HDU 1159 Common Subsequence (LCS)
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题意:给定两行字符串,求最长公共子序列。
析:dp[i][j] 表示第一串以 i 个结尾和第二个串以 j 个结尾,最长公共子序列,剩下的就简单了。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> //#include <tr1/unordered_map> #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; //using namespace std :: tr1; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const LL LNF = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e3 + 5; const LL mod = 10000000000007; const int N = 1e6 + 5; const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1}; const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1}; const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); } int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int dp[maxn][maxn]; char s1[maxn], s2[maxn]; int main(){ while(scanf("%s %s", s1+1, s2+1) == 2){ n = strlen(s1+1); m = strlen(s2+1); memset(dp, 0, sizeof dp); int ans = 0; for(int i = 1; i <= n; ++i) for(int j = 1; j <= m; ++j) if(s1[i] == s2[j]) { dp[i][j] = dp[i-1][j-1] + 1; ans = Max(ans, dp[i][j]); } else { dp[i][j] = Max(dp[i-1][j], dp[i][j-1]); ans = Max(ans, dp[i][j]); } printf("%d\n", ans); } return 0; }
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