HDU3247 AC自动机+dp

题意：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

题解：所有串建AC自动机。对以资源串结尾的结点跑bfs，求出到其他资源串结尾的最小距离。当前结点的fail结点不能入队列，因为当前结点读下一个字符可能会遇到禁止字符串，而fail结点读相同字符可能没有遇到禁止字符串...所以每次直接将nex的可行结点入队列。(fail结点不入队列，直接设为与当前结点距离为0应该也是可行的，然而WA了)很多时候我都开始质疑数据问题了...

如果当前字符串本身包含另一个资源串，那距离不应当是0吗？有一种解释是dp的时候经过 | 操作就可以了。

 

#include <bits/stdc++.h>
#include <cstdio>
#include <iostream>
#include <algorithm>
#define ll long long
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define st first
#define nd second
#define mp make_pair
#define pii pair<int, int>
#define gg puts("gg");
#define local
//#define out1
using namespace std;
const int N  = 6e4+10;
const int inf = 0x3f3f3f3f;
int id(char c){
    return c == '1';
}
struct Tire{
    int nex[N][2], fail[N],end[N];
    int root, L;
    int node[15], tot;
    int newnode(){
        nex[L][0] = nex[L][1] = -1;
        end[L] = 0;
        return L++;
    }
    void init(){
        L = tot = 0;
        root = newnode();
    }
    void insert(char* s, int tag){
        int now = root;
        for(int i = 0; s[i]; i++){
            int p = id(s[i]);
            if(nex[now][p] == -1)
                nex[now][p] = newnode();
            now = nex[now][p];
        }
        if(tag >= 0)
            end[now] |= 1<<tag;
        else end[now] = -1;
    }
    void build(){
        queue<int> Q;
        fail[root] = root;
        for(int i = 0; i < 2; i++){
            int& u = nex[root][i];
            if(u == -1)
                u = root;
            else{
                fail[u] = root;
                Q.push(u);
            }
        }
        while(!Q.empty()){
            int now = Q.front();
            Q.pop();
            for(int i = 0; i < 2; i++){
                int& u = nex[now][i];
                if(u == -1)
                    u = nex[ fail[now] ][i];
                else{
                    fail[u] = nex[ fail[now] ][i];
                    end[u] |= end[ fail[u] ];
                    Q.push(u);
                }
            }
        }
    }
};
Tire ac;
char s[N];
void gmin(int& a, int b){
    if(a > b) a = b;
}
int dis[N], ve[N], tot;
int w[555][555];

void bfs(int node, int i){
    memset(dis, 0x3f, sizeof(dis));
    queue<int> Q;
    Q.push(node); dis[node] = 0;
    int s = 0;

    #ifdef out1
    int j = node;
    while(ac.fail[j] != ac.root){
        j = ac.fail[j];
        dis[j] = 0;
    }
    dis[j] = 0;
    #endif

    while(!Q.empty()){
        int sz = Q.size();
        s++;
        while(sz--){
            int f = Q.front(); Q.pop();
            for(int j = 0; j < 2; j++){
                int to = ac.nex[f][j];
                if(dis[to] < inf||ac.end[to] < 0) continue ;
                Q.push(to);
                dis[to] = s;

                #ifdef out1
                while(ac.fail[to] != ac.root&&dis[ac.fail[to]] == inf){
                    to = ac.fail[to];
                    dis[to] = s;
                }
                #endif
            }
        }
    }
    for(int j = 0; j < tot; j++)
        w[i][j] = dis[ ve[j] ];
}
int dp[555][1<<11];
int main(){
    #ifdef locl
    freopen("in", "r", stdin);
    #ifdef out1
        freopen("out1", "w", stdout);
    #else
        freopen("out2", "w", stdout);
    #endif // out1
    #endif // locl
    int n, m, t, ca = 1;
    while(~scanf("%d%d", &n, &m), n+m){
        ac.init();

        memset(w, 0x3f, sizeof(w));
        for(int i = 0; i < n; i++){
            w[i][i] = 0;
            scanf("%s", s);
            ac.insert(s, i);
        }
        for(int i = 0; i < m; i++){
            scanf("%s", s);
            ac.insert(s, -1);
        }
        ac.build();

        tot = 0;
        for(int i = 0; i < ac.L; i++)
            if(!i||ac.end[i] > 0) ve[tot++] = i;
        memset(w, 0x3f, sizeof(w));
        for(int i = 0; i < tot; i++)
            bfs(ve[i], i);
//        for(int i = 0; i < tot; i++){
//            for(int j = 0; j < tot; j++)
//                printf("%10d ", w[i][j]);
//            puts("");
//        }

        memset(dp, 0x3f, sizeof(dp));
        for(int i = 0; i < tot; i++)
            dp[i][ ac.end[ve[i]] ] = w[0][i];
        for(int i = 0; i < (1<<n); i++){
            for(int j = 0; j < tot; j++){
                if( dp[j][i] < 0x3f3f3f3f ) for(int k = 0; k < tot; k++)
                    if(j != k&&w[j][k] < 0x3f3f3f3f)
                        gmin(dp[k][i|ac.end[ve[k]]], dp[j][i]+w[j][k]);
            }
        }

//        for(int i = 0; i < n; i++)
//            for(int j = 0; j < (1<<n); j++)
//                printf("%d %d: %d\n", i, j, dp[i][j]);

        int ans = inf;
        for(int i = 0; i < tot; i++)
            ans = min(ans, dp[i][ (1<<n)-1 ]);
        printf("%d\n", ans);
    }
    return 0;
}
/*
2 2
1110
0111
101
1001

3 3
0001000
00100
010
11
111
1111
*/