uestc oj 1217 The Battle of Chibi (dp + 离散化 + 树状数组)

Posted 2020-08-09 Recoder

题目链接：http://acm.uestc.edu.cn/#/problem/show/1217

给你一个长为n的数组，问你有多少个长度严格为m的上升子序列。

dp[i][j]表示以a[i]结尾长为j的上升子序列个数。常规是三个for。

这里用树状数组优化一下，类似前缀和的处理，两个for就好了。

 1 //#pragma comment(linker, "/STACK:102400000, 102400000")
 2 #include <algorithm>
 3 #include <iostream>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <cstdio>
 7 #include <vector>
 8 #include <cmath>
 9 #include <ctime>
10 #include <list>
11 #include <set>
12 #include <map>
13 using namespace std;
14 typedef long long LL;
15 typedef pair <int, int> P;
16 const int N = 1e3 + 5;
17 LL dp[N][N], mod = 1e9 + 7;
18 int a[N], b[N], m, n;
19 LL bit[N][N];
20 
21 void add(int pos, int i, int val) { //上升子序列长度为pos 
22     for( ; i <= n; i += (i&-i))
23         bit[pos][i] = (bit[pos][i] + val) % mod;
24 }
25 
26 LL sum(int pos, int i) {
27     LL s = 0;
28     for( ; i >= 1; i -= (i&-i))
29         s = (s + bit[pos][i]) % mod;
30     return s;
31 }
32 
33 int main()
34 {
35     int t;
36     scanf("%d", &t);
37     for(int ca = 1; ca <= t; ++ca) {
38         scanf("%d %d", &n, &m);
39         for(int i = 1; i <= n; ++i) {
40             scanf("%d", a + i);
41             b[i] = a[i];
42         }
43         sort(b + 1, b + n + 1);
44         for(int i = 1; i <= n; ++i) {
45             a[i] = lower_bound(b + 1, b + n + 1, a[i]) - b; //离散化
46         }
47         memset(dp, 0, sizeof(dp));
48         memset(bit, 0, sizeof(bit));
49         dp[1][1] = 1;
50         add(1, a[1], 1);
51         for(int i = 2; i <= n; ++i) {
52             dp[i][1] = 1;
53             for(int k = max(m - (n - i), 1); k <= min(i, m); ++k) { //这边可以优化一下
54                 dp[i][k] = (dp[i][k] + sum(k - 1, a[i] - 1)) % mod; //比a[i]小且上升子序列长度为k-1
55                 add(k, a[i], dp[i][k]);
56             }
57         }
58         LL res = 0;
59         for(int i = 1; i <= n; ++i) {
60             res = (res + dp[i][m]) % mod;
61         }
62         printf("Case #%d: %lld\n", ca, res);
63     }
64     return 0; 
65 }