【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 }

View Code

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 }

View Code