USACO Longest Prefix 【水】

用Dp的思想解决了这道题目，也就是所谓的暴力= = 

题意：给出一个集合，一个字符串，找出这个字符串的最长前缀，使得前缀可以划分为这个集合中的元素（集合中的元素可以不全部使用）。

还不会Trie 树QAQ

Source Code:

/*
ID: wushuai2
PROG: prefix
LANG: C++
*/
//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler
#include <stdio.h>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cmath>
#include <stack>
#include <string>
#include <map>
#include <set>
#include <list>
#include <queue>
#include <vector>
#include <algorithm>
#define Max(a,b) (((a) > (b)) ? (a) : (b))
#define Min(a,b) (((a) < (b)) ? (a) : (b))
#define Abs(x) (((x) > 0) ? (x) : (-(x)))
#define MOD 1000000007
#define pi acos(-1.0)
#define RV(num) ((num) > 0 ? 0 : 1)

using namespace std;

typedef long long           ll      ;
typedef unsigned long long  ull     ;
typedef unsigned int        uint    ;
typedef unsigned char       uchar   ;

template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}
template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;}

const double eps = 1e-7      ;
const int M = 660000         ;
const ll P = 10000000097ll   ;
const int INF = 0x3f3f3f3f   ;
const int MAX_N = 20         ;
const int MAXSIZE = 101000000;

string str[220];
char a[2200000];
bool vis[220000];
int str_num, len, len_a;

int main() {
    ofstream fout ("prefix.out");
    ifstream fin ("prefix.in");
    int i, j, k, l, m, n, t, s, c, w, q, num;
    for(;;){
        fin >> str[++str_num];
        if(str[str_num].compare(".") == 0)  break;
    }
    while(fin >> a[len_a++])
    vis[0] = true;
    for(i = 0; i < len_a; ++i){
        if(!vis[i]) continue;
        for(j = 1; j < str_num; ++j){
            int length_str = str[j].length();
            if(length_str + i >= len_a)   continue;
            for(k = 0; k < length_str; ++k){
                if(a[i + k] != str[j][k])   break;
            }
            if(k == length_str){
                checkmax(len, length_str + i);
                vis[length_str + i] = true;
            }
        }
    }
    fout << len << endl;


    fin.close();
    fout.close();
    return 0;
}