BZOJ1195 HNOI2006 最短母串

非主流做法 ， 此题就不提示了。

看到此题 ， Fuxey的第一感觉 ， 这玩意我以前见过啊 ， AC自动机上的动态规划。 中午回学校啪啪敲了一份AC自动机上的状压DP。

WA WA WA

然后检查 ， 代码。 起初我认为同层(即同一个状态S) 互相影响并不会更优 ， 所以不会影响答案， 然而需要考虑同层的转化。
不要紧 ， Dijkstra上

TLE……

然而n^2的Dijkstra并不能过 ， 优先队列优化一下呗。 然而A了……

（Fuxey建议大家还是写主流的状压DP写法 ， 除非想练写AC自动机 ）

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <deque>
#include <queue>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <stack>
#include <cassert>


using namespace std;
const int maxn = 12;
const int maxlen = 51;
const int sigmaSize = 26;
const int INF = 0x3f3f3f3f;
const int maxnode = maxlen*maxn;

int n;
char s[maxlen];
int d[maxnode][1<<maxn] , loop[maxnode][1<<maxn];
int pre[maxnode][1<<maxn];

__inline int id(char c) { return c-'A'; } 
struct ACaut
{
    int n;
    queue<int> q;
    int g[maxnode][sigmaSize];
    int mark[maxnode] , v[maxnode] , f[maxnode];
    vector<int> l[maxnode] , r[maxnode]; int book[maxnode];
    void init() { n = 0; }

    void add(int num)
    {
        int len = strlen(s) , u = 0;
        for(int i=0;i<len;i++)
        {
            int c = id(s[i]);
            if(!g[u][c])
            {
                g[u][c] = ++n;
                mark[n] = f[n] = 0;
                memset(g[n] , 0 , sizeof(g[n]));
            }
            u = g[u][c];
        }
        mark[u] |= 1<<(num-1);
    }

    void getFail()
    {
        mark[0] = v[0] = f[0] = 0;
        for(int i=0;i<sigmaSize;i++)
        {
            int t = g[0][i];

            if(!t) continue;
            q.push(t);
            if(mark[t]) v[t] = mark[t];
        }

        while(!q.empty())
        {
            int now = q.front(); q.pop();

            for(int i=0;i<sigmaSize;i++)
            {
                int t = g[now][i] , j;
                if(!t)
                {
                    g[now][i] = g[f[now]][i];
                    continue;
                }
                else q.push(t);

                for(j=f[now];j && !g[j][i];j = f[j]) ;
                f[t] = g[j][i]?g[j][i]:0;
                v[t] = v[f[t]];
                if(mark[t]) v[t] |= mark[t];
            }
        }
    }

    void dpInit()
    {
        for(int i=0;i<=n;i++)  memset(d[i] , -1 , sizeof(d[i]));
    } 


    void dp(int s)
    {
        if(s==(1<<::n)-1) { for(int i=0;i<=n;i++) d[i][s] = 0; return; }

        for(int i=0;i<=n;i++) l[i].clear() , r[i].clear();
        for(int u=0;u<=n;u++)
        {
            int& now = d[u][s] = INF; book[u] = 0;
            for(int i=0;i<sigmaSize;i++)
            {
                int t = g[u][i];
                if((s|v[t]) == s && t!=u) l[t].push_back(u) , r[t].push_back(i);
                else if(d[t][s|v[t]]+1 < now) now = d[t][s|v[t]]+1 , pre[u][s] = i;
            }
        }

        priority_queue<pair<int, int> > pq;
        for(int i=0;i<=n;i++) if(d[i][s]!=INF) pq.push(make_pair(-d[i][s], i));

        while(pq.size())
        {
            int now = pq.top().second , v = -pq.top().first; pq.pop();
            if(d[now][s]!=v) continue;

            for(int i=0;i<l[now].size();i++)
            {
                int k = l[now][i];
                if(d[k][s] > v+1 || (d[k][s]==v+1 && pre[k][s] > r[now][i]))
                {
                    pre[k][s] = r[now][i];  
                    if(d[k][s] == v+1)  continue;
                    d[k][s] = v+1;
                    pq.push(make_pair(-d[k][s], k));
                }
            }
        }
    }

    void print()
    {
        int s = 0;
        for(int i=0 , j=1;j<=d[0][0];j++)
        {
            printf("%c" , 'A'+pre[i][s]);
            i = g[i][pre[i][s]];
            s |= v[i];
        }
    }
}solver;

int main()
{   
    cin>>n;

    solver.init();
    for(int i=1;i<=n;i++)
    {
        scanf("%s" , s);
        solver.add(i);
    }   


    solver.getFail();   
    for(int i=(1<<n)-1;i>=0;i--) solver.dp(i);
    solver.print(  );

    return 0;
}

附上最卡时限的数据:
TestData 9:

Input:

12
AABBBABBBBBAAAAABABBBAAABBABBAABABAABAAABAAA
AAABBBAAABABBBABBBABABAABAABABBBBABBBAABBA
BBBBABAABABBBABABABBBAAABBABBBABAABBBAAABABBABB
BABBBAABBABABBABBBAAABABAABBAA
BBAABAAABAABBBBBABBBBBAABBBBABBBBBBABAAAAABAA
ABAABBBBBABBAABAABBAAABABBBAAABBBBBABAAB
ABBBBBABBAABAABBAAABABBBAAABBBBBABAABAAAAABBBBBAB
BAAABABAABBAA
BAABAAABABBBBBABBBABAAAAAABBAAABBBAAABABBB
AAABABBABBBAAAABBBBAAAAABBBBABBBBABBAABAAABAAAB
AAABABBBAAABBBBBABAABAAAAABBBBBABABABBAAABABAB
BBAABBBBABABBAABABABBBBBBBBABAAABABAAAAA

Output:

BAABAAABABBBBBABBBABAAAAAABBAAABBBAAABABBBABBBABABAABAABABBBBABBBAABBABABBABBBAAABABAABBAABAAABAABBBBBABBBBBAABBBBABBBBBBABAAAAABAABBBBBABBAABAABBAAABABBBAAABBBBBABAABAAAAABBBBBABABABBAAABABABBAABBBBABABBAABABABBBBBBBBABAAABABAAAAABBBBABAABABBBABABABBBAAABBABBBABAABBBAAABABBABBBAAAABBBBAAAAABBBBABBBBABBAABAAABAAABBBABBBBBAAAAABABBBAAABBABBAABABAABAAABAAA