hdu3247 Resource Archiver AC自动机+状态压缩DP

Resource Archiver

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 100000/100000 K (Java/Others)
Total Submission(s): 1928    Accepted Submission(s): 625

Problem Description

Great! Your new software is almost finished! The only thing left to do is archiving all your n resource files into a big one.
Wait a minute… you realized that it isn’t as easy as you thought. Think about the virus killers. They’ll find your software suspicious, if your software contains one of the m predefined virus codes. You absolutely don’t want this to happen.
Technically, resource files and virus codes are merely 01 strings. You’ve already convinced yourself that none of the resource strings contain a virus code, but if you make the archive arbitrarily, virus codes can still be found somewhere.
Here comes your task (formally): design a 01 string that contains all your resources (their occurrences can overlap), but none of the virus codes. To make your software smaller in size, the string should be as short as possible.

 

Input

There will be at most 10 test cases, each begins with two integers in a single line: n and m (2 <= n <= 10, 1 <= m <= 1000). The next n lines contain the resources, one in each line. The next m lines contain the virus codes, one in each line. The resources and virus codes are all non-empty 01 strings without spaces inside. Each resource is at most 1000 characters long. The total length of all virus codes is at most 50000. The input ends with n = m = 0.

 

Output

For each test case, print the length of shortest string.

 

Sample Input

2 21110011110110010 0

 

Sample Output

5

  把上面N个串拼起来，可重叠，不能包含下面的M个病毒串，拼出的串长度最小是多少。

  构造AC自动机，标记病毒节点和非病毒节点包含的串。由于N只有10，dp的时候只要在单词的尾节点之间进行就行了，把所有单词的尾节点存下来，BFS预处理出这些节点之间的最短路，dp[i][j]表示第i个单词尾节点已经包含的串状态为j的最小长度，进行状态转移。

#include<cstdio>#include<cstring>#include<algorithm>#include<string>#include<iostream>#include<queue>using namespace std;const int MAXN=1010;const int MAXM=15;const int MAXL=100010;const int MAXNODE=60010;const int LOGMAXN=50;const int INF=0x3f3f3f3f;const int SIGMA_SIZE=2;const int MOD=20090717;int N,M,K;int p[MAXNODE],dis[MAXM][MAXM],d[MAXNODE];int dp[MAXM][(1<<11)+10];struct AC{    int ch[MAXNODE][SIGMA_SIZE];    int val[MAXNODE];    int f[MAXNODE];    int virus[MAXNODE];    int sz;    void clear(){        memset(ch[0],0,sizeof(ch[0]));        val[0]=0;        virus[0]=0;        sz=1;    }    int idx(char c){        return c-'0';    }    int insert(char* s,int v){        int u=0;        for(int i=0;s[i];i++){            int c=idx(s[i]);            if(!ch[u][c]){                memset(ch[sz],0,sizeof(ch[sz]));                val[sz]=0;                virus[sz]=0;                ch[u][c]=sz++;            }            u=ch[u][c];        }        if(v>0) val[u]|=v;        else virus[u]=1;    }    void get_fail(){        queue<int> q;        f[0]=0;        for(int c=0;c<SIGMA_SIZE;c++){            int u=ch[0][c];            if(u){                f[u]=0;                q.push(u);            }        }        while(!q.empty()){            int r=q.front();            q.pop();            for(int c=0;c<SIGMA_SIZE;c++){                int u=ch[r][c];                if(!u){                    ch[r][c]=ch[f[r]][c];                    continue;                }                q.push(u);                f[u]=ch[f[r]][c];                val[u]|=val[f[u]];                virus[u]|=virus[f[u]];            }        }    }}ac;void get_dis(){    p[0]=0;    K=1;    memset(dis,-1,sizeof(dis));    for(int u=0;u<ac.sz;u++) if(!ac.virus[u]&&ac.val[u]){        p[K++]=u;    }    queue<int> q;    for(int i=0;i<K;i++){        memset(d,-1,sizeof(d));        d[p[i]]=0;        q.push(p[i]);        while(!q.empty()){            int u=q.front();            q.pop();            for(int c=0;c<SIGMA_SIZE;c++){                int v=ac.ch[u][c];                if(!ac.virus[v]&&d[v]<0){                    d[v]=d[u]+1;                    q.push(v);                }            }        }        for(int j=0;j<K;j++) dis[i][j]=d[p[j]];    }}void DP(){    memset(dp,INF,sizeof(dp));    dp[0][0]=0;    int MAXSTATE=(1<<N)-1;    for(int s=0;s<=MAXSTATE;s++)        for(int i=0;i<K;i++) if(dp[i][s]!=INF){            for(int j=0;j<K;j++) if(dis[i][j]!=-1){                dp[j][s|ac.val[p[j]]]=min(dp[j][s|ac.val[p[j]]],dp[i][s]+dis[i][j]);            }        }    int ans=INF;    for(int i=0;i<K;i++) if(dp[i][MAXSTATE]!=INF){        ans=min(ans,dp[i][MAXSTATE]);    }    printf("%d\n",ans);}char str[MAXN];int main(){    freopen("in.txt","r",stdin);    while(scanf("%d%d",&N,&M)!=EOF&&(N||M)){        ac.clear();        for(int i=0;i<N;i++){            scanf("%s",str);            ac.insert(str,1<<i);        }        for(int i=0;i<M;i++){            scanf("%s",str);            ac.insert(str,-1);        }        ac.get_fail();        get_dis();        DP();    }    return 0;}