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] , 0 , 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 , 0 , 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