LCS/LIS/lcis template Summary

/************************* LCS/LIS/lcis template summary: *************************//************* **************************************** LCS: the longest common subsequence calculates the LCS of the sequence a with the length of len1 and the sequence B with the length of len2. Note: The sequence subscript starts to scroll the array writing method from 0. Returns the length of LCS ************************************* * **************/int lcs (INT len1, int len2) {memset (DP, 0, sizeof (DP); For (INT I = 1; I <= len1; I ++) {for (Int J = 1; j <= len2; j ++) {If (S1 [I-1] = S2 [J-1]) DP [I % 2] [J] = DP [(I-1) % 2] [J-1] + 1; else {int M1 = DP [(I-1) % 2] [J]; int m2 = DP [I % 2] [J-1]; DP [I % 2] [J] = max (M1, M2) ;}} return DP [len1% 2] [len2];} /*************************************** * *** LIS: most Long rising sub-sequence HDU 1257 minimum interception system ******************************* * ************/# include <stdio. h ># include <algorithm> using namespace STD; const int maxn = 1000 + 10; const int INF = 30000 + 10; int A [maxn]; // missile height int H [maxn]; // H [I] indicates the current height of the I-th system interception int main () {int N; while (scanf ("% d", & N )! = EOF) {for (INT I = 0; I <n; I ++) {scanf ("% d", & A [I]); // H [I] = inf;} H [0] =-1; // guarantee that the boundary increments H [1] = A [0]; // The first int Len = 1; // The length of the For (INT I = 1; I <n; I ++) {int Index = lower_bound (H, H + Len + 1, a [I])-H; // ensure that H [Index] is the first H [Index] = A [I] in array h; If (index> Len) len = index;} printf ("% d \ n", Len);} return 0 ;} /*************************************** * *********** lcis: the longest common ascending subsequence returns the length of the LCS sequence ************* starting from 1 when the length of sequence a is n and the length of sequence B is M ************* **************************************** /int dp [maxn]; int LCS (int n, int m) {memset (DP, 0, sizeof (DP); For (INT I = 1; I <= N; I ++) {int TMP = 0; // The maximum DP [J] For (Int J = 1; j <= m; j ++) {if (a [I]> B [J] & DP [J]> TMP) TMP = DP [J]; else if (a [I] = B [J]) DP [J] = TMP + 1 ;}} int ans = 0; For (INT I = 1; I <= m; I ++) ans = max (ANS, DP [I]); Return ans ;}