Giftbox
Time Limit:4000MS? Memory Limit:131072K
Total Submit:1210 Accepted:218

Description

Bobobo and Bottle are good friends, and the birthday of Bottle is coming. So Bobobo is considering preparing Bottle an unexpected gift for his birthday. As a common sense that when a person at his or her birthday party, he or she will open the gift with the presence of the person who gives the gift and then expresses the appreciation. Bobobo knows that the characteristic of Bottle is rush when he receives something important, so he wants to play a joke on him in this way…

Input

The input file contains multiple test cases. The first line of each test case contains two numbers. The first one is a positive integer number N (1 ≤ N ≤ 500), the number of boxes in the gift shop, and the second one is a positive number d (3 ≤ d ≤ 1 000), the dimension of all the boxes. The next one line contains d positive integers ( G₁, G₂, …, G_d ) representing the dimensions of the gift. And the subsequent N lines each contain d positive integers ( X₁, X₂, …, X_d ) representing the dimensions of each box. You may assume that all the numbers you encounter are positive integers and less than 2³¹. The input data is terminated by EOF.

Output

The output of each test case will contain only one line. Output the maximum number of the boxes that Bobobo can choose. If Bobobo can not find any box which can contain the gift, output “Please look for another gift shop!”

Sample Input

5 7
4 6 8 2 7 5 3
2 8 13 6 10 9 4
80 70 12 3 6 8 2
8 7 4 6 9 10 12
100 200 300 400 500 600 700
800 800 800 800 800 800 800

Sample Output

3

Source
POJ Monthly--2006.09.29, sza

?

#include? < iostream >
#include? < algorithm >
using ? namespace ?std;

const ? int ?MAXN? = ? 510 ;
const ? int ?MAXM? = ? 1010 ;

int ?n,?m;
int ?data[MAXN][MAXM];
int ?map[MAXN][MAXN];
int ?i,?j,?k;
int ?f[MAXN];
int ?d[MAXN];
int ?ans;

int ?DP( int ?b)
{
???? if ?(d[b]? > ? 0 )? return ?d[b];
???? int ?i;
???? int ?t? = ? 0 ;
???? for ?(i = 1 ;?i <= n;?i ++ )
???? {
???????? if ?(f[i]? && ?b? != ?i? && ?map[b][i]? && ?t? < ?DP(i)? + ? 1 )?t? = ?DP(i)? + ? 1 ;
????}
????d[b]? = ?t;
???? return ?d[b];
}

int ?main()
{???
???? while ?(scanf( " %d%d " ,? & n,? & m)? != ?EOF)
???? {
???????? for ?(i = 0 ;?i <= n;?i ++ )
???????? {
???????????? for ?(j = 0 ;?j < m;?j ++ )
????????????????scanf( " %d " ,? & data[i][j]);
????????????sort(data[i],?data[i] + m);
????????}
????????
????????memset(f,? 1 ,? sizeof (f));
????????
???????? for ?(i = 1 ;?i <= n;?i ++ )
???????????? for ?(j = 0 ;?j < m;?j ++ )
???????????????? if ?(data[i][j]? <= ?data[ 0 ][j])
???????????????? {
????????????????????f[i]? = ? 0 ;
???????????????????? break ;
????????????????}
????????????????
???????? for ?(i = 0 ;?i <= n;?i ++ )
???????????? for ?(j = 1 ;?j <= n;?j ++ )
???????????? {
????????????????map[i][j]? = ? 1 ;
???????????????? for ?(k = 0 ;?k < m;?k ++ )
???????????????? {
???????????????????? if ?(data[i][k]? >= ?data[j][k])
???????????????????? {
????????????????????????map[i][j]? = ? 0 ;
???????????????????????? break ;
????????????????????}
????????????????}
????????????}
????????????
????????memset(d,? 0 ,? sizeof (d));

????????ans? = ?DP( 0 );

???????? if ?(ans? == ? 0 )
????????????printf( " Please?look?for?another?gift?shop!\n " );
???????? else
????????????printf( " %d\n " ,?ans);
???????????????????????
????}
????system( " pause " );
???? return ? 0 ;
}