hdu 5349 维护序列 OR Treap

由于只删除最小值和只查询最大值，所以我们只要维护一个maxn和一个size即可，需要注意的是删除到集合空时需要重新将maxn赋值为无穷小。

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 using namespace std;
 5
 6 int main ()
 7 {
 8     int n;
 9     while ( scanf("%d", &n) != EOF )
10     {
11         int op, tmp, maxn = -1111111111, size = 0;
12         for ( int i = 0; i < n; i++ )
13         {
14             scanf("%d", &op);
15             if ( op == 1 )
16             {
17                 scanf("%d", &tmp);
18                 size++;
19                 if ( tmp > maxn ) maxn = tmp;
20             }
21             else if ( op == 2 )
22             {
23                 if ( size > 0 )
24                 {
25                     size--;
26                     if ( size == 0 ) maxn = -1111111111;
27                 }
28             }
29             else
30             {
31                 if ( size == 0 )
32                 {
33                     printf("0\n");
34                 }
35                 else
36                 {
37                     printf("%d\n", maxn);
38                 }
39             }
40         }
41     }
42     return 0;
43 }

另外，别的平衡的数据结构也可以过，这里用的是Treap（主要是测模板），比赛推荐用multiset。

  1 #include <iostream>
  2 #include <cstring>
  3 #include <cstdlib>
  4 #include <cstdio>
  5 using namespace std;
  6
  7 struct Node
  8 {
  9     Node * ch[2];
 10     int v, r, size;
 11     int cmp( int x )
 12     {
 13         if ( x == v ) return -1;
 14         return x < v ? 0 : 1;
 15     }
 16     void maintain()
 17     {
 18         size = 1;
 19         if ( ch[0] != NULL ) size += ch[0]->size;
 20         if ( ch[1] != NULL ) size += ch[1]->size;
 21     }
 22 };
 23
 24 Node * root;
 25
 26 void rotate( Node * & o, int d )
 27 {
 28     Node * k = o->ch[d ^ 1];
 29     o->ch[d ^ 1] = k->ch[d];
 30     k->ch[d] = o;
 31     o->maintain();
 32     k->maintain();
 33     o = k;
 34 }
 35
 36 void insert( Node * & o, int x )
 37 {
 38     if ( o == NULL )
 39     {
 40         o = new Node();
 41         o->ch[0] = o->ch[1] = NULL;
 42         o->v = x;
 43         o->r = rand();
 44         o->size = 1;
 45     }
 46     else
 47     {
 48         int d = ( x <= o->v ? 0 : 1 );
 49         insert( o->ch[d], x );
 50         if ( o->ch[d]->r > o->r )
 51         {
 52             rotate( o, d ^ 1 );
 53         }
 54         else
 55         {
 56             o->maintain();
 57         }
 58     }
 59 }
 60
 61 void remove( Node * & o, int x )
 62 {
 63     int d = o->cmp(x);
 64     if ( d == -1 )
 65     {
 66         if ( o->ch[0] != NULL && o->ch[1] != NULL )
 67         {
 68             int dd = ( o->ch[0]->r > o->ch[1]->r ? 1 : 0 );
 69             rotate( o, dd );
 70             remove( o->ch[dd], x );
 71         }
 72         else
 73         {
 74             Node * u = o;
 75             if ( o->ch[0] == NULL ) o = o->ch[1];
 76             else o = o->ch[0];
 77             delete u;
 78         }
 79     }
 80     else
 81     {
 82         remove( o->ch[d], x );
 83     }
 84     if ( o != NULL ) o->maintain();
 85 }
 86
 87 int findm( Node * o, int d )
 88 {
 89     while ( o->ch[d] != NULL ) o = o->ch[d];
 90     return o->v;
 91 }
 92
 93 int main ()
 94 {
 95     int n;
 96     while ( scanf("%d", &n) != EOF )
 97     {
 98         root = NULL;
 99         for ( int i = 0; i < n; i++ )
100         {
101             int op, x;
102             scanf("%d", &op);
103             if ( op == 1 )
104             {
105                 scanf("%d", &x);
106                 insert( root, x );
107             }
108             else if ( op == 2 )
109             {
110                 if ( root != NULL )
111                 {
112                     x = findm( root, 0 );
113                     remove( root, x );
114                 }
115             }
116             else
117             {
118                 if ( root == NULL )
119                 {
120                     x = 0;
121                 }
122                 else
123                 {
124                     x = findm( root, 1 );
125                 }
126                 printf("%d\n", x);
127             }
128         }
129     }
130     return 0;
131 }