【POJ】3294 Life Forms 【后缀数组——求在超过一半串中出现的最长串】

传送门：【POJ】3294 Life Forms

题目分析：我们将所有的串合并成一个串，相邻两串中间用分隔符隔开，注意分隔符一定要用各不相同的且都不在原串中出现的。然后构造后缀数组，这个用倍增算法就好了。然后，我们二分最长串的长度k，然后扫一遍height数组，当>=k时我们统计出现在不同串的个数cnt，当<k时我们就看是否cnt>n/2，是的话返回yes，否则返回no，yes的话就调整下界，否则调整上界。然后输出的话和二分长度时用的函数类似，只不过返回的地方我们换成输出。具体细节留待各位品味。输出‘？’当且仅当长度为0。

代码如下：

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

typedef long long LL ;

#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 clr( a , x ) memset ( a , x , sizeof a )
#define cpy( a , x ) memcpy ( a , x , sizeof a )

const int MAXN = 110005 ;

char buf[1005] ;
int s[MAXN] ;
int start[105] ;
int t1[MAXN] , t2[MAXN] , c[MAXN] , xy[MAXN] ;
int sa[MAXN] , rank[MAXN] , height[MAXN] ;
int vis[105] , Time ;
int n , m ;

int cmp ( int *r , int a , int b , int d ) {
	return r[a] == r[b] && r[a + d] == r[b + d] ;
}

void getHeight ( int n , int k = 0 ) {
	For ( i , 0 , n ) rank[sa[i]] = i ;
	rep ( i , 0 , n ) {
		if ( k ) -- k ;
		int j = sa[rank[i] - 1] ;
		while ( s[i + k] == s[j + k] ) ++ k ;
		height[rank[i]] = k ;
	}
}

void da ( int n , int m = 300 ) {
	int i , d , p , *x = t1 , *y = t2 , *t ;
	for ( i = 0 ; i < m ; ++ i ) c[i] = 0 ;
	for ( i = 0 ; i < n ; ++ i ) ++ c[x[i] = s[i]] ;
	for ( i = 1 ; i < m ; ++ i ) c[i] += c[i - 1] ;
	for ( i = n - 1 ; i >= 0 ; -- i ) sa[-- c[x[i]]] = i ;
	for ( d = 1 , p = 0 ; p < n ; d <<= 1 , m = p ) {
		for ( p = 0 , i = n - d ; i < n ; ++ i ) y[p ++] = i ;
		for ( i = 0 ; i < n ; ++ i ) if ( sa[i] >= d ) y[p ++] = sa[i] - d ;
		for ( i = 0 ; i < m ; ++ i ) c[i] = 0 ;
		for ( i = 0 ; i < n ; ++ i ) ++ c[xy[i] = x[y[i]]] ;
		for ( i = 1 ; i < m ; ++ i ) c[i] += c[i - 1] ;
		for ( i = n - 1 ; i >= 0 ; -- i ) sa[-- c[xy[i]]] = y[i] ;
		for ( t = x , x = y , y = t , p = 1 , x[sa[0]] = 0 , i = 1 ; i < n ; ++ i ) {
			x[sa[i]] = cmp ( y , sa[i - 1] , sa[i] , d ) ? p - 1 : p ++ ;
		}
	}
	getHeight ( n - 1 ) ;
}

int search ( int x , int l = 0 , int r = n ) {
	while ( l < r ) {
		int mid = ( l + r ) >> 1 ;
		if ( start[mid] >= x ) r = mid ;
		else l = mid + 1 ;
	}
	return start[l] == x ? l : l - 1 ;
}

int check ( int k , int cnt = 0 ) {
	++ Time ;
	For ( i , 2 , m ) {
		if ( height[i] < k ) {
			if ( cnt > n / 2 ) return 1 ;
			++ Time ;
			cnt = 0 ;
		} else {
			int x = search ( sa[i - 1] ) ;
			int y = search ( sa[i] ) ;
			if ( vis[x] != Time ) {
				++ cnt ;
				vis[x] = Time ;
			}
			if ( vis[y] != Time ) {
				++ cnt ;
				vis[y] = Time ;
			}
		}
	}
	if ( cnt >= ( n + 1 ) / 2 ) return 1 ;
	return 0 ;
}

void print ( int k , int cnt = 0 ) {
	++ Time ;
	For ( i , 2 , m ) {
		if ( height[i] < k ) {
			if ( cnt > n / 2 ) {
				int L = sa[i - 1] ;
				int R = sa[i - 1] + k ;
				rep ( j , L , R ) printf ( "%c" , s[j] - 100 ) ;
				printf ( "\n" ) ;
			}
			++ Time ;
			cnt = 0 ;
		} else {
			int x = search ( sa[i - 1] ) ;
			int y = search ( sa[i] ) ;
			if ( vis[x] != Time ) {
				++ cnt ;
				vis[x] = Time ;
			}
			if ( vis[y] != Time ) {
				++ cnt ;
				vis[y] = Time ;
			}
		}
	}
}

void solve () {
	m = 0 ;
	start[n] = MAXN ;
	rep ( i , 0 , n ) {
		start[i] = m ;
		scanf ( "%s" , buf ) ;
		int len = strlen ( buf ) ;
		rep ( j , 0 , len ) s[m + j] = buf[j] + 100 ;
		s[m + len] = i ;//*********divide sign
		m += len + 1 ;
	}
	s[m - 1] = 0 ;
	da ( m ) ;
	int l = 0 , r = 1000 ;
	while ( l < r ) {
		int mid = ( l + r + 1 ) >> 1 ;
		if ( check ( mid ) ) l = mid ;
		else r = mid - 1 ;
	}
	if ( !l ) printf ( "?\n" ) ;
	else print ( l ) ;
}

int main () {
	int flag = 0 ;
	Time = 0 ;
	clr ( vis , 0 ) ;
	while ( ~scanf ( "%d" , &n ) && n ) {
		if ( flag ) printf ( "\n" ) ;
		flag = 1 ;
		solve () ;
	}
	return 0 ;
}