[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
", 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[8] = {0, 0, -1, 1, 1, 1, -1, -1};
const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
const int maxn = 1e3 + 7;
const int md = 1e9 + 9;
const int inf = 1e9 + 7;
const LL inf_L = 1e18 + 7;
const double pi = acos(-1.0);
const double eps = 1e-6;

template<class T>T gcd(T a, T b){return b==0?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[300010];
    void GetNext(char s[]) {
        mem0(next);
        next[0] = next[1] = 0;
        for(int i = 1; s[i]; i++) {
            int j = next[i];
            while(j && s[i] != s[j]) j = next[j];
            next[i + 1] = s[j] == s[i]? j + 1 : 0;
        }
    }
};
KMP kmp;
char s[300010], s2[300010], str[600010];
char hsh[300];
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 + 1) >> 1;
        for (int i = 0; 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 < 2 * res; i ++) str[i] = hsh[str[i % res]];
        str[2 * res] = 0;
        puts(str);
    }
    return 0;
}