文章目录
[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
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_lval_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
- 左区间仍只有左端点没连线
-
_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
- 右区间只有右端点没连线
-
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
- 左区间只有左端点没连线
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;
}