HDU 3247 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): 899    Accepted Submission(s): 238

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 2 1110 0111 101 1001 0 0

 

 

Sample Output

5

 

 

Source

2009 “NIT Cup” National Invitational Contest

 

 

Recommend

wujianhua

 

 

 

AC自动机第二题。。。

不解释。。。。慢慢体会吧

/*
HDU 3247 Resource Archiver
给你 n 个串，和 m 个病毒串。要把这 n 个串
连起来来，可以重叠，但不能包含 m 个病毒串
中的任意一个。求把 n 个串连起来的最小长度。

*/
#include <stdio.h>
#include <math.h>
#include <queue>
#include <string.h>
#include <algorithm>
using namespace std;
const int MAXN=60000;//结点的最大数，50000+1000*10
const int MAXL=(1<<10);
const int MAX=2;
const int MAXM=11;
typedef struct Trie_Node
{
    bool virus;//是否病毒
    int end,fail;
    int next[MAX];
}Trie;
Trie tree[MAXN];
char str[50010];
int dis[MAXN];
int pos[MAXM];
int G[MAXM][MAXM];
int dp[MAXL][MAXM];
int size,cnt;


void insert(char *s,int id)
{
    int p=0;
    int i=0;
    while(s[i]!='\0')
    {
        if(tree[p].next[s[i]-'0']==0)
        {
            tree[++size].end=0;
            tree[size].fail=0;
            tree[size].virus=false;
            for(int j=0;j<MAX;j++)
                 tree[size].next[j]=0;
            tree[p].next[s[i]-'0']=size;
        }
        p=tree[p].next[s[i]-'0'];
        i++;
    }
    if(id>=0) tree[p].end=(1<<id);
    else tree[p].virus=true;
}
void BFS()
{
    queue<int>q;
    int temp,p;
    q.push(0);
    while(!q.empty())
    {
        temp=q.front();
        q.pop();
        for(int i=0;i<MAX;i++)
        {
            if(tree[temp].next[i]!=0)
            {
                p=tree[temp].next[i];
                q.push(p);
                if(temp!=0)tree[p].fail=tree[tree[temp].fail].next[i];
                tree[p].end|=tree[tree[p].fail].end;
                tree[p].virus|=tree[tree[p].fail].virus;
            }
            else tree[temp].next[i]=tree[tree[temp].fail].next[i];
        }
    }
}
void Path(int k)
{
    queue<int>q;
    q.push(pos[k]);
    memset(dis,-1,sizeof(dis));
    dis[pos[k]]=0;
    int now,p;

    while(!q.empty())
    {
        now=q.front();
        q.pop();
        for(int i=0;i<MAX;i++)
        {
            p=tree[now].next[i];
            if(dis[p]<0&&!tree[p].virus)
            {
                dis[p]=dis[now]+1;
                q.push(p);
            }
        }
    }
    for(int i=0;i<cnt;i++)
        G[k][i]=dis[pos[i]];

}
inline int MIN(int x,int y)
{
    if(x<0||y<0)return x>y?x:y;
    else return x>y?y:x;
}
void DoIt(int n)
{
    memset(dp,-1,sizeof(dp));
    dp[0][0]=0;
    for(int i=0;i<(1<<n);i++)
       for(int j=0;j<cnt;j++)
       {
           if(dp[i][j]<0)continue;
           for(int k=0;k<cnt;k++)
           {
               if(G[j][k]<0)continue;
               int t=i|tree[pos[k]].end;
               dp[t][k]=MIN(dp[t][k],dp[i][j]+G[j][k]);
           }
       }
    int t=(1<<n)-1;
    int ans=-1;
    for(int i=0;i<cnt;i++)
        ans=MIN(ans,dp[t][i]);
    printf("%d\n",ans);
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n,m;
    while(scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0)break;
        tree[0].end=0;
        tree[0].fail=0;
        tree[0].virus=false;
        for(int i=0;i<MAX;i++)tree[0].next[i]=0;
        size=0;
        for(int i=0;i<n;i++)
        {
            scanf("%s",&str);
            insert(str,i);
        }
        for(int i=0;i<m;i++)
        {
            scanf("%s",&str);
            insert(str,-1);
        }
        BFS();
        cnt=1;
        pos[0]=0;
        for(int i=0;i<=size;i++)
          if(tree[i].end)
             pos[cnt++]=i;
        for(int i=0;i<cnt;i++)
          Path(i);
        DoIt(n);
    }
    return 0;
}