HDU 1503 advanced fruits (LCS deformation and output solution)

HDU 1503 advanced fruits (LCS deformation and output solution)

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1503

Question:

Give you two strings S1 and S2. You need to output them and string S. s1 is a sub-sequence of S, S2 is a sub-sequence of S, and s is the shortest string that meets the preceding requirements.

Analysis:

For example, DP [I] [J] = X indicates that the length of the LCS string consisting of the first I character of the S1 string and the first J character of the S2 string is X.

We first find the DP array value of LCS, and then follow the poj2250:

Http://blog.csdn.net/u013480600/article/details/40743953

Similar to the DFS process, the S string can be output.

The general idea is:

Assume that DFS (I, j) indicates that the output I characters (Count from 0) And a string of J characters.

If I = 0 or J = 0, output the remaining string and return.

Optional (The following three processes ensure that I> 0 and j> 0):

If s [I-1] = s [J-1], then first DFS (I-1, J-1), then output S1 [I-1] (The I character of the first string.

If s [I-1]! When = s [J-1:

If DP [I-1] [J]> DP [I] [J-1], DFS (I-1, J), then output S1 [I-1.

If DP [I-1] [J] <= DP [I] [J-1], DFS (I, J-1), then output S2 [J-1.

(Carefully understand the above process)

AC code:

# Include <cstdio> # include <algorithm> # include <cstring> using namespace STD; const int maxn = 100 + 5; int n, m; int DP [maxn] [maxn]; char S1 [maxn], S2 [maxn]; void DFS (int I, Int J) {if (I + J = 0) return; else if (I = 0) // The S1 string has been output {for (int K = 0; k <j; k ++) printf ("% C ", s2 [k]); return;} else if (j = 0) // The S2 string has been output {for (int K = 0; k <I; k ++) printf ("% C", S1 [k]); return;} // The recursion below guarantees that I> 0 and j> 0. if (S1 [I-1] = S2 [J-1]) DFS (I-1, J-1), printf ("% C", S1 [I-1]); else if (DP [I-1] [J]> DP [I] [J-1]) DFS (I-1, J), printf ("% C", S1 [I-1]); else DFS (I, J-1), printf ("% C", S2 [J-1]);} int main () {While (scanf ("% S % s", S1, s2) = 2) {// initialize n = strlen (S1); M = strlen (S2); memset (DP, 0, sizeof (DP )); // recursive for (INT I = 1; I <= N; I ++) for (Int J = 1; j <= m; j ++) {If (S1 [I-1] = S2 [J-1]) DP [I] [J] = DP [I-1] [J-1] + 1; else DP [I] [J] = max (DP [I-1] [J], DP [I] [J-1]);} // DFS + output DFS (n, m); printf ("\ n");} return 0 ;}

HDU 1503 advanced fruits (LCS deformation and output solution)