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【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 << 112 #define rson m + 1, r, rt << 1 | 113 #define lr rt << 114 #define rr rt << 1 | 115 #define eps 0.000116 #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 long21 #define DL double22 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 }
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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 }
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【HDU4991】树状数组