学步园

传送门：【CodeForces】463D Gargari and Permutations

题目分析：对于所有的数对(x,y)如果x在所有的串中都排在y的前面，则建边(x,y)，最后跑一遍最长路，最长路的长度就是答案。

代码如下：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

#define REP( i , a , b ) for ( int i = ( a ) ; i <  ( b ) ; ++ i )
#define FOR( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define REV( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )
#define CLR( a , x ) memset ( a , x , sizeof a )

typedef long long LL ;

const int MAXN = 1005 ;
const int MAXE = 1000005 ;

struct Edge {
	int v ;
	Edge* next ;
} E[MAXE] , *H[MAXN] , *cur ;

int G[MAXN][MAXN] ;
int d[MAXN] ;
int in[MAXN] ;
int Q[MAXN] , head , tail ;
int n , k ;
int s[MAXN] ;

void clear () {
	cur = E ;
	CLR ( H , 0 ) ;
	CLR ( G , 0 ) ;
	CLR ( d , 0 ) ;
	CLR ( in , 0 ) ;
	head = tail = 0 ;
}

void addedge ( int u , int v ) {
	cur -> v = v ;
	cur -> next = H[u] ;
	H[u] = cur ++ ;
}

void topo () {
	FOR ( i , 1 , MAXN ) if ( !in[i] ) d[i] = 1 , Q[tail ++] = i ;
	while ( head != tail ) {
		int u = Q[head ++] ;
		travel ( e , H , u ) {
			int v = e -> v ;
			if ( d[v] < d[u] + 1 ) d[v] = d[u] + 1 ;
			if ( -- in[v] == 0 ) Q[tail ++] = v ;
		}
	}
}

void solve () {
	int ans = 0 ;
	clear () ;
	FOR ( _ , 1 , k ) {
		FOR ( i , 1 , n ) {
			scanf ( "%d" , &s[i] ) ;
			REP ( j , 1 , i ) ++ G[s[j]][s[i]] ;
		}
	}
	FOR ( i , 1 , n ) FOR ( j , 1 , n ) if ( G[i][j] == k ) addedge ( i , j ) , ++ in[j] ;
	topo () ;
	FOR ( i , 1 , n ) if ( d[i] > ans ) ans = d[i] ;
	printf ( "%d\n" , ans ) ;
}

int main () {
	while ( ~scanf ( "%d%d" , &n , &k ) ) solve () ;
	return 0 ;
}