LuoGuP3667

╰+攻爆jí腚メ 2023-06-02 11:59 22阅读 0赞

[LuoGuP3667][]  
这题对我来说难的一批(题意理解错三遍,垃圾翻译,还是英文原题面好)  
就是给你\\(2n\\)个串,要你找一个区间,使得前\\(n\\)个串的这个区间不能和后\\(n\\)个串中的区间有任何一个相同,求一个最短长度.  
这显然可以二分,不过听取了\\(dalao\\)的建议,我选择了枚举左端点,二分右端点的方式.  
怎么\\(check\\)呢?你把所有串\\(hash\\)一下,然后你显然可以任意查询某一个区间的\\(hash\\)值.  
把后\\(n\\)个串的对应区间的\\(hash\\)值插入一个\\(set\\),直接查询就行了.  
\\(Code:\\)

#include <algorithm> #include <iostream> #include <cstdlib> #include <cstring> #include <cstdio> #include <string> #include <vector> #include <queue> #include <cmath> #include <ctime> #include <map> #include <set> #define MEM(x,y) memset ( x , y , sizeof ( x ) ) #define rep(i,a,b) for (int i = (a) ; i <= (b) ; ++ i) #define per(i,a,b) for (int i = (a) ; i >= (b) ; -- i) #define pii pair < int , int > #define X first #define Y second #define rint read<int> #define int long long #define pb push_back #define mid ( ( l + r ) >> 1 ) using std::set ; using std::pair ; using std::max ; using std::min ; using std::priority_queue ; using std::vector ; using std::swap ; using std::sort ; using std::unique ; using std::greater ; template < class T > inline T read () { T x = 0 , f = 1 ; char ch = getchar () ; while ( ch < '0' || ch > '9' ) { if ( ch == '-' ) f = - 1 ; ch = getchar () ; } while ( ch >= '0' && ch <= '9' ) { x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ; ch = getchar () ; } return f * x ; } const int inf = 1e15 ; const int N = 1e3 + 100 ; const int base = 233 ; const int mod = 1e9 + 7 ; int n , m , g[N] , ans ; char s[N] ; unsigned int hash[N][N] ; std::set < int > st ; inline int query (int x , int l , int r) { return ( hash[x][r] - hash[x][l-1] * g[r-l+1] % mod + mod ) % mod ; } inline bool check (int l , int r) { st.clear () ; rep ( i , n + 1 , ( n << 1 ) ) st.insert ( query ( i , l , r ) ) ; rep ( i , 1 , n ) if ( st.find ( query ( i , l , r ) ) != st.end () ) return false ; return true ; } signed main (int argc , char * argv[]) { n = rint () ; m = rint () ; rep ( i , 1 , ( n << 1 ) ) { scanf ("%s" , s + 1 ) ; rep ( j , 1 , m ) hash[i][j] = ( hash[i][j-1] * base % mod + s[j] ) % mod ; } g[0] = 1 ; rep ( i , 1 , m ) g[i] = g[i-1] * base % mod ; ans = inf ; rep ( i , 1 , m ) { int l = i + 1 , r = m , res = ( m << 1 ) ; while ( l <= r ) { if ( check ( i , mid ) ) res = mid , r = mid - 1 ; else l = mid + 1 ; } ans = min ( ans , res - i + 1 ) ; } printf ("%lld\n" , ans ) ; return 0 ; }

转载于:https://www.cnblogs.com/Equinox-Flower/p/11499146.html

[LuoGuP3667]: https://www.luogu.org/problem/P3667