acm 2 1025 Sequence two

1.1025

2.Sequence two

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 98 Accepted Submission(s): 46

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 (<=100) 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 subsequence by lexicographical. For example initial sequence 1 3 2 the total legal subsequences is 5. According to order is {1}; {2}; {3}; {1,2}; {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<=100, 1<p<=100000).

 

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 5

1 3 2

3 6

1 3 2

4 100

1 2 3 2

 

Sample Output

1

2

3

1 2

1 3

1

2

3

1 2

1 3

1

2

3

1 2

1 3

2 2

2 3

1 2 2

1 2 3

3.和上一题差不多*-*

4.dfs，首先排一次顺序，注意输下标，生成的子串递增，下标也是递增，不能有重复的子串。

5.#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

struct numberr

{

    int number;

    int postion;

};

bool cmp(const numberr &a,const numberr &b)

{

    if(a.number!=b.number)

        return a.number<b.number;

    return a.postion<b.postion;

}

numberr b[1005];

int a[1005];

int n,p,cou,dep;

int c[1005];

int dfs(int nowdep,int pos,int rr)

{

    if(nowdep==dep)

    {

        cou++;

        for(int i=0;i<nowdep-1;i++)

        {

            cout<<c[i]<<" ";

        }

        cout<<c[nowdep-1]<<endl;

        if(cou==p) return 1;

        return 0;

    }

    int pre,flag=0;

    for(int i=pos;i<=n;i++)

    {

        if(b[i].postion>rr)

        {

            if(!flag)

            {

                flag=1;

                pre=b[i].number;

            }

            else if(pre==b[i].number) continue;

        pre=b[i].number;

        c[nowdep]=b[i].number;

        if(dfs(nowdep+1,i+1,b[i].postion))

                return 1;

        }

    }

    return 0;

}

int main()

{

  

    while(cin>>n>>p)

    {

        for(int i=1;i<=n;i++)

        {

            cin>>b[i].number;

            b[i].postion=i;

        }

        sort(b+1,b+n+1,cmp);

        cou=0;

        for(int i=1;i<n;i++)

        {

            dep=i;

            if(dfs(0,1,0))break;

        }

        cout<<endl;

    }

    return 0;

}