POJ 3691 DNA repair(AC自动机+DP)

题目链接

能AC还是很开心的...此题没有POJ2778那么难，那个题还需要矩阵乘法，两个题有点相似的。

做题之前，把2778代码重新看了一下，回忆一下当时做题的思路，回忆AC自动机是干嘛的...

状态表示dp[i][j]长度为i的以j串为结束的最小改变数目。AC自动机预处理一下，然后DP。

卡内存+不知道状态数，MLE+RE+WA了多次。

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <queue>
 #include <algorithm>
 #include <cstdlib>
 using namespace std;
 #define N 2222222
 #define INF 10000000
 int trie[N][];
 int o[N];
 int que[N];
 int fail[N];
 int dp[][];
 int t;
 void CL()
 {
     memset(trie,-,sizeof(trie));
     memset(o,,sizeof(o));
     t = ;
 }
 int judge(char s)
 {
     switch(s)
     {
         case'A':return ;
         case'C':return ;
         case'G':return ;
         case'T':return ;
     }
     return ;
 }
 void insert(char *str)
 {
     int i,len,root;
     root = ;
     len = strlen(str);
     for(i = ;i < len;i ++)
     {
         if(trie[root][judge(str[i])] == -)
         trie[root][judge(str[i])] = t ++;
         root = trie[root][judge(str[i])];
     }
     o[root] = ;
 }
 void build_ac()
 {
     int head,tail,front,i;
     head = tail = ;
     for(i = ;i < ;i ++)
     {
         if(trie[][i] != -)
         {
             fail[trie[][i]] = ;
             que[tail++] = trie[][i];
         }
         else
         {
             trie[][i] = ;
         }
     }
     while(head != tail)
     {
         front = que[head++];
         if(o[fail[front]])
         o[front] = ;
         for(i = ;i < ;i ++)
         {
             if(trie[front][i] != -)
             {
                 que[tail++] = trie[front][i];
                 fail[trie[front][i]] = trie[fail[front]][i];
             }
             else
             {
                 trie[front][i] = trie[fail[front]][i];
             }
         }
     }
 }
 int main()
 {
     int n,i,j,k,len,flag,cas = ;
     char str[];
     while(scanf("%d",&n)!=EOF)
     {
         if(n == ) break;
         CL();
         for(i = ;i <= n;i ++)
         {
             scanf("%s",str);
             insert(str);
         }
         build_ac();
         scanf("%s",str);
         len = strlen(str);
         for(i = ;i <= len;i ++)
         {
             for(j = ;j <= t;j ++)
             dp[i][j] = INF;
         }
         dp[][] = ;
         for(i = ;i < len;i ++)
         {
             for(j = ;j < t;j ++)
             {
                 if(o[j]) continue;
                 for(k = ;k < ;k ++)
                 {
                     if(o[trie[j][k]]) continue;
                     flag = !(k == judge(str[i]));
                     dp[i+][trie[j][k]] = min(dp[i+][trie[j][k]],dp[i][j] + flag);
                 }
             }
         }
         int ans = INF;
         for(i = ;i < t;i ++)
         {
             ans = min(ans,dp[len][i]);
         }
         printf("Case %d: ",cas ++);
         if(ans == INF)
         printf("-1\n");
         else
         printf("%d\n",ans);
     }
     return ;
 }