poj1159 dp最長公用子串

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 1 //Accepted    204 KB    891 ms 2 //dp最長公用子串 3 //dp[i][j]=max(dp[i-1][j],dp[i][j-1]) 4 //dp[i][j]=max(dp[i][j],dp[i-1][j-1]+1) (s1[i]==s2[j]) 5 #include <cstdio> 6 #include <cstring> 7 #include <iostream> 8 using namespace std; 9 const int imax_n = 5002;10 int dp[2][imax_n];11 char s1[imax_n];12 char s2[imax_n];13 int n;14 int max(int a,int b)15 {16     return a>b?a:b;17 }18 void Dp()19 {20     memset(dp,0,sizeof(dp));21     for (int i=1;i<=n;i++)22     {23         for (int j=1;j<=n;j++)24         {25             dp[i%2][j]=max(dp[(i-1)%2][j],dp[i%2][j-1]);26             if (s1[i-1]==s2[j-1])27             dp[i%2][j]=max(dp[i%2][j],dp[(i-1)%2][j-1]+1);28         }29     }30     printf("%d\n",n-dp[n%2][n]);31 }32 int main()33 {34     while (scanf("%d",&n)!=EOF)35     {36         scanf("%s",s1);37         for (int i=0;i<n;i++)38         s2[i]=s1[n-1-i];39         s2[n]=‘\0‘;40         Dp();41     }42     return 0;43 }

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