CF 346B. Lucky Common Subsequence(DP+KMP)

这题确实很棒。。又是无想法。。其实是AC自动机+DP的感觉，但是只有一个串，用kmp就行了。

dp[i][j][k]，k代表前缀为virus[k]的状态，len表示其他所有状态串，处理出Ac[len][26]数组来，DP就可以了。状态转移那里一直没想清楚，wa了很多次，记录路径倒是不复杂，瞎搞搞就行。

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 #include<algorithm>
 using namespace std;
 char s1[],s2[],virus[];
 int dp[][][];
 int pre[][][];
 int pre1[][][];
 int Ac[][];
 int next[];
 char ans[];
 void kmp()
 {
     int i,j,len,temp;
     len = strlen(virus);
     next[] = -;
     j = -;
     for(i = ; i < len; i ++)
     {
         while(j >= &&virus[j+] != virus[i])
             j = next[j];
         if(virus[j+] == virus[i]) j ++;
         next[i] = j;
     }
     for(i = ; i < len; i ++)
     {
         for(j = ; j < ; j ++)
         {
             temp = i;
             while(temp >= &&virus[temp+] != 'A'+j)
                 temp = next[temp];
             if(virus[temp+] == 'A' + j) temp ++;
             if(temp == -)
                 Ac[i][j] = len;
             else
                 Ac[i][j] = temp;
         }
     }
     for(i = ; i < ; i ++)
     {
         if(i + 'A' == virus[])
             Ac[len][i] = ;
         else
             Ac[len][i] = len;
     }
 }
 int main()
 {
     int i,j,k,len1,len2,len,maxz,a,b,kk;
     scanf("%s%s%s",s1,s2,virus);
     len1 = strlen(s1);
     len2 = strlen(s2);
     len = strlen(virus);
     kmp();
     for(i = ; i <= len1; i ++)
     {
         for(j = ; j <= len2; j ++)
         {
             for(k = ; k <= len; k ++)
             {
                 if(k == len-) continue;
                 if(dp[i][j][k] < dp[i-][j][k])
                 {
                     dp[i][j][k] = dp[i-][j][k];
                     pre[i][j][k] = ;
                     pre1[i][j][k] = k;
                 }
                 if(dp[i][j][k] < dp[i][j-][k])
                 {
                     dp[i][j][k] = dp[i][j-][k];
                     pre[i][j][k] = ;
                     pre1[i][j][k] = k;
                 }
                 if(s1[i-] == s2[j-])
                 {
                     if(Ac[k][s1[i-]-'A'] == len-) continue;
                     else if(dp[i][j][Ac[k][s1[i-]-'A']] < dp[i-][j-][k] + )
                     {
                         dp[i][j][Ac[k][s1[i-]-'A']] = dp[i-][j-][k] + ;
                         pre[i][j][Ac[k][s1[i-]-'A']] = ;
                         pre1[i][j][Ac[k][s1[i-]-'A']] = k;
                     }
                 }
             }
         }
     }
     maxz = ;
     for(i = ; i <= len1; i ++)
     {
         for(j = ; j <= len2; j ++)
         {
             for(k = ; k <= len; k ++)
             {
                 if(maxz < dp[i][j][k])
                 {
                     maxz = dp[i][j][k];
                     a = i;
                     b = j;
                     kk = k;
                 }
             }
         }
     }
     if(maxz == )
     {
         printf("0\n");
         return ;
     }
     int num = ;
     //printf("%d\n",maxz);
     while(a != &&b != )
     {
         if(pre[a][b][kk] == )
         {
             ans[num++] = s1[a-];
             kk = pre1[a][b][kk];
             a --;
             b --;
         }
         else if(pre[a][b][kk] == )
         {
             kk = pre1[a][b][kk];
             a --;
         }
         else if(pre[a][b][kk] == )
         {
             kk = pre1[a][b][kk];
             b --;
         }
         else
             break;
     }
     for(i = num-; i >= ; i --)
     {
         printf("%c",ans[i]);
     }
     printf("\n");
     return ;
 }