Uvalive 4978 fields and farmers

Uvalive_4978

The difficulty of this question lies in how to convert it into a mathematical model. For detailed conversion ideas, refer to the official question: http://cepc10.ii.uni.wroc.pl/solutions.html.

After the transformation of the mathematical model, the problem becomes that a number of 8-bit binary numbers are given, and a number of numbers and these numbers are calculated as 11111111, this can be solved using the exclusion principle.

First, we agree that a [I] indicates that the final result has only the number of solutions where I bit is 0, so what we need at last is a [0]. then, we agree that f [I] indicates the total number of methods in which any one of the eight bits in the result is 0, and the other BITs do not matter. In this way, F [0] is 2 ^ n. When calculating f [I], first enumerate which I bit is 0, and then find that the number of I bit 0 has a total of X, so the result of the last or the number of I-Bit 0 is 2 ^ X-1, accumulate these 2 ^ X-1 values obtained when I bit is 0 and then observe the features of the expression f [I.

Obviously f [0] contains all the situations, so f [0] = A [0] + A [1] + A [2] +... + A [8] And then f [1] are considered. First, there is only one number of solutions whose bit is 0 in the final result, after we calculate the f [I] method, each scheme is calculated only once. Therefore, one of F [1] is a [1], then, for the number of solutions with only two bits 0 in the final result, we calculate the number once when the first bit is 0, and the number is calculated once when the second bit is 0, then, in F [1], there will be a C () * A [2], that is, 2 * A [2]. Then, the expression f [1] = C () * A [1] + C () is obtained) * A [2] +... + C (8, 1) * A [8].

Similarly, we can get f [2] = C () * A [2] + C () * A [3] +... + C (8, 2) * A [8], you can push the remaining f [I.

In this case, we can find that a [0] = f [0]-f [1] + F [2]-f [3] +... -F [7] + F [8]. Then, F [I] is calculated based on the above method to obtain a [0.

# Include <stdio. h> # Include < String . H> # Include <Algorithm># Define Maxd 1000010 # Define INF 0x3f3f3f # Define Mod4 1000000007 Typedef  Long   Long  Ll;  Int DX [] = { 1 , 1 , 0 , 0 , 1 , 1 ,1 , 1 }, Dy [] = { 0 , 0 , 1 , 1 , 1 , 1 ,- 1 ,- 1 }, D [ 8  ];  Int N, H [ 260 ], X [maxd], Y [maxd];  Char  Ch;  Void  Init (){  Int  I, J; scanf (  "  % D  " ,& N); D [  0 ] = D [ 2 ] = D [ 4 ] = D [ 6 ] = Inf, d [1 ] = D [ 3 ] = D [ 5 ] = D [ 7 ] =- INF;  For (I = 0 ; I <n; I ++ ) {Scanf (  "  % D  " , & X [I], & Y [I]); D [  0 ] = STD: min (d [ 0 ], X [I]), d [1 ] = STD: max (d [ 1  ], X [I]); D [  2 ] = STD: min (d [ 2 ], Y [I]), d [ 3 ] = STD: max (d [ 3  ], Y [I]); D [  4 ] = STD: min (d [ 4 ], X [I] + Y [I]), d [ 5 ] = STD: max (d [ 5 ], X [I] + Y [I]); D [ 6 ] = STD: min (d [ 6 ], X [I]-y [I]), d [ 7 ] = STD: max (d [ 7 ], X [I]- Y [I]);} memset (H,  0 , Sizeof  (H ));  For (I = 0 ; I <n; I ++ ){  Int T = 0  ; For (J = 0 ; J < 8 ; J ++ ){  If (Dx [J] * X [I] + dy [J] * Y [I] = d [J]) t = T < 1 | 1  ;  Else T <= 1  ;} ++ H [T] ;}} ll powmod (ll,  Int N) {ll ans = 1  ;  While  (N ){  If (N & 1 ) Ans = ans * A % MOD; n >>= 1 , A = A * A % MOD ;}  Return  Ans ;}  Void DFS ( Int Cur,Int St, Int N, Int & Ans ){  If (Cur = 8  ){  Int I, num = 0  ;  For (I = 0 ; I < 256 ; I ++ ) If ((~ St & I) = 0 ) Num + = H [I];  If (N & 1 ) Ans = (ANS-powmod ( 2 , Num) % MOD;  Else Ans = (ANS + powmod ( 2 , Num) % MOD;  Return  ;} DFS (cur + 1 , St, N +1  , ANS); DFS (cur + 1 , St | 1 < Cur, N, ANS );}  Void  Solve (){  Int Ans = 0  ; DFS (  0 , 0 , 0  , ANS );  If (ANS <0 ) Ans + = MOD; printf (  "  % D \ n  "  , ANS );}  Int  Main (){  Int  T; scanf (  "  % D  " ,& T );  While (T --) {Init (); solve ();}  Return   0  ;}