首页 > 其他分享> > splay/fhq-treap 问卷调查反馈—— [JSOI2008]火星人prefix（splay），Strange Queries（fhq-treap）

splay/fhq-treap 问卷调查反馈—— [JSOI2008]火星人prefix（splay），Strange Queries（fhq-treap）

2021-09-07 20:02:02 作者：互联网

文章目录

[JSOI2008]火星人prefix
Strange Queries

[JSOI2008]火星人prefix

BZOJ1014
思路很好想，哈希字符串即可

只是平衡树的码量大

注意因为splay加入哨兵的原因，每个点在平衡树内的排名比真实排名大 1 1 1（有极小值的占位）

~~考虑的时候就只在询问的时候考虑，修改的时候忘了就挂了~~

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define maxn 100005
#define ull unsigned long long
struct node {
	int son[2], f, val, siz;
	ull hash;
}t[maxn];
int root, n, Q;
ull mi[maxn];
int val[maxn];
char s[maxn];

void read( int &x ) {
	x = 0; char c = getchar();
	while( c < '0' or c > '9' ) c = getchar();
	while( '0' <= c and c <= '9' ) { x = ( x << 1 ) + ( x << 3 ) + ( c ^ 48 ); c = getchar(); }
}

void pushup( int x ) {
	if( ! x ) return;
	t[x].siz = 1;
	if( t[x].son[0] ) t[x].siz += t[t[x].son[0]].siz;
	if( t[x].son[1] ) t[x].siz += t[t[x].son[1]].siz;
	t[x].hash = mi[t[t[x].son[0]].siz] * t[x].val + t[t[x].son[0]].hash + t[t[x].son[1]].hash * mi[t[t[x].son[0]].siz + 1];
}

void rotate( int x ) {
	int fa = t[x].f;
	int Gfa = t[fa].f;
	int k = t[fa].son[1] == x;
	t[Gfa].son[t[Gfa].son[1] == fa] = x;
	t[x].f = Gfa;
	t[fa].son[k] = t[x].son[k ^ 1];
	if( t[x].son[k ^ 1] ) t[t[x].son[k ^ 1]].f = fa;
	t[x].son[k ^ 1] = fa;
	t[fa].f = x;
	pushup( fa );
	pushup( x );
}

void splay( int x, int goal ) {
	while( t[x].f ^ goal ) {
		int fa = t[x].f, Gfa = t[fa].f;
		if( Gfa ^ goal )	
			( t[Gfa].son[0] == fa ) ^ ( t[fa].son[0] == x ) ? rotate( x ) : rotate( fa );
		rotate( x );
	}
	if( ! goal ) root = x;
}

int build( int now, int l, int r ) {
	if( l > r ) return 0;
	int mid = ( l + r ) >> 1;
	t[mid].hash = val[mid];
	t[mid].val = val[mid];
	t[mid].siz = 1;
	t[mid].f = now;
	if( l == r ) return l;
	t[mid].son[0] = build( mid, l, mid - 1 );
	t[mid].son[1] = build( mid, mid + 1, r );
	pushup( mid );
	return mid;
}

int find( int x ) {
	int now = root;
	while( 1 ) {
		if( t[t[now].son[0]].siz >= x ) now = t[now].son[0];
		else if( t[t[now].son[0]].siz + 1 == x ) return now;
		else x -= t[t[now].son[0]].siz + 1, now = t[now].son[1];
	}
}

int Hash( int x, int y ) {
	int l = find( x - 1 );
	int r = find( y + 1 );
	splay( l, 0 );
	splay( r, l );
	return t[t[r].son[0]].hash;
}

int main() {
	mi[0] = 1;
	for( int i = 1;i < maxn;i ++ ) mi[i] = mi[i - 1] * 27ull;
	scanf( "%s", s ); read( Q ); 
	val[++ n] = 0;
	int len = strlen( s );
	for( int i = 0;i < len;i ++ )
		val[++ n] = s[i] - 'a' + 1;
	val[++ n] = 0;
	root = build( root, 1, n );
	char opt[5], ch[5]; int x, y;
	while( Q -- ) {
		scanf( "%s", opt ); read( x ); x ++;
		switch( opt[0] ) {
			case 'Q' : {
				read( y ); y ++;
				if( x > y ) swap( x, y );
				int ans = 0, l = 1, r = n - y;
				while( l <= r ) {
					int mid = ( l + r ) >> 1;
					if( Hash( x, x + mid - 1 ) == Hash( y, y + mid - 1 ) )
						ans = mid, l = mid + 1;
					else
						r = mid - 1;
				}
				printf( "%d\n", ans );
				break;
			}
			case 'R' : {
				scanf( "%s", ch );
				splay( find( x ), 0 );
				t[root].val = ch[0] - 'a' + 1;
				pushup( root );
				break;
			}
			case 'I' : {
				scanf( "%s", ch );
				int l = find( x );
				int r = find( x + 1 );
				splay( l, 0 );
				splay( r, l );
				t[r].son[0] = ++ n;
				t[n].hash = ch[0] - 'a' + 1;
				t[n].val = ch[0] - 'a' + 1;
				t[n].siz = 1;
				t[n].f = r;
				splay( n, 0 );
				break;
			}
		}
	}
	return 0;
}

Strange Queries

CODECHEF

显然每个点只会与其左右相邻点连边

这种区间内合并的问题，是非常常见的类似线段树维护区间最大子段和的感觉

但是这是动态维护，所以用平衡树就行

具体而言，对于每一段区间

维护区间只左端点还没有连线的最小值l_
只右端点还没有连线的最小值_r
只左右端点都没有连线的最小值l__r
区间每个点都至少有一条连线的最小值ans
维护区间左右端点的权值val_l val_r

考虑两个区间的合并

ans
- 两个区间的答案相加
- 左区间的右端点和右区间的左端点都没连线，此时在两点之间连线
  
  花费为右区间左端点的权值与左区间右端点的权值差
l_
- 左区间仍只有左端点没连线l_，右区间均连线ans
- 左区间两端都没连线l__r，右区间左端点没连线l_，同样两区间可以连线
_r
- 右区间只有右端点没连线_r，左区间均连线ans
- 右区间两端都没连线l__r，左区间右端点没连线_r，两区间连线
l__r
- 左区间只有左端点没连线l_，右区间只有右端点没连线_r
- 左区间和右区间两端都没连线l__r，此时左区间右端点和右区间左端点连线

用fhq-treap维护即可

#include <cstdio>
#include <algorithm>
using namespace std;
#define int long long
#define maxn 200005
#define inf 1e9
struct node {
	int key, lson, rson, val, val_l, val_r, l_, _r, l__r, ans;
	node(){}
	node( int v ) {
		ans = inf;
		key = rand();
		val = val_l = val_r = v;
		lson = rson = l_ = _r = l__r = 0;
	}
}t[maxn];
int T, n, Q, root, cnt;
int a[maxn];

void pushup( node lst, node nxt, int &ans, int &l_, int &_r, int &l__r ) {
	if( lst.val_l <= -inf and lst.val_r >= inf ) {
		ans = nxt.ans;
		l_ = nxt.l_;
		_r = nxt._r;
		l__r = nxt.l__r;
		return;
	}
	if( nxt.val_l <= -inf and nxt.val_r >= inf ) {
		ans = lst.ans;
		l_ = lst.l_;
		_r = lst._r;
		l__r = lst.l__r;
		return;
	}
	int w = nxt.val_l - lst.val_r;
	ans = min( lst.ans + nxt.ans, lst._r + nxt.l_ + w );
	l_ = min( lst.l_ + nxt.ans, lst.l__r + nxt.l_ + w );
	_r = min( nxt._r + lst.ans, nxt.l__r + lst._r + w );
	l__r = min( lst.l_ + nxt._r, lst.l__r + nxt.l__r + w );
}

void pushup( int x ) {
	t[x].val_l = t[x].val_r = t[x].val;
	t[x].l_ = t[x]._r = t[x].l__r = 0;
	t[x].ans = inf;
	if( t[x].lson ) {
		pushup( t[t[x].lson], t[x], t[x].ans, t[x].l_, t[x]._r, t[x].l__r );
		t[x].val_l = t[t[x].lson].val_l;
	}
	if( t[x].rson ) {
		pushup( t[x], t[t[x].rson], t[x].ans, t[x].l_, t[x]._r, t[x].l__r );
		t[x].val_r = t[t[x].rson].val_r;
	}
}

void split( int now, int val, int &x, int &y ) {
	if( ! now ) x = y = 0;
	else {
		if( t[now].val <= val ) {
			x = now;
			split( t[now].rson, val, t[now].rson, y );
			pushup( x );
		}
		else {
			y = now;
			split( t[now].lson, val, x, t[now].lson );
			pushup( y );
		}
	}
}

int merge( int x, int y ) {
	if( ! x or ! y ) return x + y;
	if( t[x].key < t[y].key ) {
		t[x].rson = merge( t[x].rson, y );
		pushup( x );
		return x;
	}
	else {
		t[y].lson = merge( x, t[y].lson );
		pushup( y );
		return y;
	}
}

signed main() {
	t[0].val_l = -inf, t[0].val_r = inf;
	scanf( "%lld", &T );
	while( T -- ) {
		scanf( "%lld %lld", &n, &Q );
		root = 0;
		for( int i = 1;i <= n;i ++ ) 
			scanf( "%lld", &a[i] );
		sort( a + 1, a + n + 1 );
		for( int i = 1;i <= n;i ++ ) {
			t[i] = node( a[i] );
			root = merge( root, i );
		}
		cnt = n;
		int opt, x, l, r, L, R;
		while( Q -- ) {
			scanf( "%lld %lld", &opt, &x );
			if( opt & 1 ) {
				split( root, x, l, r );
				t[++ cnt] = node( x );
				root = merge( l, merge( cnt, r ) );
			}
			else {
				split( root, x, l, r );
				split( l, x - 1, L, R );
				root = merge( L, r );
			}
			printf( "%lld\n", t[root].ans >= inf ? 0 : t[root].ans );
		}
	}
	return 0;
}

标签：__,val,int,mid,son,splay,treap,ans,fhq
来源： https://blog.csdn.net/Emm_Titan/article/details/120165942