hdu 2610 Sequence one ( dfs+可行性剪枝 )

Sequence one

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 483    Accepted Submission(s): 179

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.
 

 

Author

yifenfei

题目分析:注意几点,判重,可以根据它是子序列的性质,只需判断当前元和上一个元素间在原始串中有没有相同的元,有则重复,如果是当前串的第一位,就判断和之前所有的元中有没有相同的,然后是当前短串找不到的情况下,是找不到更长的符合条件的串的,所以剪枝时要减去,还有剩余原始串长度如果小于剩余当前的串的长度,也是一定无解的,所以要跟据这些有解的必要条件进行剪枝,也就是可行性剪枝
 

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <map>#define MAX 1007using namespace std;int n,p;int s[MAX];int a[MAX];int cnt;bool check ( int s ,int e ){    for ( int i = s+1 ; i < e ; i++ )        if ( a[i] == a[e] ) return false;    return true;}void dfs ( int cur , int now , int len ){    if ( n - cur < len - now ) return;    if ( cnt >= p ) return;    if ( now == len )    {        cnt++;        printf ( "%d" , s[0] );        for ( int i = 1 ; i < len ; i++ )            printf ( " %d" , s[i] );        puts ( "" );        return;    }    for ( int i = cur+1 ; i <= n ; i++ )    {        if ( now > 0 && a[i] < s[now-1] ) continue;        bool mark = false;        for ( int j = cur+1 ; j < i ; j++ )           if ( a[j] == a[i] )           {               mark = true;               break;           }         //if ( now == 0 && !check ( 0 , i ) ) continue;        //if ( now != 0 && !check ( cur , i ) ) continue;        if ( mark ) continue;        s[now] = a[i];        dfs ( i , now+1 , len );        if ( cnt == p ) return;    }}int main ( ){    //bool flag = false;    while ( ~scanf ( "%d%d" , &n , &p ) )    {        //if ( flag ) puts ("");        //else flag = true;        for ( int i = 1 ; i <= n ; i++ )             scanf ( "%d" , &a[i] );        cnt = 0;        for ( int i = 1 ; i < n ; i++ )        {            int judge = cnt;            dfs ( 0 , 0 , i );            if ( cnt == judge || cnt >= p ) break;        }        puts("");    }}