UVa 1401 Remember the word

方法：dp + trie

比较明显可以想出一个dp，对于给定的string str，d[i] ＝ 表示str.substr(i) 的方法数， d[str.length()] = 1, d[i] = sum(d[i+x.length()], x 是 str.substr(i) 的prefix)，最后答案是d[0]。状态转移的时候，可以通过走由可行x组成的trie来判断。

code:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#include <fstream>
#include <cassert>
#include <unordered_map>
#include <unordered_set>
#include <cmath>
#include <sstream>
#include <time.h>
#include <complex>
#include <iomanip>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a))
#define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a))
#define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a))
#define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a))
#define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
#define FOREACH(a,b) for (auto &(a) : (b))
#define rep(i,n) FOR(i,0,n)
#define repn(i,n) FORN(i,1,n)
#define drep(i,n) DFOR(i,n-1,0)
#define drepn(i,n) DFOR(i,n,1)
#define MAX(a,b) a = Max(a,b)
#define MIN(a,b) a = Min(a,b)
#define SQR(x) ((LL)(x) * (x))
#define Reset(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(v) v.begin(),v.end()
#define ALLA(arr,sz) arr,arr+sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(all(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr,sz) sort(ALLA(arr,sz))
#define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
#define PERMUTE next_permutation
#define TC(t) while(t--)
#define forever for(;;)
#define PINF 1000000000000
#define newline '
'

#define test if(1)if(0)cerr
using namespace std;
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;
typedef pair<double,double> dd;
typedef pair<char,char> cc;
typedef vector<ii> vii;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> l4;
const double pi = acos(-1.0);

const int maxnode = 400000+5;
const int sigma_size = 26;


struct Trie
{
    int g[maxnode][sigma_size];
    int val[maxnode];
    int sz;
    int newnode()
    {
        Reset(g[sz], 0);
        val[sz] = 0;
        return sz++;
    }
    void init() { sz = 1; Reset(g[0], 0); };
    int idx(char c) { return c-'a'; };

    void insert(const string &str, int v)
    {
        int u = 0, n = str.length();
        for (int i = 0; i < n; ++i)
        {
            int c = idx(str[i]);
            if (!g[u][c]) g[u][c] = newnode();
            u = g[u][c];
        }
        val[u] = v;
    }
};

Trie solver;
string str;
string input[4000];
int n;
const int mod = 20071027;
int d[300001];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int kase = 0;
    while (cin >> str >> n)
    {
        solver.init();
        for (int i = 0; i < n; ++i) cin >> input[i];
        for (int i = 0; i < n; ++i) solver.insert(input[i], 1);
        int len = str.length();
        Reset(d, 0);
        d[len] = 1;
        for (int i = len-1; i >= 0; --i)
        {
            int u = 0;
            for (int j = i; j < len; ++j)
            {
                int c = solver.idx(str[j]);
                if (!solver.g[u][c])
                {
                    break;
                }
                u = solver.g[u][c];
                if (solver.val[u]) d[i] = (d[i] + d[j+1]) % mod;
            }
        }
        cout << "Case " << ++kase << ": " << d[0] << newline;
    }
}