[hdu4300] next数组的应用

题意：给你一个密文和明文的对应表以及一个密文+明文的字符串，明文可能只出现前面的一部分（也就是说是原明文的前缀），求最短的明文。

思路：首先密文的长度至少占到一半，所以先把那一半解密，问题转化为找一个最长的后缀使得和前缀相等，并且满足后缀长度不超过原串的一半，显然用next数组即可解决。

 #pragma comment(linker, "/STACK:10240000,10240000")

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstdlib>
 #include <cstring>
 #include <map>
 #include <queue>
 #include <deque>
 #include <cmath>
 #include <vector>
 #include <ctime>
 #include <cctype>
 #include <set>
 #include <bitset>
 #include <functional>
 #include <numeric>
 #include <stdexcept>
 #include <utility>

 using namespace std;

 #define mem0(a) memset(a, 0, sizeof(a))
 #define mem_1(a) memset(a, -1, sizeof(a))
 #define lson l, m, rt << 1
 #define rson m + 1, r, rt << 1 | 1
 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
 #define rep_down1(a, b) for (int a = b; a > 0; a--)
 #define all(a) (a).begin(), (a).end()
 #define lowbit(x) ((x) & (-(x)))
 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 #define pchr(a) putchar(a)
 #define pstr(a) printf("%s", a)
 #define sstr(a) scanf("%s", a)
 #define sint(a) scanf("%d", &a)
 #define sint2(a, b) scanf("%d%d", &a, &b)
 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
 #define pint(a) printf("%d\n", a)
 #define test_print1(a) cout << "var1 = " << a << endl
 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
 #define mp(a, b) make_pair(a, b)
 #define pb(a) push_back(a)

 typedef unsigned int uint;
 typedef long long LL;
 typedef pair<int, int> pii;
 typedef vector<int> vi;

 const int dx[] = {, , -, , , , -, -};
 const int dy[] = {-, , , , , -, , - };
 const int maxn = 1e3 + ;
 const int md = 1e9 + ;
 const int inf = 1e9 + ;
 const LL inf_L = 1e18 + ;
 const double pi = acos(-1.0);
 const double eps = 1e-;

 template<class T>T gcd(T a, T b){return b==?a:gcd(b,a%b);}
 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
 template<class T>T condition(bool f, T a, T b){return f?a:b;}
 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
 int make_id(int x, int y, int n) { return x * n + y; }

 struct KMP {
     int next[];
     void GetNext(char s[]) {
         mem0(next);
         next[] = next[] = ;
         for(int i = ; s[i]; i++) {
             int j = next[i];
             while(j && s[i] != s[j]) j = next[j];
             next[i + ] = s[j] == s[i]? j +  : ;
         }
     }
 };
 KMP kmp;
 char s[], s2[], str[];
 char hsh[];
 int main() {
     //freopen("in.txt", "r", stdin);
     int T;
     cin >> T;
     rep_up0(cas, T) {
         cin >> s;
         int len = strlen(s);
         rep_up0(i, len) {
             hsh[s[i]] = 'a' + i;
         }
         cin >> s2;
         len = strlen(s2);
         int pos = (len + ) >> ;
         for (int i = ; i < pos; i ++) s2[i] = hsh[s2[i]];
         kmp.GetNext(s2);
         int res = kmp.next[len];
         while (len - res < pos) res = kmp.next[res];
         res = len - res;
         rep_up0(i, pos) str[i] = s[s2[i] - 'a'];
         for (int i = pos; i < res; i ++) str[i] = s2[i];
         for (int i = res; i <  * res; i ++) str[i] = hsh[str[i % res]];
         str[ * res] = ;
         puts(str);
     }
     return ;
 }