51Nod 1092 回文字符串
最开始毫无头绪,然后参照了一位dalao的博客,思路是一个正序的字符串将其逆序,然后求最长公共子序列(LCS),emm也属于动态规划。
1 #include <iostream> 2 #include <algorithm> 3 #include <cstdio> 4 #include <cstring> 5 using namespace std; 6 7 const int maxn = 1105; 8 char s1[maxn]; 9 char s2[maxn]; 10 11 int dp[maxn][maxn]; 12 13 int main(){ 14 cin >> s1; 15 int len = strlen(s1); 16 for (int i = 0,j = len - 1; i < len; i++,j--){ 17 s2[i] = s1[j]; 18 } 19 memset(dp, 0, sizeof(dp)); 20 for (int i = 1; i <= len; i++){ 21 for (int j = 1; j <= len; j++){ 22 if (s1[i - 1] == s2[j - 1]){ 23 dp[i][j] = max(dp[i - 1][j - 1] + 1, max(dp[i - 1][j], dp[i][j - 1])); 24 } 25 else{ 26 dp[i][j] = max(dp[i - 1][j - 1], max(dp[i - 1][j], dp[i][j - 1])); 27 } 28 } 29 } 30 cout << len - dp[len][len] << endl; 31 return 0; 32 }