DP+BIT(优化复杂度) UESTC 1217 The Battle of Chibi

题目传送门

题意：问n长度的序列，找出长度m的上升子序列的方案数。

分析：这个问题就是问：dp[i][j] = sum (dp[i-1][k]) (1 <= k <= n, a[k] < a[j])，当前长度i一层，最后下标j一层，之前的最后下标k一层，这样是n^3的复杂度。然后树状数组可以优化j，k的复杂度，就是j扫描一遍的同时，将a[j]的信息更新到树上，那么扫描就可以用 logn时间统计出k的信息，在这之前先对a[]离散化。

/************************************************
* Author        :Running_Time
* Created Time  :2015/10/21 星期三 13:20:36
* File Name     :C.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;
struct BIT  {
    int c[N], SZ;
    void add(int &a, int b) {
        a += b;
        while (a >= MOD)    a -= MOD;
    }
    void init(int n)    {
        memset (c, 0, sizeof (c));
        SZ = n;
    }
    void updata(int i, int x)   {
        while (i <= SZ) {
            add (c[i], x);  i += i & (-i);
        }
    }
    int query(int i)    {
        int ret = 0;
        while (i)   {
            add (ret, c[i]);    i -= i & (-i);
        }
        return ret;
    }
}bit;
int a[N], A[N];
int dp[N][N];

void compress(int n) {
    sort (A, A+n);
    int nn = unique (A, A+n) - A;
    for (int i=0; i<n; ++i)    {
        a[i] = lower_bound (A, A+n, a[i]) - A + 1;
    }
}

int main(void)    {
    int T, cas = 0; scanf ("%d", &T);
    while (T--) {
        int n, m;  scanf ("%d%d", &n, &m);
        bit.init (n);
        for (int i=0; i<n; ++i)    {
            scanf ("%d", &a[i]);
            A[i] = a[i];
        }
        compress (n);
        memset (dp, 0, sizeof (dp));
        for (int i=0; i<n; ++i) dp[1][i] = 1;
        int ans = 0;
        for (int i=2; i<=m; ++i)    {
            bit.init (n);
            for (int j=0; j<n; ++j) {
                dp[i][j] = bit.query (a[j] - 1);
                bit.updata (a[j], dp[i-1][j]);
            }
        }
        for (int i=0; i<n; ++i) bit.add(ans, dp[m][i]);
        printf ("Case #%d: %d\n", ++cas, ans);
    }

    return 0;
}