HDU
.php?pid=4125
题意：
给你N个数字，先按照给数的顺序建一棵二叉查找树，然后按中序遍历树的结点，并记录访问结点的奇偶顺序，这样就会得到一个序列，再给你一个0,1序列，现在要你求给出的序列在遍历出来的序列中出现的次数，序列允许相互覆盖。
思路：
暴力建树如果二叉树退化成一个树链的话，时间复杂度就会变成O(N^2)，可能会超时。这样我们就要求找一种logn的建树策略，很容易就会想到线段树，对于每个元素x，我们找到在它前面的比x小的最大值y和比x大的最小值z，可以证明的是x要么插在y的右边，要么插在z的左边。 做一我们可以用线段树在O( n*logn )的时间内将树建好，然后就是dfs求遍历序列， 最后kmp就ok 了。
代码：
#include <stdio.h>
#include <string.h>
#include <stack>
using namespace std;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
const int inf = (1 << 30) ;
const int NN = 600010 ;
int N ;
int lft[NN<<2] ,rgh[NN<<2] ;void build(int l ,int r, int idx){lft[idx] = rgh[idx] = -1 ;if(l == r) return ;int mid = (l + r) >> 1 , ls = idx<<1 , rs = idx<<1|1 ;build(l , mid ,ls) ;build(mid+1 , r , rs) ;
}int left[NN] , right[NN] ;
char seq[NN * 3] , str[7010];
int cnt , len1 , len2 ;
int next[7010] ;struct Node{int now , f;Node(){}Node( int _now , int _f):now( _now ) , f( _f ){}
} ;
stack< Node > ss ;void dfs(int now){
aaa :seq[cnt++] = ( now&1 ) + '0' ;int l = left[now] ;if(l != -1 ){ss.push( Node(now , 0) ) ;now = l ;goto aaa ;
bbb1 :seq[ cnt++ ] = ( now&1 ) + '0' ;}int r = right[now] ;if( r!=-1 ){ss.push( Node(now , 1) ) ;now = r ;goto aaa ;
bbb2 :seq[ cnt++ ] = ( now&1 ) + '0' ;}if( ss.empty() ) return ;now = ss.top().now ;int ff = ss.top().f ; ss.pop() ;if( ff ) goto bbb2 ;else goto bbb1 ;
}void get_next(){len1 = strlen( str ) ;int i , j ;next[0] = -1 ;i = 1 ; j = -1 ;for( ; i<len1 ;i ++ ){while( j!=-1 && str[i]!=str[j+1] ) j = next[j] ;if( str[i] == str[j+1] ) j++ ;next[i] = j ;}
}int kmp(){get_next() ;len2 = strlen( seq ) ;int i, j ; i = 0 ; j = -1 ;int cc = 0 ;for( ;i<len2 ;i++ ){while( j!=-1 && str[j+1]!=seq[i] ) j = next[j] ;if( seq[i] == str[j+1]) j++ ;if( j==len1-1 ){cc ++ ;j = next[j] ;}}return cc ;
}inline int MIN(int a, int b){ return a < b ? a : b ;}
inline int MAX(int a, int b){ return a > b ? a : b ;}void up(int l , int r, int idx){int ls = idx<<1 ,rs = idx<<1|1 ;if( lft[ls] == -1 ) lft[idx] = lft[rs] ;else lft[idx] = lft[ls] ;if( rgh[rs] == -1 ) rgh[idx] = rgh[ls] ;else rgh[idx] = rgh[rs] ;
}
void update(int l , int r, int idx, int pos ){if(l == r){lft[idx] = rgh[idx] = pos ;return ;}int mid = (l + r) >> 1 , ls = idx<<1 ,rs = idx<<1|1 ;if( pos<=mid ) update(l , mid , ls , pos) ;else update(mid+1 , r , rs , pos) ;up(l , r , idx);
}int ask_small(int l , int r, int idx , int a, int b){if( rgh[idx] == -1 ) return -1;if(l==a && r==b) return rgh[idx] ;int mid = (l + r) >> 1 , ls = idx<<1 , rs = idx<<1|1 ;if( b<=mid ) return ask_small(l , mid , ls , a , b) ;else if( mid<a ) return ask_small( mid+1 ,r ,rs , a ,b) ;else{int aa = ask_small( mid+1 , r , rs , mid+1 , b) ;if( aa != -1 ) return aa ;aa = ask_small( l , mid , ls , a , mid ) ;return aa ;}
}int ask_big(int l , int r, int idx, int a, int b){if( lft[idx] == -1 ) return -1 ;if(l==a && r==b) return lft[idx] ;int mid = (l + r) >> 1 , ls = idx<<1 , rs = idx<<1|1 ;if( b<=mid ) return ask_big(l , mid , ls , a , b) ;else if( mid<a ) return ask_big( mid+1 ,r ,rs , a ,b) ;else{int aa = ask_big( l , mid , ls , a , mid ) ;if( aa != -1 ) return aa ;aa = ask_big( mid+1 , r , rs , mid+1 , b ) ;return aa ;}
}int main(){int T , a , b; scanf("%d",&T) ;int s_pos , m_pos , cas = 0 ;while( T-- ){scanf("%d",&N) ;build(1 , N ,1);for(int i=1;i<=N;i++) left[i] = right[i] = -1 ;for(int i=1;i<=N;i++){scanf("%d",&a);if( i == 1 ) b = a ;if( a == 1 ) s_pos = -1 ;else s_pos = ask_small( 1 , N , 1 , 1 , a-1 ) ;if( a == N ) m_pos = -1 ;else m_pos = ask_big( 1 , N , 1 , a+1 , N ) ;if( s_pos == -1 && m_pos == -1 ){;}else if( s_pos==-1 && m_pos!=-1 ){left[ m_pos ] = a ;}else if( s_pos!=-1 && m_pos==-1 ){right[ s_pos ] = a ;}else{if( right[ s_pos ] == -1 ) right[ s_pos ] = a ;else left[ m_pos ] = a ;}update(1 , N , 1 , a ) ;}cnt = 0 ;while( !ss.empty() ) ss.pop() ;dfs( b ) ;seq[cnt] = 0 ;scanf("%s",str);printf("Case #%d: %d\n",++cas,kmp() );}return 0 ;
}