ACM--steps--4.3.7--Sequence one

Sequence one

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 271 Accepted Submission(s): 77 

Problem Description

Search is important in the acm algorithm. When you want to solve a problem by using the search method, try to cut is very important.
Now give you a number sequence, include n (<=1000) integers, each integer not bigger than 2^31, you want to find the first P subsequences that is not decrease (if total subsequence W is smaller than P, than just give the first W subsequences). The order of subsequences is that: first order the length of the subsequence. Second order the sequence of each integer’s position in the initial sequence. For example initial sequence 1 3 2 the total legal subsequences is 5. According to order is {1}; {3}; {2}; {1,3}; {1,2}. {1,3} is first than {1,2} because the sequence of each integer’s position in the initial sequence are {1,2} and {1,3}. {1,2} is smaller than {1,3}. If you also can not understand , please see the sample carefully. 

 

Input

The input contains multiple test cases.
Each test case include, first two integers n, P. (1<n<=1000, 1<p<=10000). 

 

Output

For each test case output the sequences according to the problem description. And at the end of each case follow a empty line.

 

Sample Input

3 51 3 23 61 3 24 1001 2 3 2

 

Sample Output

1321 31 21321 31 21231 21 32 32 21 2 31 2 2
Hint
Hint : You must make sure each subsequence in the subsequences is unique.
#include<iostream>#include<cstring>using namespace std;int dyx[1009];int n,p;//n表示母列中所包含的数字的个数,而p则表示需要求的是几个固定个数的递增子列，如果不够则将全部输出.int len,cont;//len表示进行搜索的长度，cont表示已经搜索了几个串了.struct name{    int now,pos;//now记录在子列中的数，pos记录对应的数字在母列当中的位置.};name wyx[10009];bool flag;//做出一个标记，假如在搜索长度为3的子串时没有符合标准的，那么长度为4的子串自然也不存在.bool check(int st,int ed){    for(int i=st+1;i<ed;i++)    {        if(dyx[i]==dyx[ed])        return false;    }    return true;}void set(int length){    for(int i=0;i<length-1;i++)    cout<<wyx[i].now<<" ";    cout<<wyx[length-1].now<<endl;//最后一个后面没有空格。}void dfs(int dep,int pos){    //dep表示当前搜索的深度,当前搜索的是第几个    //pos表示在当前搜索的子列中的第几个位置.    if(cont>=p)//子列个数到了    return;    if(dep==len)//搜索到了    {        cont++;//有符合标准的字串长度加1；        flag=true;        set(len);        return;    }    //首先这里面也有两个剪枝    //第一个剪枝，如果当前搜索的是子列的第一个位置，那么从母列中的第一个元素到当前位置的元素是否有相等的情况。    //如果出现了相等的情况的话，则舍弃，因为之前肯定搜索过该元素。    //第二个剪枝，如果搜索的不是当前子列的第一个位置的话，那么从当前搜索的子列的前一个元素在母列当中的位置开始    //一直搜索到当前位置之前的一个元素，看是否出现过相同的元素，如果出现过当前的元素，则放弃该元素的搜索.    for(int i=pos;i<n;i++)    {        //判断是否递增；        //可以出现相等的情况        if((dep!=0&&wyx[dep-1].now<=dyx[i])||dep==0)        {             //第一个剪枝的情况。        if(dep==0&&!check(-1,i))        continue;        if(dep!=0&&!check(wyx[dep-1].pos,i))        continue;        wyx[dep].now=dyx[i];        wyx[dep].pos=i;        dfs(dep+1,i+1);        }    }    return;}int main(){    while(cin>>n>>p)    {        for(int i=0;i<n;i++)        cin>>dyx[i];        cont=0;        for(int i=1;i<n;i++)        {            flag=false;            len=i;            dfs(0,0);            if(cont>=p||!flag)            break;        }        cout<<endl;    }    return 0;}


 

 

Author

yifenfei

 

Source

奋斗的年代

 

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