BZOJ3173 [Tjoi2013]最长上升子序列

可以称为，模拟题、、、

我们发现，由于是从小到大插入的，所以后插入的数不会影响先插入的数的ans

于是只要对最后的序列求一次LIS即可。

问题就集中在如何求最后的序列：

方法一：treap无脑模拟插入操作

就当是treap的练手吧。。。结果RE了一版，后来突然一拍脑袋发现。。bz上不让调用time()函数。。。各种蛋疼

 

  1 /**************************************************************
  2     Problem: 3173
  3     User: rausen
  4     Language: C++
  5     Result: Accepted
  6     Time:660 ms
  7     Memory:3936 kb
  8 ****************************************************************/
  9  
 10 #include <cstdlib>
 11 #include <cstdio>
 12 #include <cstring>
 13 #include <algorithm>
 14  
 15 using namespace std;
 16 const int N = 100005;
 17 const int inf = 1e9;
 18  
 19 struct treap_node {
 20   treap_node *son[2];
 21   int pri, sz, v;
 22 } *null, *root, mempool[N], *cnt_treap = mempool;
 23  
 24 int n, cnt, len;
 25 int a[N], ans[N], mn[N];
 26  
 27 int read() {
 28   int x = 0;
 29   char ch = getchar();
 30   while (ch < '0' || '9' < ch)
 31     ch = getchar();
 32   while ('0' <= ch && ch <= '9')
 33     (x *= 10) += ch - '0', ch = getchar();
 34   return x;
 35 }
 36  
 37 #define Ls (p -> son[0])
 38 #define Rs (p -> son[1])
 39 #define Pri (p -> pri)
 40 #define Sz (p -> sz)
 41 #define V (p -> v)
 42 inline void treap_update(treap_node *p) {
 43   Sz = Ls -> sz + Rs -> sz + 1;
 44 }
 45  
 46 void treap_rotate(treap_node *&p, int ch) {
 47   treap_node *tmp = p -> son[ch];
 48   p -> son[ch] = tmp -> son[!ch];
 49   tmp -> son[!ch] = p;
 50   treap_update(p), treap_update(tmp);
 51   p = tmp;
 52 }
 53  
 54 void treap_insert(treap_node *&p, int rank) {
 55   if (p == null) {
 56     p = ++cnt_treap, Ls = Rs = null;
 57     Pri = rand(), Sz = 1, V = cnt;
 58     return;
 59   }
 60   ++Sz;
 61   if (Ls -> sz < rank) {
 62     treap_insert(Rs, rank - Ls -> sz - 1);
 63     if (Rs -> pri > Pri) treap_rotate(p, 1);
 64   } else {
 65     treap_insert(Ls, rank);
 66     if (Ls -> pri > Pri) treap_rotate(p, 0);
 67   }
 68 }
 69  
 70 void get_seq(treap_node *p) {
 71   if (p == null) return;
 72   get_seq(Ls);
 73   a[++cnt] = V;
 74   get_seq(Rs);
 75 }
 76 #undef Ls
 77 #undef Rs
 78 #undef Pri
 79 #undef Sz
 80 #undef Num
 81  
 82 int main() {
 83   int i, t;
 84   n = read();
 85   null = ++cnt_treap;
 86   null -> son[0] = null -> son[1] = null;
 87   null -> pri = null -> sz = null -> v = 0;
 88   root = null;
 89   for (cnt = 1; cnt <= n; ++cnt)
 90     treap_insert(root, read());
 91   cnt = 0;
 92   get_seq(root);
 93   memset(mn, 127, sizeof(mn));
 94   for (mn[0] = -inf, i = 1; i <= n; ++i) {
 95     t = upper_bound(mn, mn + len + 1, a[i]) - mn;
 96     if (mn[t - 1] <= a[i]) {
 97       mn[t] = min(mn[t], a[i]);
 98       ans[a[i]] = t;
 99       len = max(t, len);
100     }
101   }
102   for (i = 1; i <= n; ++i)
103     printf("%d\n", ans[i] = max(ans[i], ans[i - 1]));
104   return 0;
105 }
View Code

方法二：倒过来做，从最后开始一个个删，用树状数组维护，求序列的前缀第k大

BIT的这种应用方式也是蛮神的。。。

 

 1 /**************************************************************
 2     Problem: 3173
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:248 ms
 7     Memory:3540 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <cstring>
12 #include <algorithm>
13  
14 #define lowbit(x) (x & -x)
15 using namespace std;
16 const int N = 100005;
17 const int inf = 1e9;
18 const int Maxlen = N * 8;
19    
20 int n, len;
21 int a[N], ans[N], mn[N];
22 int bit[N], b[N];
23 char buf[Maxlen], *c = buf;
24 int Len;
25  
26 inline int read() {
27   int x = 0;
28   while (*c < '0' || '9' < *c) ++c;
29   while ('0' <= *c && *c <= '9')
30     x = x * 10 + *c - '0', ++c;
31   return x;
32 }
33  
34 void print(int x) {
35   if (x >= 10) print(x / 10);
36   putchar(x % 10 + '0');
37 }
38  
39 inline int get_kth(int k) {
40   int res = 0, cnt = 0, i;
41   for (i = 20; ~i; --i) {
42     res += 1 << i;
43     if (res >= n || cnt + bit[res] >= k) res -= 1 << i;
44     else cnt += bit[res];
45   }
46   return res + 1;
47 }
48  
49 inline void bit_del(int x) {
50   while (x <= n)
51     --bit[x], x += lowbit(x);
52 }
53  
54 int main() {
55   Len = fread(c, 1, Maxlen, stdin);
56   buf[Len] = '\0';
57   int i, t, w, mx;
58   n = read();
59   for (i = 1; i <= n; ++i) {
60     b[i] = read(), ++bit[i];
61     if (i + lowbit(i) <= n)
62       bit[i + lowbit(i)] += bit[i];
63   }
64   for (i = n; i; --i) {
65     a[w = get_kth(b[i] + 1)] = i;
66     bit_del(w);
67   }
68   memset(mn, 127, sizeof(mn));
69   for (mn[0] = -inf, i = 1; i <= n; ++i) {
70     t = upper_bound(mn, mn + len + 1, a[i]) - mn;
71     if (mn[t - 1] <= a[i]) {
72       mn[t] = min(mn[t], a[i]);
73       ans[a[i]] = t;
74       len = max(t, len);
75     }
76   }
77   for (i = 1, mx = 0; i <= n; ++i) {
78     if (ans[i] > mx) mx = ans[i];
79     print(mx);
80     putchar('\n');
81   }
82   return 0;
83 }
View Code

(p.s. 在我的各种读入/输出/常数优化后，终于成为rank.1 yeah!)

posted on 2015-01-30 23:09  Xs酱~  阅读(1458)  评论(0)  编辑  收藏  举报