【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）：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>
#include <stack>
#include <string>
#define mem0(a) memset(a, 0, sizeof(a))
#define mem(a, b) memset(a, b, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define lr rt << 1
#define rr rt << 1 | 1
#define eps 0.0001
#define lowbit(x) ((x) & -(x))
#define memc(a, b) memcpy(a, b, sizeof(b))
typedef char Str[120];
using namespace std;
#define LL long long
#define DL double

map<int, int> Hash;
int mol = 123456789;
int a[12000], f[12000][120], b[12000], R[12000], c[120000], N, n, m;

void init()
{
        sort(a + 1, a + 1 + n);
        N = 0;
        for(int i = 1; i <= n; i++) {
                if(a[i] == a[i - 1] && i > 1) continue;
                Hash[a[i]] =  ++N;
        }
}
int query(int p)
{
        int ans = 0;
        while(p) {
                ans += c[p];
                if(ans >= mol) ans -= mol;
                p -= lowbit(p);
        }
        return ans;
}
void update(int p, int x)
{
        while(p <= N) {
                c[p] += x;
                if(c[p] >= mol) c[p] -= mol;
                p += lowbit(p);
        }
}
int main()
{
        //freopen("input.txt", "r", stdin);
        //freopen("output.txt", "w", stdout);
        while(~scanf("%d%d", &n, &m)) {
                for(int i = 1; i <= n; i++) {
                        scanf("%d", a + i);
                }
                memc(b, a);
                init();
                mem0(f);
                //puts("d");
                for(int i = 1; i <= n; i++) f[i][1] = 1;
                for(int i = 1; i <= n; i++) R[i] = Hash[b[i]];
                for(int j = 2; j <= m; j++) {
                        mem0(c);
                        for(int i = 1; i <= n; i++) {
                                f[i][j] = query(R[i] - 1);
                                update(R[i], f[i][j - 1]);
                                //cout<< i<< " "<< j<< " "<< f[i][j]<< endl;
                        }
                }
                int ans = 0;
                for(int i = m; i <= n; i++) {
                        ans += f[i][m];
                        if(ans >= mol) ans -= mol;
                }
                cout<< ans<< endl;
        }
        return 0;
}

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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>
#include <stack>
#include <string>
#define mem0(a) memset(a, 0, sizeof(a))
#define mem(a, b) memset(a, b, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define lr rt << 1
#define rr rt << 1 | 1
#define eps 0.0001
#define lowbit(x) ((x) & -(x))
#define memc(a, b) memcpy(a, b, sizeof(b))
typedef char Str[120];
using namespace std;
#define LL long long
#define DL double
struct seg{
        int sum;
}tree[12000 << 2];
map<int, int> Hash;
int mol = 123456789;
int a[12000], f[12000][120], b[12000], R[12000], N, n, m;

void init()
{
        sort(a + 1, a + 1 + n);
        N = 0;
        for(int i = 1; i <= n; i++) {
                if(a[i] == a[i - 1] && i > 1) continue;
                Hash[a[i]] =  ++N;
        }
}
void build(int l, int r, int rt)
{
        tree[rt].sum = 0;
        if(l == r) return;
        int m = (l + r) >> 1;
        build(lson);
        build(rson);
}
void pushUP(int rt)
{
        tree[rt].sum = tree[rt << 1].sum + tree[rt<< 1 | 1].sum;
        if(tree[rt].sum >= mol) tree[rt].sum -= mol;
}
void update(int p, int x, int l, int r, int rt)
{
        if(l == r) {
                tree[rt].sum += x;
                if(tree[rt].sum >= mol) tree[rt].sum -= mol;
                return;
        }
        int m = (l + r) >> 1;
        if(p <= m) update(p, x, lson);
        else update(p, x, rson);
        pushUP(rt);
}
int query(int L, int R, int l, int r, int rt)
{
        if(L <= l && r <= R) {
                return tree[rt].sum;
        }
        int m = (l + r) >> 1, res = 0;
        if(L <= m) res+= query(L, R, lson);
        if(R > m) res += query(L, R, rson);
        if(res >= mol) res -= mol;
        return res;
}
int main()
{
        //freopen("input.txt", "r", stdin);
        //freopen("output.txt", "w", stdout);
        while(~scanf("%d%d", &n, &m)) {
                for(int i = 1; i <= n; i++) {
                        scanf("%d", a + i);
                }
                memc(b, a);
                init();
                mem0(f);
                //puts("d");
                for(int i = 1; i <= n; i++) f[i][1] = 1;
                for(int i = 1; i <= n; i++) R[i] = Hash[b[i]];
                for(int j = 2; j <= m; j++) {
                        build(1, N, 1);
                        for(int i = 1; i <= n; i++) {
                                if(R[i] > 1) f[i][j] = query(1, R[i] - 1, 1, N, 1);
                                update(R[i], f[i][j - 1], 1, N, 1);
                        }
                }
                int ans = 0;
                for(int i = m; i <= n; i++) {
                        ans += f[i][m];
                        if(ans >= mol) ans -= mol;
                }
                cout<< ans<< endl;
        }
        return 0;
}