AC自动机 + 二维最短路 HDU 4511 小明系列故事――女友的考验

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这个题还是比较好想的。

首先将所有不可行方案建立AC自动机，然后跑最短路。

首先将小明放在(sta = 0,pos = 0)处，sta表示AC自动机上点的编号，pos表示坐标点的编号。

根据pos枚举下一次可以到达的地方[pos+1,n],然后sta在自动机上移动，如果某一步会使sta位于有标记的节点，那么这一步是不可行。

#include <iostream>
#include<time.h>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<string>
#include<map>
#include<vector>
#include <algorithm>
#include <queue>
#include <cmath>

#define LL long long
#define ULL unsigned long long

using namespace std;

const int MAXS = 55,MAXN  = 510;
const int MATN = 70;

struct MAT
{
    int row,col;
    ULL mat[MATN][MATN];

    void Init(int R,int C,int val)
    {
        row = R,col = C;

        for(int i = 1;i <= row; ++i)
            for(int j = 1;j <= col; ++j)
                    mat[i][j] = (i == j ? val : 0);
    }

    MAT Multi(MAT c,LL MOD)
    {
        MAT tmp;
        tmp.Init(this->row,c.col,0);

        int i,j,k;
       for(k = 1;k <= this->col; ++k)
            for(i = 1;i <= tmp.row; ++i)
                for(j = 1;j <= tmp.col; ++j)
                    tmp.mat[i][j] += (this->mat[i][k]*c.mat[k][j]);

        return tmp;
    }

    MAT Quick(LL n,LL MOD)
    {
        MAT res,tmp = *this;

        res.Init(row,col,1);

        while(n)
        {
            if(n&1)
                res = res.Multi(tmp,MOD);
            tmp = tmp.Multi(tmp,MOD);
            n >>= 1;
        }

        return res;
    }

    void Output()
    {
        cout<<"         ****************        "<<endl;
        int i,j;
        for(i = 1;i <= row; ++i)
        {
            for(j = 1;j <= col; ++j)
                    printf("%3lld ",mat[i][j]);
            puts("");
        }
        cout<<"         &&&&&&&&&&&&&       "<<endl;
    }

};

struct Trie
{
    int next[MAXS];
    int fail;
    int flag;
} st[MAXN];

queue<int> q;

struct Pos
{
    int x,y;
}pos[55];

struct ACautomaton
{
    int Top,root;

    int  Creat()
    {
        memset(st[Top].next,-1,sizeof(st[Top].next));
        st[Top].flag = 0;
        st[Top].fail = -1;
        return Top++;
    }

    void Init()
    {
        Top = 0;
        root = Creat();
    }

    inline int CalIndex(int c)
    {
        return c;
    }

    void Insert(int *s)
    {
        int i = 0,tr = root,tmp;

        while(s[i] != -1)
        {
            tmp = CalIndex(s[i]);
            if(st[tr].next[tmp] == -1)
                st[tr].next[tmp] = Creat();
            tr = st[tr].next[tmp],++i;
        }
        st[tr].flag = 1;
    }

    void GetFail()
    {
        st[root].fail = -1;
        q.push(root);

        int f,t;

        while(q.empty() == false)
        {
            f = q.front();
            q.pop();

            for(int i = 0; i < MAXS; ++i)
            {
                if(st[f].next[i] != -1)
                {
                    t = st[f].fail;

                    while(t != -1 &&  st[t].next[i] == -1)
                        t = st[t].fail;

                    if(t == -1)
                        st[st[f].next[i]].fail = root;
                    else
                        st[st[f].next[i]].fail = st[t].next[i];

                    q.push(st[f].next[i]);
                }
            }
        }
    }

    int Match(char *s)
    {
        int i ,tr = root,tmp;
        int ans = 0;
        for(i = 0; s[i] != '\0'; ++i)
        {
            if(s[i] < 'A' || 'Z' < s[i])
            {
                tr = root;
                continue;
            }
            tmp = CalIndex(s[i]);

            while(tr != -1 && st[tr].next[tmp] == -1 )
                tr = st[tr].fail;

            if(tr == -1)
            {
                tr = root;
                continue;
            }

            tr = st[tr].next[tmp];

            tmp = tr;

            while(tmp != root && st[tmp].flag != -1)
            {
                if(st[tmp].flag)
                    ans++,st[tmp].flag = -1;
            }
        }
        return ans;
    }
};

int num[10];

double dis[510][55];

double val[55][55];

struct Q
{
    int s,p;
};

queue<Q> que;

double Cal(int P1,int P2)
{
    if(P1 == 0 && P2 == 1)
        return 0;
    Pos p1 = pos[P1];
    Pos p2 = pos[P2];
    return sqrt((p1.x+0.0-p2.x)*(p1.x+0.0-p2.x) + (p1.y+0.0-p2.y)*(p1.y+0.0-p2.y) + 0.0);
}

int main()
{
    
    int n,m;
    int i,k,j;

    ACautomaton AC;

    while(scanf("%d %d",&n,&m)  && (n||m))
    {
        for(i = 1;i <= n; ++i)
            scanf("%d %d",&pos[i].x,&pos[i].y);

        AC.Init();

        while(m--)
        {
            scanf("%d",&k);
            for(i = 0;i < k; ++i)
                scanf("%d",&num[i]);
            num[k] = -1;
            AC.Insert(num);
        }

        AC.GetFail();

        for(i = 1;i <= n; ++i)
        {
            for(j = 1;j <= n; ++j)
                val[i][j] = sqrt((pos[i].x-pos[i].x)*(pos[i].x-pos[i].x) + (pos[i].y-pos[i].y)*(pos[i].y-pos[i].y));
        }

        for(i = 1;i <= n; ++i)
            val[i][0] = val[0][i] = 0;

        for(i = 0;i < AC.Top; ++i)
            for(j = 1;j <= n; ++j)
                dis[i][j] = -1;

        dis[0][0] = 0;

        que.push((Q){0,0});

        Q f,t;

        while(que.empty() == false)
        {
           f = que.front();
           que.pop();

           for(i = f.p+1;i <= (f.p == 0 ? 1 : n); ++i)
            {
                int tr = f.s;

                while(tr != -1)
                {
                    if(st[tr].next[i] != -1 && st[st[tr].next[i]].flag != 0)
                        break;
                    tr = st[tr].fail;
                }

                if(tr != -1)
                    continue;

                tr = f.s;

                while(tr != -1)
                {
                    if(st[tr].next[i] != -1)
                        break;
                    tr = st[tr].fail;
                }

                if(tr == -1)
                    tr = 0;
                else
                    tr = st[tr].next[i];

                if(dis[tr][i] == -1 || dis[tr][i] > dis[f.s][f.p] + Cal(f.p,i))
                {
                    dis[tr][i] = dis[f.s][f.p] + Cal(f.p,i);

                    que.push((Q){tr,i});
                }
            }
        }
        double Min = -1;
        for(i = 0;i < AC.Top; ++i)
        {
                if(dis[i][n] != -1)
                {
                    if(Min == -1)
                        Min = dis[i][n];
                    else
                        Min = min(Min,dis[i][n]);
                }
        }
        if(Min == -1)
            printf("Can not be reached!\n");
        else
            printf("%.2lf\n",Min);
    }

    return 0;
}

AC自动机 + 二维最短路 HDU 4511 小明系列故事――女友的考验

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原文地址：http://blog.csdn.net/zmx354/article/details/44487977