2016SDAU编程练习二1025

Sequence two 


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.<br>Now give you a number sequence, include n (&lt;=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. <br>
 


Input
The input contains multiple test cases.<br>Each test case include, first two integers n, P. (1<n<=100, 1<p<=100000). <br>
 


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<br>1 3 2<br>3 6<br>1 3 2<br>4 100<br>1 2 3 2<br> 


Sample Output
1<br>2<br>3<br>1 2<br>1 3<br><br>1<br>2<br>3<br>1 2<br>1 3<br><br>1<br>2<br>3<br>1 2<br>1 3<br>2 2<br>2 3<br>1 2 2<br>1 2 3<br><br><br><div style='font-family:Times New Roman;font-size:14px;background-color:F4FBFF;border:#B7CBFF 1px dashed;padding:6px'><div style='font-family:Arial;font-weight:bold;color:#7CA9ED;border-bottom:#B7CBFF 1px dashed'><i>Hint</i></div>Hint : You must make sure each subsequence in the subsequences is unique.</div> 


Author
yifenfei
 


Source

奋斗的年代

题意：给一个数列，还是求递增子序列

思路：感觉和24差不多吧

感想：完全不想写了。。。

AC代码：

#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()
{
    //freopen("r.txt","r",stdin);
    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;
}