AOJ1370: Hidden Anagrams(hash)

Posted 2021-02-12 zwfymqz

题意

题目链接

Sol

直接对出现的次数hash即可，复杂度(O(26n^2))

一开始没判长度条件疯狂wa

#include<bits/stdc++.h> 
//#define int long long 
#define LL long long 
#define ull unsigned long long 
using namespace std;
const int MAXN = 4001, mod = 1e9 + 7, INF = 1e9 + 10;
const double eps = 1e-9;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename A> inline void debug(A a){cout << a << '
';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, M;
char s1[MAXN], s2[MAXN];
map<ull, bool> mp;
ull base = 29;
int num[27];
ull get() {
    ull now = 0;
    for(int i = 0; i < 26; i++) 
        now = now * base + num[i];
    return now;
}
signed main() {
    scanf("%s", s1 + 1);
    scanf("%s", s2 + 1);
    N = strlen(s1 + 1); M = strlen(s2 + 1);
    for(int len = min(N, M); len >= 1; len--) {
        mp.clear();
        memset(num, 0, sizeof(num));
        for(int i = 1; i <= N; i++) {
            num[s1[i] - 'a']++;
            if(i > len) num[s1[i - len] - 'a']--;
            if(i >= len) mp[get()] = 1;
        }
        memset(num, 0, sizeof(num));
        for(int i = 1; i <= M; i++) {
            num[s2[i] - 'a']++;
            if(i > len) num[s2[i - len] - 'a']--;
            if(i >= len && mp[get()]) {
                cout << len << '
';
                return 0;
            }
        }
    }
    puts("0");
    return 0;
}