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  • zoj 3765 块状链表 OR splay

    各种操作o(╯□╰)o...不过都挺简单,不需要lazy标记。

    方法1:块状链表

    块状链表太强大了,区间操作实现起来简单暴力,效率比splay稍微慢一点,内存开销小很多。

      1 #include <iostream>
      2 #include <cstring>
      3 #include <cstdio>
      4 using namespace std;
      5 
      6 const int N = 900;
      7 const int LOW = 260;
      8 const int UP = 320;
      9 int sz[N];
     10 int next[N];
     11 int g[N][2];
     12 int n, m, ps, bl;
     13 
     14 struct B
     15 {
     16     int v, s;
     17 } b[N][N];
     18 
     19 int gcd( int x, int y )
     20 {
     21     if ( y ) return gcd( y, x % y );
     22     return x;
     23 }
     24 
     25 void update( int cur, int stu )
     26 {
     27     g[cur][stu] = 0;
     28     for ( int i = 0; i < sz[cur]; i++ )
     29     {
     30         if ( b[cur][i].s == stu )
     31         {
     32             g[cur][stu] = gcd( g[cur][stu], b[cur][i].v );
     33         }
     34     }
     35 }
     36 
     37 void spilt( int cur )
     38 {
     39     int tmp = next[cur];
     40     int ncur = bl++;
     41     next[cur] = ncur;
     42     next[ncur] = tmp;
     43     for ( int i = sz[cur] / 2; i < sz[cur]; i++ )
     44     {
     45         b[ncur][sz[ncur]++] = b[cur][i];
     46     }
     47     sz[cur] = sz[cur] / 2;
     48     update( cur, 0 );
     49     update( cur, 1 );
     50     update( ncur, 0 );
     51     update( ncur, 1 );
     52 }
     53 
     54 void insert( int pos, int val, int stu )
     55 {
     56     int cur = 0, p = sz[cur];
     57     while ( p < pos + 1 && next[cur] != -1 )
     58     {
     59         cur = next[cur];
     60         p += sz[cur];
     61     }
     62     if ( p < pos + 1 )
     63     {
     64         pos = p;
     65     }
     66     p -= sz[cur];
     67     pos -= p;
     68     for ( int j = sz[cur] - 1; j >= pos; j-- )
     69     {
     70         b[cur][j + 1] = b[cur][j];
     71     }
     72     b[cur][pos].v = val;
     73     b[cur][pos].s = stu;
     74     sz[cur]++;
     75     if ( sz[cur] > UP )
     76     {
     77         spilt(cur);
     78     }
     79     else
     80     {
     81         g[cur][stu] = gcd( g[cur][stu], val );
     82     }
     83 }
     84 
     85 void remove( int pos )
     86 {
     87     int cur = 0, p = sz[cur];
     88     while ( p < pos + 1 && next[cur] != -1 )
     89     {
     90         cur = next[cur];
     91         p += sz[cur];
     92     }   
     93     if ( p < pos + 1 )
     94     {
     95         return ;
     96     }
     97     else
     98     {
     99         p -= sz[cur];
    100         pos -= p;
    101         int tt = b[cur][pos].s;
    102         for ( int j = pos; j < sz[cur] - 1; j++ )
    103         {
    104             b[cur][j] = b[cur][j + 1];
    105         }
    106         sz[cur]--;
    107         update( cur, tt );
    108     }
    109 }
    110 
    111 void reverse( int pos )
    112 {
    113     int cur = 0, p = sz[cur];
    114     while ( p < pos + 1 && next[cur] != -1 )
    115     {
    116         cur = next[cur];
    117         p += sz[cur];
    118     }   
    119     if ( p < pos + 1 )
    120     {
    121         return ;
    122     }
    123     else
    124     {
    125         p -= sz[cur];
    126         pos -= p;
    127         int tt = b[cur][pos].s;
    128         b[cur][pos].s ^= 1;
    129         update( cur, tt );
    130         g[cur][tt ^ 1] = gcd( g[cur][tt ^ 1], b[cur][pos].v );
    131     }
    132 }
    133 
    134 void modify( int pos, int val )
    135 {
    136     int cur = 0, p = sz[cur];
    137     while ( p < pos + 1 && next[cur] != -1 )
    138     {
    139         cur = next[cur];
    140         p += sz[cur];
    141     }   
    142     if ( p < pos + 1 )
    143     {
    144         return ;
    145     }
    146     else
    147     {
    148         p -= sz[cur];
    149         pos -= p;
    150         b[cur][pos].v = val;
    151         update( cur, b[cur][pos].s );
    152     }
    153 }
    154 
    155 int query( int l, int r, int stu )
    156 {
    157     int cur = 0, p = sz[cur];
    158     while ( p < l + 1 && next[cur] != -1 )
    159     {
    160         cur = next[cur];
    161         p += sz[cur];
    162     }        
    163     p -= sz[cur];
    164     l -= p;  
    165     int ncur = 0, q = sz[ncur];
    166     while ( q < r + 1 && next[ncur] != -1 )
    167     {
    168         ncur = next[ncur];
    169         q += sz[ncur];
    170     }        
    171     q -= sz[ncur];
    172     r -= q;  
    173     int ans = 0;
    174     if ( cur != ncur )
    175     {
    176         for ( int i = next[cur]; i != ncur; i = next[i] )
    177         {
    178             if ( g[i][stu] == 0 ) continue;
    179             ans = gcd( ans, g[i][stu] );
    180         }
    181         for ( int j = l; j < sz[cur]; j++ )
    182         {
    183             if ( b[cur][j].s == stu )
    184             {
    185                 ans = gcd( ans, b[cur][j].v ); 
    186             }
    187         }
    188         for ( int j = 0; j <= r; j++ )
    189         {
    190             if ( b[ncur][j].s == stu )
    191             {
    192                 ans = gcd( ans, b[ncur][j].v );
    193             }
    194         }
    195     }
    196     else
    197     {
    198         for ( int j = l; j <= r; j++ )
    199         {
    200             if ( b[cur][j].s == stu )
    201             {
    202                 ans = gcd( ans, b[cur][j].v );
    203             }
    204         }
    205     }
    206     if ( ans == 0 ) ans = -1;
    207     return ans;
    208 }
    209 
    210 int main ()
    211 {
    212     while ( scanf("%d%d", &n, &m) != EOF )
    213     {
    214         ps = LOW;
    215         memset( sz, 0, sizeof(sz) );
    216         memset( next, -1, sizeof(next) );
    217         memset( g, 0, sizeof(g) );
    218         for ( int i = 0; i < n; i++ )
    219         {
    220             int la = i / ps, lb = i % ps;
    221             scanf("%d%d", &b[la][lb].v, &b[la][lb].s);
    222         }
    223         int k = 0;
    224         while ( ( k + 1 ) * ps < n ) k++;
    225         for ( int i = 0; i < k; i++ )
    226         {
    227             sz[i] = ps;
    228             next[i] = i + 1;
    229         }
    230         sz[k] = n - k * ps;
    231         for ( int i = 0; i <= k; i++ )
    232         {
    233             g[i][0] = g[i][1] = 0;
    234             for ( int j = 0; j < sz[i]; j++ )
    235             {
    236                 int ss = b[i][j].s;
    237                 g[i][ss] = gcd( g[i][ss], b[i][j].v );
    238             }
    239         }
    240         bl = k + 1;
    241         char op[2];
    242         int num1, num2, num3;
    243         while ( m-- )
    244         {
    245             scanf("%s", op);
    246             if ( op[0] == 'Q' )
    247             {
    248                 scanf("%d%d%d", &num1, &num2, &num3);
    249                 num1--;
    250                 num2--;
    251                 printf("%d
    ", query( num1, num2, num3 ));
    252             }
    253             else if ( op[0] == 'I' )
    254             {
    255                 scanf("%d%d%d", &num1, &num2, &num3);
    256                 insert( num1, num2, num3 );
    257             }
    258             else if ( op[0] == 'D' )
    259             {
    260                 scanf("%d", &num1);
    261                 num1--;
    262                 remove(num1);
    263             }
    264             else if ( op[0] == 'R' )
    265             {
    266                 scanf("%d", &num1);
    267                 num1--;
    268                 reverse(num1);      
    269             }
    270             else if ( op[0] == 'M' )
    271             {
    272                 scanf("%d%d", &num1, &num2);
    273                 num1--;
    274                 modify( num1, num2 );
    275             }
    276         }
    277     }
    278     return 0;
    279 }

    方法2:splay

    写起来也不算特别长。

      1 #include <iostream>
      2 #include <cstring>
      3 #include <cstdio>
      4 using namespace std;
      5 
      6 const int N = 200002;
      7 int v[N];
      8 int s[N];
      9 int n, m;
     10 
     11 int gcd( int x, int y )
     12 {
     13     if ( y ) return gcd( y, x % y );
     14     return x;
     15 }
     16 
     17 struct Node 
     18 {
     19     Node * ch[2];
     20     int g[2];
     21     int val;
     22     int stu;
     23     int rank;
     24     int sz;
     25 
     26     int cmp( int x ) const 
     27     {
     28         if ( x == rank ) return -1;
     29         return x < rank ? 0 : 1;
     30     }
     31 
     32     void maintain()
     33     {
     34         sz = rank = 1;
     35         g[0] = g[1] = 0;
     36         g[stu] = val;
     37         if ( ch[0] != NULL )
     38         {
     39             sz += ch[0]->sz;
     40             rank += ch[0]->sz;
     41             if ( ch[0]->g[0] ) g[0] = gcd( g[0], ch[0]->g[0] );
     42             if ( ch[0]->g[1] ) g[1] = gcd( g[1], ch[0]->g[1] );
     43         }
     44         if ( ch[1] != NULL )
     45         {
     46             sz += ch[1]->sz;
     47             if ( ch[1]->g[0] ) g[0] = gcd( g[0], ch[1]->g[0] );
     48             if ( ch[1]->g[1] ) g[1] = gcd( g[1], ch[1]->g[1] );
     49         }
     50     }
     51 } * root;
     52 
     53 void rotate( Node * & o, int d )
     54 {
     55     Node * k = o->ch[d ^ 1];
     56     o->ch[d ^ 1] = k->ch[d];
     57     k->ch[d] = o;
     58     o->maintain();
     59     k->maintain();
     60     o = k;
     61 }
     62 
     63 void splay( Node * & o, int k )
     64 {
     65     int d = o->cmp(k);
     66     if ( d != -1 )
     67     {
     68         if ( d == 1 ) k -= o->rank;
     69         Node * p = o->ch[d];
     70         int d2 = p->cmp(k);
     71         if ( d2 != -1 )
     72         {
     73             int k2 = ( d2 == 0 ? k : k - p->rank );    
     74             splay( p->ch[d2], k2 );
     75             if ( d == d2 )
     76             {
     77                 rotate( o, d ^ 1 );
     78             }
     79             else
     80             {
     81                 rotate( o->ch[d], d );
     82             }
     83         }
     84         rotate( o, d ^ 1 );
     85     }
     86 }
     87 
     88 Node * build( int l, int r )
     89 {
     90     if ( l > r ) return NULL;
     91     Node * o = new Node();
     92     int mid = ( l + r ) >> 1;
     93     o->sz = r - l + 1;
     94     o->val = v[mid];
     95     o->stu = s[mid];
     96     o->rank = mid - l + 1;
     97     o->g[0] = o->g[1] = 0;
     98     o->g[o->stu] = o->val;
     99     o->ch[0] = build( l, mid - 1 );
    100     o->ch[1] = build( mid + 1, r );
    101     o->maintain();
    102     return o;
    103 }
    104 
    105 int query( int l, int r, int stu )
    106 {
    107     splay( root, l );
    108     splay( root->ch[1], r - l + 2 );
    109     return root->ch[1]->ch[0]->g[stu];
    110 }
    111 
    112 void insert( int k, int vv, int ss )
    113 {
    114     splay( root, k );
    115     splay( root->ch[1], 1 );
    116     Node * & o = root->ch[1]->ch[0];
    117     o = new Node();
    118     o->sz = o->rank = 1;
    119     o->g[0] = o->g[1] = 0;
    120     o->val = vv;
    121     o->stu = ss;
    122     o->g[ss] = vv;
    123     root->ch[1]->maintain();
    124     root->maintain();
    125 }
    126 
    127 void remove( int k )
    128 {
    129     splay( root, k );
    130     splay( root->ch[1], 2 );
    131     root->ch[1]->ch[0] = NULL;
    132     root->ch[1]->maintain();
    133     root->maintain();
    134 }
    135 
    136 void reverse( int k )
    137 {
    138     splay( root, k );
    139     root->stu ^= 1;
    140     root->maintain();
    141 }
    142 
    143 void modify( int k, int vv )
    144 {
    145     splay( root, k );
    146     root->val = vv;
    147     root->maintain();
    148 }
    149 
    150 int main ()
    151 {
    152     while ( scanf("%d%d", &n, &m) != EOF )
    153     {
    154         for ( int i = 1; i <= n; i++ )
    155         {
    156             scanf("%d%d", v + i, s + i);
    157         }
    158         root = build( 0, n + 1 );
    159         char cmd[11];
    160         int a, b, c;
    161         while ( m-- )
    162         {
    163             scanf("%s", cmd);
    164             if ( cmd[0] == 'Q' )
    165             {
    166                 scanf("%d%d%d", &a, &b, &c);
    167                 int tmp = query( a, b, c );
    168                 if ( !tmp ) tmp = -1;
    169                 printf("%d
    ", tmp);
    170             }
    171             else if ( cmd[0] == 'I' )
    172             {
    173                 scanf("%d%d%d", &a, &b, &c);
    174                 a++;
    175                 insert( a, b, c );
    176             }
    177             else if ( cmd[0] == 'D' )
    178             {
    179                 scanf("%d", &a);
    180                 remove(a);
    181             }
    182             else if ( cmd[0] == 'R' )
    183             {
    184                 scanf("%d", &a);
    185                 a++;
    186                 reverse(a);
    187             }
    188             else if ( cmd[0] == 'M' )
    189             {
    190                 scanf("%d%d", &a, &b);
    191                 a++;
    192                 modify( a, b );
    193             }
    194         }
    195     }
    196     return 0;
    197 }
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  • 原文地址:https://www.cnblogs.com/huoxiayu/p/4736490.html
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