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