POJ 1159 Palindrome

来源：互联网发布：js除以10000四舍五入编辑：程序博客网时间：2024/06/17 05:37

　　寻找串与其逆串的最长公共子序列。

　　因为此子序列必是回文串，剩下的字符就是需要插入的。

 1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstring> 5  6 #define Max(a,b) a > b ? a : b 7  8 using namespace std; 9 10 char s1[5010],s2[5010];11 12 short int CountLen[5010][5010];13 14 int main()15 {16     int len;17     while(~scanf("%d",&len))18     {19     scanf("%s",(s1+1));20 21     int mid = len/2;22 23     int i,j;24 25     for(j = 1,i = len;i >= 1; --i,++j)26     {27         s2[j] = s1[i];28     }29 30     s2[j] = '\0';31 32     for(i = 0;i <= len; ++i)33     {34         for(j = 0;j <= len; ++j)35         {36             if(i == 0 || j == 0)37             {38                     CountLen[i][j] == 0;39             }40             else if(s1[i] == s2[j])41             {42                 CountLen[i][j] = CountLen[i-1][j-1]+1;43             }44             else45             {46                 CountLen[i][j] = Max(CountLen[i][j-1],CountLen[i-1][j]);47             }48         }49     }50 51 52     printf("%d\n",len-CountLen[len][len]);53     }54     return 0;55 }

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　　用 int 有可能会MLE　可以用short也可以用滚动数组。

　　下面是用滚动数组优化的代码

 1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstring> 5  6 #define Max(a,b) a > b ? a : b 7  8 using namespace std; 9 10 char s1[5010],s2[5010];11 12 short int CountLen[2][5010],temp[5010];13 14 int main()15 {16     int len;17     while(~scanf("%d",&len))18     {19     scanf("%s",(s1+1));20 21     int i,j;22 23     for(j = 1,i = len;i >= 1; --i,++j)24     {25         s2[j] = s1[i];26     }27 28     s2[j] = '\0';29 30     memset(CountLen,0,sizeof(CountLen));31     memset(temp,0,sizeof(temp));32 33     for(i = 0;i <= len; ++i)34     {35         for(j = 0;j <= len; ++j)36         {37             if(i == 0 || j == 0)38             {39                 CountLen[i&1][j] = 0;40             }41             else if(s1[i] == s2[j])42             {43                 CountLen[i&1][j] = CountLen[(i+1)&1][j-1]+1;44             }45             else46             {47                 CountLen[i&1][j] = Max(CountLen[i&1][j-1],CountLen[(i+1)&1][j]);48             }49         }50     }51 52     printf("%d\n",len-CountLen[len&1][len]);53     }54     return 0;55 }

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