Resource Archiver

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 100000/100000 K (Java/Others)
Total Submission(s): 1429    Accepted Submission(s): 418


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

題意：給定n個子串，m個病毒串，然後讓合併，使得串儘可能的短，但是不能含有病毒串，

這個題好難，n不超過10，看了別人的思路,先把子串與病毒串插進ac自動機去，子串標記為對應的狀態，病毒串末尾標記為-1，然後把合法的子串節點取出來，然後通過bfs求兩兩之間的

最短路，然後狀態壓縮dp。

程式碼：

/* ***********************************************
Author :xianxingwuguan
Created Time :2014-2-5 17:31:21
File Name :4.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
struct node{
	int next[60010][2],end[60010],fail[60030];
	int root,L;
	int newnode(){
		next[L][0]=next[L][1]=-1;
		end[L++]=0;
		return L-1;
	}
	void init(){
		L=0;
		root=newnode();
	}
	void insert(char *str,int id){
		int len=strlen(str);
		int now=root;
		for(int i=0;i<len;i++){
			int p=str[i]-'0';
			if(next[now][p]==-1)
				next[now][p]=newnode();
			now=next[now][p];
		}
		if(id==-1)
			end[now]=-1;
		else end[now]|=(1<<id);
	}
	void build(){
		queue<int> q;
		fail[root]=root;
		for(int i=0;i<2;i++){
			if(next[root][i]==-1)
				next[root][i]=root;
			else {
				fail[next[root][i]]=root;
				q.push(next[root][i]);
			}
		}
		while(!q.empty()){
			int now=q.front();q.pop();
			if(end[fail[now]]==-1)end[now]=-1;
			else end[now]|=end[fail[now]];
			for(int i=0;i<2;i++){
				if(next[now][i]==-1)
					next[now][i]=next[fail[now]][i];
				else {
					fail[next[now][i]]=next[fail[now]][i];
					q.push(next[now][i]);
				}
			}
		}
	}
	int dp[1<<10][60],dis[60][60],pnt[100],cnt,que[100010],dist[100000];
	void bfs(int s){
		for(int i=0;i<L;i++)
			dist[i]=INF;
		dist[pnt[s]]=0;
		int front=0,rear=0;
		que[rear++]=pnt[s];
		while(front<rear){
			int now=que[front++];
			for(int i=0;i<2;i++){
				int u=next[now][i];
				if(end[u]==-1||dist[u]!=INF)
					continue;
				dist[u]=dist[now]+1;
				que[rear++]=u;
			}
		}
		for(int i=0;i<cnt;i++)dis[s][i]=dist[pnt[i]];
	}
	void F1(int n){
         cnt=0;
		 for(int i=0;i<L;i++)
			 if(i==0||end[i]>0)
				 pnt[cnt++]=i;
	//	 cout<<"haha:"<<L<<" "<<cnt<<endl;
		 for(int i=0;i<cnt;i++)bfs(i);
         for(int i=0;i<(1<<n);i++)
			 for(int j=0;j<cnt;j++)
				 dp[i][j]=INF;
		 dp[0][0]=0;
		 for(int i=0;i<(1<<n);i++)
			 for(int j=0;j<cnt;j++){
				 if(dp[i][j]==INF)continue;
				 for(int k=0;k<cnt;k++)
					 dp[i|end[pnt[k]]][k]=min(dp[i|end[pnt[k]]][k],dp[i][j]+dis[j][k]);
			 }
		 int ans=INF;
		 for(int i=0;i<cnt;i++)
			 ans=min(ans,dp[(1<<n)-1][i]);
		 cout<<ans<<endl;
	}

}ac;
char str[10030];
int main()
{
     //freopen("data.in","r",stdin);
     //freopen("data.out","w",stdout);
     int i,j,k,m,n;
     while(~scanf("%d%d",&n,&m)){
		 if(n==0&&m==0)break;
		 ac.init();
		 for(i=0;i<n;i++){
			 scanf("%s",str);
			 ac.insert(str,i);
		 }
		 while(m--){
			 scanf("%s",str);
			 ac.insert(str,-1);
		 }
		 ac.build();
		 ac.F1(n);
	 }
     return 0;
}