[AC + dp] HDU 2457 DNA Repair

Question:

Give n words, and then a string of STR (containing only A, G, C, and T)

Question: How many characters can be modified for STR without containing N words?

Ideas:

Create a trie graph and create an automatic DP

DP [I] [J] indicates that at least a few characters must be modified on the J node for I letters that pass through Str.

Trie*P=Node[J]->Next[K]; If (P->Mark) Continue; // inaccessibleDP[I] [P->ID] =Min(DP[I] [P->ID],DP[I-1] [J] + (GETID(Fuck[I])! =K));

That is, step I goes from node J to the corresponding K. If K is not the letter in this step, modify it and take the minimum value.

When the traversal length is Len, the minimum value is used.

Code:

#include"cstdlib"#include"cstdio"#include"cstring"#include"cmath"#include"queue"#include"algorithm"#include"iostream"#include"map"#include"string"#define inf 9999999using namespace std;int triecont;struct trie{    int mark,id;    trie *next[5],*fail;    trie()    {        memset(next,0,sizeof(next));        fail=NULL;        mark=id=0;    }};trie *root,*node[1234];int getid(char x){    if(x=='A') return 0;    else if(x=='C') return 1;    else if(x=='G') return 2;    return 3;}void init(char *v){    trie *p=root;    for(int i=0;v[i];i++)    {        int tep=getid(v[i]);        if(p->next[tep]==NULL)        {            p->next[tep]=new trie();            node[triecont]=p->next[tep];            p->next[tep]->id=triecont++;        }        p=p->next[tep];    }    p->mark=1;}void getac(){    queue<trie*>q;    q.push(root);    while(!q.empty())    {        trie *p=q.front();        q.pop();        for(int i=0;i<4;i++)        {            if(p->next[i]==NULL)            {                if(p==root) p->next[i]=root;                else p->next[i]=p->fail->next[i];            }            else            {                if(p==root) p->next[i]->fail=root;                else p->next[i]->fail=p->fail->next[i];                q.push(p->next[i]);                trie *tep=p;                while(tep!=root && tep->mark!=1)                    tep=tep->fail;                p->mark=tep->mark;            }        }    }}int dp[1234][1234];char fuck[1234];int main(){    int n;    int cas=1;    while(scanf("%d",&n),n)    {        triecont=0;        root=new trie();        node[triecont]=root;        root->id=triecont++;        while(n--)        {            char x[33];            scanf("%s",x);            init(x);        }        getac();        //for(int i=0;i<triecont;i++) printf("%d:%d\n",i,node[i]->mark);        scanf("%s",fuck+1);        int len=strlen(fuck+1);        for(int i=0;i<=len;i++)  for(int j=0;j<triecont;j++) dp[i][j]=inf;        dp[0][0]=0;        for(int i=1;i<=len;i++)        {            for(int j=0;j<triecont;j++)            {                if(dp[i-1][j]==inf) continue;                for(int k=0;k<4;k++)                {                    trie *p=node[j]->next[k];                    if(p->mark) continue;                    dp[i][p->id]=min(dp[i][p->id],dp[i-1][j]+(getid(fuck[i])!=k));                }            }        }        int ans=inf;        for(int i=0;i<triecont;i++) ans=min(ans,dp[len][i]);        printf("Case %d: %d\n",cas++,ans==inf?-1:ans);    }    return 0;}

[AC + dp] HDU 2457 DNA Repair