URAL - 1158 Censored! AC自动机+dp - ProLightsfx的技术成长

Censored! AC自动机+dp

Source

URAL - 1158

ACM ICPC 2001. Northeastern European Region, Northern Subregion

My Solution

题意：给出n个不同的字符，用这n个字符构成长度为m的字符串，要求每个串的子串都不出现给定的p个串中的任一个，求满足要求的字符串的个数。

AC自动机+dp

因为构成的最终串是由一个字符一个字符添加到字符串尾部构成的，那么如果一个串的后缀如果恰好是某个给定串的前缀时，这个串就可能最终成为非法串。

用k个给定串建立AC自动机，然后从根节点开始递推，

dpij表示递推到第j个字符当前在自动机上的i号节点时的方案数，如果下一个节点是k，且不是危险节点，则把dpij加到dp[k][i+1]里，

跑一遍，然后答案就是所有非危险节点的方案数的和(其实危险节点上都是0)。因为危险节点是给定串的终点或者其后缀节点是危险节点的点，遍历到危险节点的点上的方案必定是包含了给定串的方案，故不能记录这些。

此外这里dpij会很多，故要用高精度整数。笔者自己收藏的大整数类的版，可行长太长了，所以MLE了一发，调小了才过。⊙﹏⊙‖∣

复杂度 O(n^3)

#include 
#include 
#include 
#include 
#include 


using namespace std;
const int CHAR_SIZE = 51;
const int MAX_SIZE = 105;
map<char, int> mp;

struct AC_Machine{
    int ch[MAX_SIZE][CHAR_SIZE], danger[MAX_SIZE], fail[MAX_SIZE];
    int sz;

    void init(){
        sz = 1;
        memset(ch[0], 0, sizeof ch[0]);
        memset(danger, 0, sizeof danger);
    }

    void _insert(char *s){
        int n = strlen(s);
        int u = 0, c;
        for(int i = 0; i < n; i++){
            c = mp[s[i]];
            if(!ch[u][c]){
                memset(ch[sz], 0, sizeof ch[sz]);
                danger[sz] = 0;
                ch[u][c] = sz++;
            }
            u = ch[u][c];
        }
        danger[u] = 1;
    }

    void _build(){
        queue Q;
        fail[0] = 0;
        for(int c = 0, u; c < CHAR_SIZE; c++){
            u = ch[0][c];
            if(u){Q.push(u); fail[u] = 0;}
        }
        int r;
        while(!Q.empty()){
            r = Q.front();
            Q.pop();
            danger[r] |= danger[fail[r]];
            for(int c = 0, u; c < CHAR_SIZE; c++){
                u = ch[r][c];
                if(!u){ch[r][c] = ch[fail[r]][c]; continue; }
                fail[u] = ch[fail[r]][c];
                Q.push(u);
            }
        }
    }
}ac;

char s[MAX_SIZE];

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define MAX_L 205 //最大长度，可以修改
using namespace std;

class bign
{
public:
    int len, s[MAX_L];//数的长度，记录数组
//构造函数
    bign();
    bign(const char*);
    bign(int);
    bool sign;//符号 1正数 0负数
    string toStr() const;//转化为字符串，主要是便于输出
    friend istream& operator>>(istream &,bign &);//重载输入流
    friend ostream& operator<<(ostream &,bign &);//重载输出流 //重载复制 bign operator=(const char*); bign operator=(int); bign operator=(const string); //重载各种比较 bool operator>(const bign &) const;
    bool operator>=(const bign &) const;
    bool operator<(const bign &) const;
    bool operator<=(const bign &) const; bool operator==(const bign &) const; bool operator!=(const bign &) const; //重载四则运算 bign operator+(const bign &) const; bign operator++(); bign operator++(int); bign operator+=(const bign&); bign operator-(const bign &) const; bign operator--(); bign operator--(int); bign operator-=(const bign&); bign operator*(const bign &)const; bign operator*(const int num)const; bign operator*=(const bign&); bign operator/(const bign&)const; bign operator/=(const bign&); //四则运算的衍生运算 bign operator%(const bign&)const;//取模（余数） bign factorial()const;//阶乘 bign Sqrt()const;//整数开根（向下取整） bign pow(const bign&)const;//次方 //一些乱乱的函数 void clean(); ~bign(); }; #define max(a,b) a>b ? a : b
#define min(a,b) a>(istream &in, bign &num)
{
    string str;
    in>>str;
    num=str;
    return in;
}

ostream &operator<<(ostream &out, bign &num)
{
    out<<num.toStr();
    return out;
}

bign bign::operator=(const char *num)
{
    memset(s, 0, sizeof(s));
    char a[MAX_L] = "";
    if (num[0] != '-')
        strcpy(a, num);
    else
        for (int i = 1; i < strlen(num); i++)
            a[i - 1] = num[i];
    sign = !(num[0] == '-');
    len = strlen(a);
    for (int i = 0; i < strlen(a); i++)
        s[i] = a[len - i - 1] - 48;
    return *this;
}

bign bign::operator=(int num)
{
    char temp[MAX_L];
    sprintf(temp, "%d", num);
    *this = temp;
    return *this;
}

bign bign::operator=(const string num)
{
    const char *tmp;
    tmp = num.c_str();
    *this = tmp;
    return *this;
}

bool bign::operator<(const bign &num) const
{
    if (sign^num.sign)
        return num.sign;
    if (len != num.len)
        return len < num.len; for (int i = len - 1; i >= 0; i--)
        if (s[i] != num.s[i])
            return sign ? (s[i] < num.s[i]) : (!(s[i] < num.s[i])); return !sign; } bool bign::operator>(const bign&num)const
{
    return num < *this;
}

bool bign::operator<=(const bign&num)const { return !(*this>num);
}

bool bign::operator>=(const bign&num)const
{
    return !(*this<num); } bool bign::operator!=(const bign&num)const { return *this > num || *this < num;
}

bool bign::operator==(const bign&num)const
{
    return !(num != *this);
}

bign bign::operator+(const bign &num) const
{
    if (sign^num.sign)
    {
        bign tmp = sign ? num : *this;
        tmp.sign = 1;
        return sign ? *this - tmp : num - tmp;
    }
    bign result;
    result.len = 0;
    int temp = 0;
    for (int i = 0; temp || i < (max(len, num.len)); i++)
    {
        int t = s[i] + num.s[i] + temp;
        result.s[result.len++] = t % 10;
        temp = t / 10;
    }
    result.sign = sign;
    return result;
}

bign bign::operator++()
{
    *this = *this + 1;
    return *this;
}

bign bign::operator++(int)
{
    bign old = *this;
    ++(*this);
    return old;
}

bign bign::operator+=(const bign &num)
{
    *this = *this + num;
    return *this;
}

bign bign::operator-(const bign &num) const
{
    bign b=num,a=*this;
    if (!num.sign && !sign)
    {
        b.sign=1;
        a.sign=1;
        return b-a;
    }
    if (!b.sign)
    {
        b.sign=1;
        return a+b;
    }
    if (!a.sign)
    {
        a.sign=1;
        b=bign(0)-(a+b);
        return b;
    }
    if (a<b)
    {
        bign c=(b-a);
        c.sign=false;
        return c;
    }
    bign result;
    result.len = 0;
    for (int i = 0, g = 0; i < a.len; i++)
    {
        int x = a.s[i] - g;
        if (i < b.len) x -= b.s[i]; if (x >= 0) g = 0;
        else
        {
            g = 1;
            x += 10;
        }
        result.s[result.len++] = x;
    }
    result.clean();
    return result;
}

bign bign::operator * (const bign &num)const
{
    bign result;
    result.len = len + num.len;

    for (int i = 0; i < len; i++)
        for (int j = 0; j < num.len; j++)
            result.s[i + j] += s[i] * num.s[j];

    for (int i = 0; i < result.len; i++)
    {
        result.s[i + 1] += result.s[i] / 10;
        result.s[i] %= 10;
    }
    result.clean();
    result.sign = !(sign^num.sign);
    return result;
}

bign bign::operator*(const int num)const
{
    bign x = num;
    bign z = *this;
    return x*z;
}
bign bign::operator*=(const bign&num)
{
    *this = *this * num;
    return *this;
}

bign bign::operator /(const bign&num)const
{
    bign ans;
    ans.len = len - num.len + 1;
    if (ans.len < 0) { ans.len = 1; return ans; } bign divisor = *this, divid = num; divisor.sign = divid.sign = 1; int k = ans.len - 1; int j = len - 1; while (k >= 0)
    {
        while (divisor.s[j] == 0) j--;
        if (k > j) k = j;
        char z[MAX_L];
        memset(z, 0, sizeof(z));
        for (int i = j; i >= k; i--)
            z[j - i] = divisor.s[i] + '0';
        bign dividend = z;
        if (dividend < divid) { k--; continue; }
        int key = 0;
        while (divid*key <= dividend) key++;
        key--;
        ans.s[k] = key;
        bign temp = divid*key;
        for (int i = 0; i < k; i++)
            temp = temp * 10;
        divisor = divisor - temp;
        k--;
    }
    ans.clean();
    ans.sign = !(sign^num.sign);
    return ans;
}

bign bign::operator/=(const bign&num)
{
    *this = *this / num;
    return *this;
}

bign bign::operator%(const bign& num)const
{
    bign a = *this, b = num;
    a.sign = b.sign = 1;
    bign result, temp = a / b*b;
    result = a - temp;
    result.sign = sign;
    return result;
}

bign bign::pow(const bign& num)const
{
    bign result = 1;
    for (bign i = 0; i < num; i++)
        result = result*(*this);
    return result;
}

bign bign::factorial()const
{
    bign result = 1;
    for (bign i = 1; i <= *this; i++) result *= i; return result; } void bign::clean() { if (len == 0) len++; while (len > 1 && s[len - 1] == '�')
        len--;
}

bign bign::Sqrt()const
{
    if(*this<0)return -1;
    if(*this<=1)return *this; bign l=0,r=*this,mid; while(r-l>1)
    {
        mid=(l+r)/2;
        if(mid*mid>*this)
            r=mid;
        else
            l=mid;
    }
    return l;
}

bign::~bign()
{
}

bign dp[MAX_SIZE][CHAR_MAX];

int main()
{
    #ifdef LOCAL
    freopen("1.in", "r", stdin);
    //freopen("1.out", "w", stdout);
    int T = 1;
    while(T--){
    #endif // LOCAL
    //ios::sync_with_stdio(false); cin.tie(0);

    int n, m, p;
    scanf("%d%d%d", &n, &m, &p);
    scanf("%s", s);
    for(int i = 0; i < n; i++){mp[s[i]] = i;}
    ac.init();
    while(p--){
        scanf("%s", s);
        ac._insert(s);
    }
    int i, j, k;
    for(i = 0; i < ac.sz; i++){
        for(j = 0; j <= m; j++){
            dp[i][j] = 0;
        }
    }
    ac._build();
    dp[0][0] = 1;
    for(i = 1; i <= m; i++){
        for(j = 0; j < ac.sz; j++){
            for(k = 0; k < n; k++){
                if(!ac.danger[ac.ch[j][k]]){
                    dp[ac.ch[j][k]][i] += dp[j][i-1];
                }
            }
        }
    }
    bign ans = 0;
    for(i = 0; i < ac.sz; i++){ans += dp[i][m];}
    cout << ans << endl;

    #ifdef LOCAL
    cout << endl;
    }
    #endif // LOCAL
    return 0;
}

非特殊说明，本博所有文章均为博主原创，未经许可不得转载。

https://www.prolightsfxjh.com/

Thank you!

------from ProLightsfx

URAL – 1158 Censored! AC自动机+dp

Censored! AC自动机+dp

相关推荐

HDU – 2825 Wireless Password AC自动机+状压dp

UVa 1218 Perfect Service dp：树上dp 状态转移方程的优化

UESTC 1131 男神的礼物 dp：最优矩阵链乘 triangulation 双dp

Codeforces Round #420 (Div. 2) E. Okabe and El Psy Kongroo dp+矩阵快速幂

共有 0 条评论

公告

游戏服务器开发 最新文章

1转行游戏开发第二年，开始修炼内功

2搜撤类游戏全服装备防复制方案

3从通用后台开发到游戏服务器开发...

公众号

推荐系统开发 最新文章

1时序数据库与NB时序特征

2从零搭建NB特征平台

3NB特征分类标准

4浅谈NB推荐系统架构-ProLightsfx

5一次推荐系统线上事故的排查与修...

标签云

栏目分类