Gdfzoj Sub-sequences

Tag: Represents the stream's location end strlen condition out CPP

Test instructions

Give two strings, A, B, find the LCS of A, B, and then extract a sub-sequence of length LCS, q How many are also the sub-sequences?

Exercises

First, it is clear that LCS must be required.

Now consider the record,

Now consider the definition: the first routine is defined according to the answer, the definition state sum[i][j] denotes the first I characters in a string, how many sub-sequences of length dp[i][j] have appeared in B.

Now consider the transfer: if dp[i][j] = = Dp[i-1][j], then sum[i][j] + = Sum[i-1][j] Indicates that the I-character is not selected

Now consider the required I-character, then you need to find the position of the last character in string B that is equal to a[i], and if dp[i-1][pre-1] + 1 = = Dp[i][j],sum[i][j] + + sum[i-1][pre-1]

Code:

#include <cstdio> #include <iostream> #include <cstring> #include <algorithm>using namespace std;const int N = 1e3 + 7;const int mod = 1e9 + 7;char ch1[n], Ch2[n];int dp[n][n], sum[n][n], N, M, Pre[n];int main () {g ETS (CH1 + 1), gets (CH2 + 1), n = strlen (ch1 + 1), M = strlen (CH2 + 1), for (int i = 0; I <= N; ++i) sum[i][0] = 1;for ( int j = 0; J <= M;  ++J) Sum[0][j] = 1;for (int i = 1; I <= n; ++i) {for (int j = 1; j <= m; ++j) {Dp[i][j] = max (Dp[i-1][j], dp[i][j -1]); if (ch1[i] = = Ch2[j]) dp[i][j] = max (Dp[i][j], dp[i-1][j-1] + 1);}} for (int j = 1; j <= m; ++j) {pre[ch2[j]] = j;for (int i = 1; I <= n; ++i) {if (dp[i-1][j] = = Dp[i][j]) sum[i][j] = (Sum[i][j] + sum[i-1][j])% mod;if (Pre[ch1[i]] && dp[i-1][pre[ch1[i]]-1] + 1 = = Dp[i][j]) {sum[i][j] = (sum I [j] + Sum[i-1][pre[ch1[i]]-1])% MoD;}} cout << sum[n][m] << endl;return 0;}

　　

Summarize:

The basic condition of this problem is on the basis of LCS, first DP out all the conditions of the LCS, and then based on the basis of the DP to take the calculation scheme number.

Gdfzoj Sub-sequences