标签:EDA ++i gif fine 算法 时间 math ide dash
对于序列a和序列b,求其最长公共子序列
通过动态规划的方式
dp[i][j] 前i个字符的x和前j个字符的y的最长公共子序列
当a[i] = b[j] 的时候
dp[i][j] = max(dp[i][j] , dp[i - 1][j - 1] + 1)
如果第i位和第j位相等的话,那么最长序列长度 + 1
当 a[i] != b[j] 的时候
dp[i][j] = max(dp[i][j] , dp[i - 1][j] , dp[i][j - 1])
如果第i位和第j不相等的话 , 那么最长公共子序列的长度则从 dp[i - 1][j] 和 dp[i][j - 1] 转移过来
1 void lcs() 2 { 3 for (int i = 0; i <= len1; ++i) dp[i][0] = 0; 4 for (int i = 0; i <= len2; ++i) dp[0][i] = 0; 5 for (int i = 1; i <= len1; ++i) 6 { 7 for (int j = 1; j <= len2; ++j) 8 { 9 if (arr[i] == brr[j]) 10 dp[i][j] = dp[i - 1][j - 1] + 1; 11 else 12 { 13 if (dp[i - 1][j] > dp[i][j - 1]) 14 { 15 res[i][j] = 1; 16 dp[i][j] = dp[i - 1][j]; 17 } 18 else 19 { 20 res[i][j] = 2; 21 dp[i][j] = dp[i][j - 1]; 22 } 23 } 24 } 25 } 26 } 27 string getlcs() 28 { 29 int i = len1, j = len2; 30 string ans = ""; 31 while (i > 0 && j > 0) 32 { 33 if (!res[i][j]) 34 { 35 ans = arr[i] + ans; 36 i--, j--; 37 } 38 else if (res[i][j] == 1) 39 { 40 i--; 41 } 42 else 43 { 44 j--; 45 } 46 } 47 return ans; 48 }
时间复杂度:O(n * m)
1 #include<cstdio> 2 #include<string.h> 3 #include<algorithm> 4 #include<cmath> 5 #include<iostream> 6 #include<vector> 7 #include<queue> 8 #include<set> 9 #include<stack> 10 #include<map> 11 #include<cctype> 12 #define ios ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) 13 #define mem(a,x) memset(a,x,sizeof(a)) 14 #define lson rt<<1,l,mid 15 #define rson rt<<1|1,mid + 1,r 16 #define P pair<int,int> 17 #define ull unsigned long long 18 using namespace std; 19 typedef long long ll; 20 const int maxn = 2e5 + 10; 21 const ll mod = 998244353; 22 const int inf = 0x3f3f3f3f; 23 const long long INF = 0x3f3f3f3f3f3f3f3f; 24 const double eps = 1e-7; 25 26 inline ll read() 27 { 28 ll X = 0, w = 0; char ch = 0; 29 while (!isdigit(ch)) { w |= ch == ‘-‘; ch = getchar(); } 30 while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar(); 31 return w ? -X : X; 32 } 33 34 int len1, len2; 35 int dp[1000][1000]; 36 string arr, brr; 37 int res[1000][1000]; 38 void lcs() 39 { 40 for (int i = 0; i <= len1; ++i) dp[i][0] = 0; 41 for (int i = 0; i <= len2; ++i) dp[0][i] = 0; 42 for (int i = 1; i <= len1; ++i) 43 { 44 for (int j = 1; j <= len2; ++j) 45 { 46 if (arr[i] == brr[j]) 47 dp[i][j] = dp[i - 1][j - 1] + 1; 48 else 49 { 50 if (dp[i - 1][j] > dp[i][j - 1]) 51 { 52 res[i][j] = 1; 53 dp[i][j] = dp[i - 1][j]; 54 } 55 else 56 { 57 res[i][j] = 2; 58 dp[i][j] = dp[i][j - 1]; 59 } 60 } 61 } 62 } 63 } 64 string getlcs() 65 { 66 int i = len1, j = len2; 67 string ans = ""; 68 while (i > 0 && j > 0) 69 { 70 if (!res[i][j]) 71 { 72 ans = arr[i] + ans; 73 i--, j--; 74 } 75 else if (res[i][j] == 1) 76 { 77 i--; 78 } 79 else 80 { 81 j--; 82 } 83 } 84 return ans; 85 } 86 87 int main() 88 { 89 cin >> arr >> brr; 90 len1 = arr.size(), len2 = brr.size(); 91 arr = " " + arr; 92 brr = " " + brr; 93 lcs(); 94 cout << "最长公共字串长度: " << dp[len1][len2] << endl; 95 cout << "最长公共字串: " << getlcs() << endl; 96 return 0; 97 }
https://github.com/BambooCertain/Algorithm.git
标签:EDA ++i gif fine 算法 时间 math ide dash
原文地址:https://www.cnblogs.com/DreamACMer/p/12799216.html