【HDU4991】树状数组

http://acm.hdu.edu.cn/showproblem.php?pid=4991

用f[i][j] 表示 前i个数以第i个数结尾的合法子序列的个数，则递推式不难写出：

f[i][j] = sum(f[k][j - 1]); 其中 k < i, 且a[k] < a[i]; 边界：f[i][1] = 1; 显然需要用数据结构来优化查询

如果不考虑离散的话，用一个数据结构，记录节点为a[i]的f值，同时维护一个区间f值之和，那么f[i][j] = query(0, a[i] - 1)；然后考虑特定的顺序用f值更新数据结构中的信息

具体见代码（树状数组）：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <map>
 7 #include <stack>
 8 #include <string>
 9 #define mem0(a) memset(a, 0, sizeof(a))
10 #define mem(a, b) memset(a, b, sizeof(a))
11 #define lson l, m, rt << 1
12 #define rson m + 1, r, rt << 1 | 1
13 #define lr rt << 1
14 #define rr rt << 1 | 1
15 #define eps 0.0001
16 #define lowbit(x) ((x) & -(x))
17 #define memc(a, b) memcpy(a, b, sizeof(b))
18 typedef char Str[120];
19 using namespace std;
20 #define LL long long
21 #define DL double
22
23 map<int, int> Hash;
24 int mol = 123456789;
25 int a[12000], f[12000][120], b[12000], R[12000], c[120000], N, n, m;
26
27 void init()
28 {
29         sort(a + 1, a + 1 + n);
30         N = 0;
31         for(int i = 1; i <= n; i++) {
32                 if(a[i] == a[i - 1] && i > 1) continue;
33                 Hash[a[i]] =  ++N;
34         }
35 }
36 int query(int p)
37 {
38         int ans = 0;
39         while(p) {
40                 ans += c[p];
41                 if(ans >= mol) ans -= mol;
42                 p -= lowbit(p);
43         }
44         return ans;
45 }
46 void update(int p, int x)
47 {
48         while(p <= N) {
49                 c[p] += x;
50                 if(c[p] >= mol) c[p] -= mol;
51                 p += lowbit(p);
52         }
53 }
54 int main()
55 {
56         //freopen("input.txt", "r", stdin);
57         //freopen("output.txt", "w", stdout);
58         while(~scanf("%d%d", &n, &m)) {
59                 for(int i = 1; i <= n; i++) {
60                         scanf("%d", a + i);
61                 }
62                 memc(b, a);
63                 init();
64                 mem0(f);
65                 //puts("d");
66                 for(int i = 1; i <= n; i++) f[i][1] = 1;
67                 for(int i = 1; i <= n; i++) R[i] = Hash[b[i]];
68                 for(int j = 2; j <= m; j++) {
69                         mem0(c);
70                         for(int i = 1; i <= n; i++) {
71                                 f[i][j] = query(R[i] - 1);
72                                 update(R[i], f[i][j - 1]);
73                                 //cout<< i<< " "<< j<< " "<< f[i][j]<< endl;
74                         }
75                 }
76                 int ans = 0;
77                 for(int i = m; i <= n; i++) {
78                         ans += f[i][m];
79                         if(ans >= mol) ans -= mol;
80                 }
81                 cout<< ans<< endl;
82         }
83         return 0;
84 }

另外用线段树写了一下， 不过超时（2000+ms）了，线段树才不到500ms，由此可见能用树状数组决不用线段树，递归线段树常数太大了

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cmath>
  5 #include <algorithm>
  6 #include <map>
  7 #include <stack>
  8 #include <string>
  9 #define mem0(a) memset(a, 0, sizeof(a))
 10 #define mem(a, b) memset(a, b, sizeof(a))
 11 #define lson l, m, rt << 1
 12 #define rson m + 1, r, rt << 1 | 1
 13 #define lr rt << 1
 14 #define rr rt << 1 | 1
 15 #define eps 0.0001
 16 #define lowbit(x) ((x) & -(x))
 17 #define memc(a, b) memcpy(a, b, sizeof(b))
 18 typedef char Str[120];
 19 using namespace std;
 20 #define LL long long
 21 #define DL double
 22 struct seg{
 23         int sum;
 24 }tree[12000 << 2];
 25 map<int, int> Hash;
 26 int mol = 123456789;
 27 int a[12000], f[12000][120], b[12000], R[12000], N, n, m;
 28
 29 void init()
 30 {
 31         sort(a + 1, a + 1 + n);
 32         N = 0;
 33         for(int i = 1; i <= n; i++) {
 34                 if(a[i] == a[i - 1] && i > 1) continue;
 35                 Hash[a[i]] =  ++N;
 36         }
 37 }
 38 void build(int l, int r, int rt)
 39 {
 40         tree[rt].sum = 0;
 41         if(l == r) return;
 42         int m = (l + r) >> 1;
 43         build(lson);
 44         build(rson);
 45 }
 46 void pushUP(int rt)
 47 {
 48         tree[rt].sum = tree[rt << 1].sum + tree[rt<< 1 | 1].sum;
 49         if(tree[rt].sum >= mol) tree[rt].sum -= mol;
 50 }
 51 void update(int p, int x, int l, int r, int rt)
 52 {
 53         if(l == r) {
 54                 tree[rt].sum += x;
 55                 if(tree[rt].sum >= mol) tree[rt].sum -= mol;
 56                 return;
 57         }
 58         int m = (l + r) >> 1;
 59         if(p <= m) update(p, x, lson);
 60         else update(p, x, rson);
 61         pushUP(rt);
 62 }
 63 int query(int L, int R, int l, int r, int rt)
 64 {
 65         if(L <= l && r <= R) {
 66                 return tree[rt].sum;
 67         }
 68         int m = (l + r) >> 1, res = 0;
 69         if(L <= m) res+= query(L, R, lson);
 70         if(R > m) res += query(L, R, rson);
 71         if(res >= mol) res -= mol;
 72         return res;
 73 }
 74 int main()
 75 {
 76         //freopen("input.txt", "r", stdin);
 77         //freopen("output.txt", "w", stdout);
 78         while(~scanf("%d%d", &n, &m)) {
 79                 for(int i = 1; i <= n; i++) {
 80                         scanf("%d", a + i);
 81                 }
 82                 memc(b, a);
 83                 init();
 84                 mem0(f);
 85                 //puts("d");
 86                 for(int i = 1; i <= n; i++) f[i][1] = 1;
 87                 for(int i = 1; i <= n; i++) R[i] = Hash[b[i]];
 88                 for(int j = 2; j <= m; j++) {
 89                         build(1, N, 1);
 90                         for(int i = 1; i <= n; i++) {
 91                                 if(R[i] > 1) f[i][j] = query(1, R[i] - 1, 1, N, 1);
 92                                 update(R[i], f[i][j - 1], 1, N, 1);
 93                         }
 94                 }
 95                 int ans = 0;
 96                 for(int i = m; i <= n; i++) {
 97                         ans += f[i][m];
 98                         if(ans >= mol) ans -= mol;
 99                 }
100                 cout<< ans<< endl;
101         }
102         return 0;
103 }