modified: 2013.10.26
**/
int LIS_1(int A[], int size){ int *LIS = new int[size];
vector<int> *vec = new vector<int>[size]; /* Compute optimized LIS values in bottom up manner */
for(int i=; i < size; i++){
LIS[i]=; //初始化默认长度
int max_j=, flag=;
for(int j=; j < i; j++){ //查表,找出前面最长的序列, 若将A[i]加入LIS[j](LIS[j]+1的含义)的递增子序列比当前的LIS[i]更长, 则更新LIS[i]
if(A[i] > A[j] && LIS[i] < LIS[j]+){
LIS[i] = LIS[j]+;
max_j=j;
flag=;
}
}
if(flag) //copy前面最长子序列到vec[i]
vec[i].insert(vec[i].end(), vec[max_j].begin(), vec[max_j].end());
vec[i].push_back(A[i]); //最后放入A[i]
} /*Show LIS of the current state*/
vector<int>::iterator it;
cout<<left;
for(int i=; i<size; i++){
cout<<setw()<<A[i]<< " --> ";
for(it = vec[i].begin(); it!=vec[i].end(); it++)
cout<<*it<<" ";
cout<<endl;
} /* Pick maximum of all LIS values, namely max{LIS[i]} */
int max_len=;
for(int i = ; i < size; i++ )
if( max_len < LIS[i] )
max_len = LIS[i]; delete[] LIS;
delete[] vec; return max_len;
}
/**
@description: Longest Increasing Subsequence
@author: seiyagoo
@create: 2013.10.25
@modified: 2013.10.26
**/ // Binary search (note boundaries in the caller)
// A[] is ceilIndex in the caller
int CeilIndex(int A[], int l, int r, int key) {
int m; while( r - l > ) {
m = l + (r - l)/;
(A[m] >= key ? r : l) = m; // ternary expression returns an l-value
} return r;
} int LIS_2(int A[], int size) {
// boundary case: when array size is one
if( == size ) return ; int *tailTable = new int[size];
vector<int> *vec = new vector<int>[size];
int len; // always points empty slot //memset(tailTable, INT_MAX, sizeof(tailTable[0])*size); @bug for(int i = ; i < size; i++)
tailTable[i] = INT_MAX; tailTable[] = A[]; //tailTable[0] store the smallest value
vec[].push_back(A[]); len = ;
for( int i = ; i < size; i++ ) {
if( A[i] < tailTable[] ) { //case 1: new smallest value
tailTable[] = A[i]; /*discard and create*/
vec[].clear();
vec[].push_back(A[i]);
}
else if( A[i] > tailTable[len-] ) { //case 2: A[i] wants to extend largest subsequence
tailTable[len++] = A[i]; /*clone and extend*/
vec[len-] = vec[len-];
vec[len-].push_back(A[i]);
}
else { //case 3: A[i] wants to be current end candidate of an existing subsequence, It will replace ceil value in tailTable
int ceilIndex = CeilIndex(tailTable, -, len-, A[i]);
tailTable[ceilIndex] = A[i]; /*discard, clone and extend*/
vec[ceilIndex].clear();
vec[ceilIndex] = vec[ceilIndex-];
vec[ceilIndex].push_back(A[i]);
} /*Printf all the active lists*/
vector<int>::iterator it;
cout<<left;
cout<<"A["<<i<<"] = "<<A[i]<<endl<<endl;
cout<<"active lists:"<<endl;
for(int i=; i<len; i++){
for(it = vec[i].begin(); it!=vec[i].end(); it++)
cout<<*it<<" ";
cout<<endl;
} /*Printf end elements of all the active lists*/
cout<<endl<<"end elements array:"<<endl;
for(int i = ; i < size; i++)
if(tailTable[i] != INT_MAX)
cout<<tailTable[i]<<" ";
cout<<endl;
cout<<"-------------------------"<<endl;
} delete[] tailTable;
delete[] vec; return len;
}
五、运行结果
example:





Longest Increasing Subsequences(最长递增子序列)的两种DP实现的更多相关文章
- leetcode300. Longest Increasing Subsequence 最长递增子序列 、674. Longest Continuous Increasing Subsequence
Longest Increasing Subsequence 最长递增子序列 子序列不是数组中连续的数. dp表达的意思是以i结尾的最长子序列,而不是前i个数字的最长子序列. 初始化是dp所有的都为1 ...
- [LintCode] Longest Increasing Subsequence 最长递增子序列
Given a sequence of integers, find the longest increasing subsequence (LIS). You code should return ...
- [LeetCode] Longest Increasing Subsequence 最长递增子序列
Given an unsorted array of integers, find the length of longest increasing subsequence. For example, ...
- 673. Number of Longest Increasing Subsequence最长递增子序列的数量
[抄题]: Given an unsorted array of integers, find the number of longest increasing subsequence. Exampl ...
- [LeetCode] 300. Longest Increasing Subsequence 最长递增子序列
Given an unsorted array of integers, find the length of longest increasing subsequence. Example: Inp ...
- [leetcode]300. Longest Increasing Subsequence最长递增子序列
Given an unsorted array of integers, find the length of longest increasing subsequence. Example: Inp ...
- poj 2533 Longest Ordered Subsequence 最长递增子序列
作者:jostree 转载请注明出处 http://www.cnblogs.com/jostree/p/4098562.html 题目链接:poj 2533 Longest Ordered Subse ...
- [LeetCode] Number of Longest Increasing Subsequence 最长递增序列的个数
Given an unsorted array of integers, find the number of longest increasing subsequence. Example 1: I ...
- POJ 2533 - Longest Ordered Subsequence - [最长递增子序列长度][LIS问题]
题目链接:http://poj.org/problem?id=2533 Time Limit: 2000MS Memory Limit: 65536K Description A numeric se ...
- [LeetCode] 673. Number of Longest Increasing Subsequence 最长递增序列的个数
Given an unsorted array of integers, find the number of longest increasing subsequence. Example 1: I ...
随机推荐
- Ansible学习记录二:命令
0.ansible 命令参数详解: [root@localhost ~]# ansible Usage: ansible <host-pattern> [options] Options: ...
- python hmac 加密
python2 : key 是秘钥 类型为 str msg 要加密的文件 str digestmod 要加密的方式 python3: key 是秘钥 类型为 byte msg 要加密的文件 byte ...
- 【例题 8-15 UVA - 12174】Shuffle
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 枚举第一段的范围[0..i] (0<=i<s) 然后看看[i+1..i+s-1],[i+s,i+s+s-1]..这些区间 ...
- [Angular] How to get Store state in ngrx Effect
For example, what you want to do is navgiate from current item to next or previous item. In your com ...
- Android Studio配置SVN 以及使用代码管理
一.Android Studio配置SVN Android Studio关联配置SVN非常easy,在Settings里面.找到Version Control->Subversion.在这个页面 ...
- 高性能计算机传奇(vamei)
高性能计算机是用网络将多台计算机连接在一起.并构成一个统一的系统,从而拥有远超个人电脑的计算能力.这样利用网络,让计算机合作工作的并行系统又称为集群(cluster).server.分布式计算机.超级 ...
- Logstash读写性能调整优化
继续
- echarts tooltip提示框 自定义小圆点(颜色、形状和大小等等)
项目是拿 echarts + 百度地图 来做可视化界面,现在到收尾阶段慢慢优化. 先附代码: formatter: function(params) { var result = '' params. ...
- C# Aspect-Oriented Programming(AOP) 利用多种模式实现动态代理
什么是AOP(Aspect-Oriented Programming)? AOP允许开发者动态地修改静态的OO模型,构造出一个能够不断增长以满足新增需求的系统,就象现实世界中的对象会在其生命周期中不断 ...
- Linux系列-安装经常使用软件
安装JDK: 理论篇: 一.下载JDK 二.安装 ①复制到/usr/java/路径下 [plain] view plaincopy #mkdir /usr/java/ #cp jdk-7u25-lin ...