Uva10723-cyborg genes (LIS)

Question: uva10723-cyborg genes (LIS)

Give two strings to a new string, which maintains the character characteristics of the two strings, that is, the subsequences of the first two strings can be found in the string, evaluate the shortest length of such a string and the number of different types of strings.

Solution: Lis. First, we need to find the longest common subsequence, so that the length of the new string will be the minimum: L1 + L2-L [1] [N ];

L [I] [J]: The longest length of the first I character of the first string and the first J character of the second string.

N [I] [J]: makes the first I character of the first string and the first J character of the second string form the shortest Number of new strings.

S [I] = s [J], L [I] [J] = L [I-1] [J-1] + 1; N [I] [J] = N [I-1] [J-1];

S [I]! = S [J], L [I] [J] = max (L [I-1] [J], L [I] [J-1]); if (L [I-1] [J]> L [I] [J-1]) N [I] [J] = N [I-1] [J]; (because it is the type of the string to be shortest), and vice versa. N [I] [J] = N [I-1] [J] + N [I] [J-1];

Initialization: L [I] [0] = L [0] [J] = 0; n [0] [J] = N [I] [0] = 1;

Code:

#include <cstdio>#include <cstring>const int N = 35;typedef long long ll;int l[N][N];ll n[N][N];char s1[N], s2[N];int l1, l2;int Max (const int a, const int b) { return a > b? a: b; }void init () {l1 = strlen (s1);l2 = strlen (s2);memset (l, 0, sizeof (l));for (int i = 0; i <= l1; i++)n[i][0] = 1;for (int i = 0; i <= l2; i++)n[0][i] = 1;}int main () {int t;scanf ("%d%*c", &t);for (int k = 1; k <= t; k++) {printf ("Case #%d:", k);gets (s1);gets (s2);init ();for (int i = 1; i <= l1; i++) {for (int j = 1; j <= l2; j++) {if (s1[i - 1] == s2[j - 1]) {l[i][j] = l[i - 1][j - 1] + 1;n[i][j] = n[i - 1][j - 1];} else {l[i][j] = Max (l[i - 1][j], l[i][j - 1]); //printf ("%d %d %d %d %d\n", i, j, l[i - 1][j], l[i][j - 1], l[i][j]); if (l[i - 1][j] > l[i][j - 1])n[i][j] = n[i - 1][j];else if (l[i - 1][j] < l[i][j - 1])n[i][j] = n[i][j - 1];elsen[i][j] = n[i][j - 1] + n[i - 1][j];}}}printf (" %d %lld\n", l1 + l2 - l[l1][l2], n[l1][l2]);}return 0;}