hdu3247Resource Archiver（ac自动机+spfa)

链接

这题没想到怎么做，问了下p队长，大悟。。

先求出任意两串的在trie树上的最短距离，期间是不能走到不合法的地方，我是用spfa求得，在更新和加入节点时判断一下是不是合法位置。

求出最短距离之后，找出一条从0出发遍历所有串的最短距离，可以dp出，dp[i][j]表示当前状态以节点j串结尾的最短距离。

枚举一下最后结尾的为哪一个串时距离最短。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
using namespace std;
#define N 60010
#define M 10010
#define LL long long
#define INF 0xfffffff
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;
const int child_num = 2;
int w[15][15];
bool vis[N];
int dis[N];
class AC
{
    private:
    int ch[N][child_num];
    int fail[N];
    int Q[N];
    int val[N],vv[N];
    int id[127];
    int sz ;
    int po[12];
    int dp[2110][12];
    public:
    void init()
    {
        fail[0] = 0;
        id['1'] = 1;
        id['0'] = 0;
    }
    void reset()
    {
        memset(ch[0],0,sizeof(ch[0]));
        memset(val,0,sizeof(val));
        memset(vv,0,sizeof(vv));
        sz=1;
    }
    void insert(char *a ,int key)
    {
        int p =0 ;
        for( ; *a ; a++)
        {
            int d = id[*a];
            if(ch[p][d]==0)
            {
                memset(ch[sz],0,sizeof(ch[sz]));
                ch[p][d] = sz++;
            }
            p = ch[p][d];
        }
        if(key!=-1)
        {
            val[p] = 1<<(key-1);
            po[key] = p;
        }
        else
        {
            vv[p] = 1;
        }
    }
    void construct()
    {
        int i,head=0,tail = 0;
        for(i = 0; i < child_num ; i++)
        if(ch[0][i])
        {
            fail[ch[0][i]] = 0;
            Q[tail++] = ch[0][i];
        }
        while(head!=tail)
        {
            int u = Q[head++];
            vv[u]|=vv[fail[u]];
            val[u]|=val[fail[u]];
            for(i = 0 ;i < child_num ; i++)
            {
                if(ch[u][i])
                {
                    fail[ch[u][i]] = ch[fail[u]][i];
                    Q[tail++] = ch[u][i];
                }
                else ch[u][i] = ch[fail[u]][i];
            }
        }
    }
    int spfa(int st,int ed)
    {
        int i;
        memset(vis,0,sizeof(vis));
        for(i = 0; i <= sz ;i++)
        dis[i] = INF;
        dis[st] = 0;
        vis[st] = 1;
        queue<int>q;
        q.push(st);
        while(!q.empty())
        {
            int u = q.front();
            q.pop();
            for(i = 0 ; i < child_num; i++)
            {
                int v = ch[u][i];
                if(vv[v]) continue;
                dis[v] = min(dis[v],dis[u]+1);
                //if(st == 4&&ed==8)
                //cout<<v<<" "<<dis[v]<<" "<<u<<endl;
                if(!vis[v])
                {
                    vis[v] = 1;
                    q.push(v);
                }
            }
        }
        return dis[ed];
    }
    void find_dis(int n)
    {
        int i,j;
        po[0] = 0;
        for(i = 1; i <= n ;i++)
        {
            w[0][i] = w[i][0] = spfa(0,po[i]);
            //cout<<w[0][i]<<endl;
        }
        for(i = 1; i <= n; i++)
            for(j = 1 ; j <= n ;j++)
            {
                w[i][j] = spfa(po[i],po[j]);
               // cout<<w[i][j]<<" "<<i<<" "<<j<<endl;
            }
    }
    void work(int n)
    {
        int i,j,g;
        for(i = 0 ; i < (1<<n) ; i++)
            for(j = 0;j <= n; j++)
            dp[i][j] = INF;
        dp[0][0] = 0;
        for(i = 0 ;i < (1<<n) ; i++)
        {
            for(j = 0 ;j <= n ;j++)
            {
                for(g = 0 ; g <= n ;g++)
                {
                    if((1<<(g-1))&i) continue;
                    dp[i+(1<<(g-1))][g] = min(dp[i+(1<<(g-1))][g],dp[i][j]+w[j][g]);
                }
            }
        }
        int ans = INF;
        for(i = 1 ; i <= n ; i++)
        ans = min(ans,dp[(1<<n)-1][i]);
        cout<<ans<<endl;
    }
}ac;
char vir[50010],re[1010];
int main()
{
    int n,m,i;
    ac.init();
    while(scanf("%d%d",&n,&m)&&n&&m)
    {
        ac.reset();
        for(i = 1; i <= n ;i++)
        {
            scanf("%s",re);
            ac.insert(re,i);
        }
        while(m--)
        {
            scanf("%s",vir);
            ac.insert(vir,-1);
        }
        ac.construct();
        ac.find_dis(n);
        ac.work(n);
    }
    return 0;
}