hdoj1159 Common Subsequence
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题意:
找两个字符串的最长公共子序列(Loggest Common Subsequence, LCS)。用“表格法”(姑且这么叫吧)就可以很快地做出来,至于表格法的原理,以后有时间再整理吧,一时半会也整理不出来。
比如给出的第一组输入:
具体的转移方程是:
// dp[i][j]表示在到了字符串sa的i位置和字符串sb的j位置时,可以达到的最大值,因为这里做了一个第一行列为0的初始化,所以对应到字符串时,下标要减去1if (str_a[i-1] == str_b[j-1]) { dp[i][j] = 1 + dp[i-1][j-1];} else { dp[i][j] = max(dp[i-1][j], dp[i][j-1]);}
代码(78ms, 5692KB):
#include <iostream>#include <cstring> // memset#include <string>#include <algorithm> // maxusing namespace std;const int maxn = 1001;int dp[maxn][maxn];int main() {// freopen("in.txt", "r", stdin); string sa, sb; while (cin >> sa >> sb) { memset(dp, 0, sizeof dp); int la = sa.size(), lb = sb.size(); for (int i = 1; i <= la; ++i) { for (int j = 1; j <= lb; ++j) { if (sa[i-1] == sb[j-1]) { dp[i][j] = 1 + dp[i-1][j-1]; } else { dp[i][j] = max(dp[i-1][j], dp[i][j-1]); } } } cout << dp[la][lb] << endl; } return 0;}
思考:
如果是多个字符串,又怎么处理呢?
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