POJ 1625 Censored! （AC自动机+DP+大数加法）

题目大意

• 给出包含N个字符的字典，组成长度为M的句子，求无禁止单词的句子有多少个。禁止单词有P个，单词长度不超过10，所以字符的ASCII码均大于32(1 <= N <= 50,1 <= M <= 50,0 <= P <= 10)。

分析

• 状态dp[i][j]表示第i步时，到达状态j的字符串数，途中不能经过危险结点

• 则状态转移方程为

  dp[i][j] = sum(dp[i-1][k])，其中j是k的后继状态

• 很明显，所以结果加起来超过long long。所以需要用到大数的加法

• 构造AC自动机时，失败结点为危险的结点也要变成危险结点。
• 输入字符的ASCII码可能超过128，所以需要用的映射，不然RE

代码

#include <iostream>
#include <cstring>
#include <queue>
#include <cstdio>
#include <algorithm>

using namespace std;
const int maxn = 110;
const int N = 51;
const int sigma_size = 50;
const int M = 100;

struct AC {
    int ch[maxn][sigma_size];
    int val[maxn];
    int sz;
    char s[N];
    int idx(char c) {
        int L = 0,R = strlen(s) - 1;
        while(L <= R)
        {
            int M = (L + R) / 2;
            if(c == s[M]) return M;
            if(c > s[M]) L = M + 1;
            else R = M - 1;
        }
        return L;
    }
    void init() {
        sz=1; val[0]=0;
        cin >> s;
        int len = strlen(s);
        sort(s,s + len);
        memset(ch[0],0,sizeof(ch[0]));
    }
    void insert(string const &s,int v)
    {
        int cur = 0,len = s.length();
        for(int i = 0; i < len; i++)
        {
            int u = idx(s[i]);
            if(!ch[cur][u]) {
                memset(ch[sz],sizeof(ch[sz]));
                val[sz] = 0;
                ch[cur][u] = sz++;
            }
            cur = ch[cur][u];
        }
        val[cur] = v;
    }
    int f[maxn];
    void getfail()
    {
        queue<int> Q; f[0] = 0;
        for(int i = 0; i < sigma_size; i++) {
            int v = ch[0][i];
            if(v) {f[v] = 0; Q.push(v);}
        }
        while(!Q.empty())
        {
            int cur = Q.front(); Q.pop();
            for(int i = 0; i < sigma_size; i++){
                int u = ch[cur][i];
                if(!u) {ch[cur][i] = ch[f[cur]][i]; continue;}
                Q.push(u);
                int j = f[cur];
                while(j && !ch[j][i]) j = f[j];
                f[u] = ch[j][i];
                if(val[ch[j][i]]) val[u] = 1;
            }
        }
    }
    void ADD(int *A,int *B) //大数加法A=A+B
    {
        int len1 = 0,len2 = 0;
        for(int i = M-1; i >= 0; i--) if(A[i]) {len1 = i; break;}
        for(int i = M-1; i >= 0; i--) if(B[i]) {len2 = i; break;}
        int len = max(len1,len2);
        for(int i = 0; i <= len; i++) {
            A[i] += B[i];
            A[i+1] += A[i] / 10;
            A[i] %= 10;
        }
    }
    int dp[N][maxn][M]; //dp[i][j]表示第i步时，到达状态j的字符串数
    void DP(int m,int n)
    {
        memset(dp,sizeof(dp)); dp[0][0][0] = 1;
        for(int i = 1; i <= m; i++) for(int j = 0; j < sz; j++) {
            if(val[j]) continue;
            for(int k = 0; k < n; k++)
            {
                int cur = ch[j][idx(s[k])];
                if(val[cur]) continue;
                ADD(dp[i][cur],dp[i-1][j]);
            }
        }
        int ans[M] = {0},len = 0;
        for(int i = 0; i < sz; i++) if(!val[i]) ADD(ans,dp[m][i]);
        for(int i = M-1; i >= 0; i--) if(ans[i]) {len = i; break;}
        for(int i = len; i >= 0; i--) cout << ans[i];
        cout << endl;
    }
};

AC ac;
int main()
{
    //cout << sizeof(ac.dp) / 1024 << endl;
    int n,m,p;
    while(cin >> n >> m >> p)
    {
        getchar(); ac.init();
        for(int i = 1; i <= p; i++) {
            string str; cin >> str;
            ac.insert(str,i);
        }
        ac.getfail();
        ac.DP(m,n);
    }
}

数据

• 50 50 10
  qwertyuiop[]asdfghjkl;’zxcvbnm,./QWERTYUIOP{}|AS
  aegaegu
  etoijqt
  tquqi
  witowwt
  zxcjnc
  oeit
  potieq
  iojge
  nvoq
  piqper
  88816479228348672447914159817057714124274948616722531360571673747292358424682407
  63290

• 32 50 10 、￥ウЖ┆???q忏溴骁栝觌祉铒? Β 驿? ?馥? 癌ē? く．? く．铽 Β‘飒 ウq憝? 驿Γ飒 隘ルΒ°? 1721978265757417133897998525905218434633371225724869729302087935758854628748