题目:给出n个资源,m个病毒,将资源串拼接成一个串,必须包含所有的资源串,可以重叠,但是不能包含病毒
问最小的长度为多少
将所有的资源串和病毒串都放在Trie树里建立起来,当然作上相应的标记。然后建立fail指针之后
从资源串的结尾出发,BFS,记录能到达其它资源串结尾的步数。得到所有资源串结尾状态的距离邻接阵。
之后是状态压缩DP
dp[i][j]表示资源串的状态为i时,最后一个已取资源串是j时,最短长度。
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define L(i) i<<1
#define R(i) i<<1|1
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-9
#define maxn 60010
#define MOD 1000000007
int n,m;
int dp[1050][13];
int path[13][13];
int pos[13];
struct Trie
{
int next[maxn][2],fail[maxn],en[maxn];
int root,L,cnt;
void init()
{
L = 0;
pos[0] = 0;
root = newnode();
}
int newnode()
{
for(int i = 0; i <= 1; i++)
next[L][i] = -1;
en[L++] = 0;
return L-1;
}
void Insert(char a[],int flag)
{
int now = root;
int len = strlen(a);
for(int i = 0; i < len; i++)
{
int x;
if(a[i] == '0')
x = 0;
else
x = 1;
if(next[now][x] == -1)
next[now][x] = newnode();
now = next[now][x];
}
en[now] = flag;
}
void build()
{
queue<int> Q;
fail[root] = root;
for(int i = 0; i <= 1; 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(en[fail[now]] == -1)
en[now] = en[fail[now]];
else
en[now] |= en[fail[now]];
for(int i = 0; i <= 1; 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]);
}
}
}
}
void bfs(int u,int cnt)
{
queue<int> q;
q.push(pos[u]);
int dis[maxn];
memset(dis,-1,sizeof(dis));
dis[pos[u]] = 0;
while(!q.empty())
{
int v = q.front();
q.pop();
for(int i = 0; i < 2; i++)
{
int k = next[v][i];
if(dis[k] < 0 && en[k] != -1)
{
dis[k] = dis[v] + 1;
q.push(k);
}
}
}
for(int i = 0; i < cnt; i++)
path[u][i] = dis[pos[i]];
}
void solve()
{
int cnt = 1;
for(int i = 0; i < L; i++)
if(en[i] > 0)
pos[cnt++] = i;
for(int i = 0; i < cnt; i++)
bfs(i,cnt);
// for(int i = 0; i <= n; i++)
// {for(int j = 0; j <= n; j++)
// printf("%d ",path[i][j]);printf("\n");}
memset(dp,INF,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] < INF)
{
for(int k = 0; k < cnt; k++)
{
if(path[j][k] < 0 || j == k)
continue;
dp[i|en[pos[k]]][k] = min(dp[i|en[pos[k]]][k],dp[i][j]+path[j][k]);
}
}
int ans = INF;
for(int i = 0; i < cnt; i++)
ans = min(ans,dp[(1<<n)-1][i]);
printf("%d\n",ans);
}
} ac;
int main()
{
int t,C = 1;
//scanf("%d",&t);
while(scanf("%d%d",&n,&m) && (n+m))
{
ac.init();
char s[1010];
for(int i = 0; i < n; i++)
{
scanf("%s",s);
ac.Insert(s,1<<i);
}
for(int i = 0; i < m; i++)
{
scanf("%s",s);
ac.Insert(s,-1);
}
ac.build();
ac.solve();
}
return 0;
}