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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; }
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原文地址:http://www.cnblogs.com/wushuaiyi/p/4296969.html