【BNU】33943 Super Rooks on Chessboard 【FFT】
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【BNU】33943 Super Rooks on Chessboard
UVA上的题,然而我怎么会蠢到去UVA呢!(其实是百度首先跳出来的是BNU
题目分析:
设
首先我们考虑没有主对角线覆盖这一条件时,总共的没有被覆盖的面积就是
现在我们考虑主对角线影响。
考虑没有被车覆盖的行的集合
考虑没有被车覆盖的列的集合
考虑被车覆盖的对角线的集合
那么对角线
考虑
做一遍
心得:
my code:
#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std ;typedef long long LL ;#define clr( a , x ) memset ( a , x , sizeof a )#define cpy( a , x ) memcpy ( a , x , sizeof a )const int MAXN = 100005 ;const double pi = acos ( -1.0 ) ;struct Complex { double r , i ; Complex () {} Complex ( double r , double i ) : r ( r ) , i ( i ) {} Complex operator + ( const Complex& t ) const { return Complex ( r + t.r , i + t.i ) ; } Complex operator - ( const Complex& t ) const { return Complex ( r - t.r , i - t.i ) ; } Complex operator * ( const Complex& t ) const { return Complex ( r * t.r - i * t.i , r * t.i + i * t.r ) ; }} ;void FFT ( Complex y[] , int n , int rev ) { for ( int i = 1 , j , k , t ; i < n ; ++ i ) { for ( j = 0 , k = n >> 1 , t = i ; k ; k >>= 1 , t >>= 1 ) j = j << 1 | t & 1 ; if ( i < j ) swap ( y[i] , y[j] ) ; } for ( int s = 2 , ds = 1 ; s <= n ; ds = s , s <<= 1 ) { Complex wn ( cos ( rev * 2 * pi / s ) , sin ( rev * 2 * pi / s ) ) , w ( 1 , 0 ) , t ; for ( int k = 0 ; k < ds ; ++ k , w = w * wn ) { for ( int i = k ; i < n ; i += s ) { y[i + ds] = y[i] - ( t = w * y[i + ds] ) ; y[i] = y[i] + t ; } } } if ( rev == -1 ) for ( int i = 0 ; i < n ; ++ i ) y[i].r /= n ;}bool visx[MAXN] , visy[MAXN] , visd[MAXN << 2] ;Complex x1[MAXN << 2] , x2[MAXN << 2] ;int r , c , q , n ;void solve ( int T ) { int x , y , numx = 0 , numy = 0 ; clr ( visx , 0 ) ; clr ( visy , 0 ) ; clr ( visd , 0 ) ; scanf ( "%d%d%d" , &r , &c , &q ) ; for ( int i = 0 ; i < q ; ++ i ) { scanf ( "%d%d" , &x , &y ) ; visx[x] = 1 ; visy[y] = 1 ; visd[x - y + c] = 1 ; } int n = 1 ; while ( n < r + c + 1 ) n <<= 1 ; for ( int i = 1 ; i <= r ; ++ i ) numx += visx[i] == 0 ; for ( int i = 1 ; i <= c ; ++ i ) numy += visy[i] == 0 ; for ( int i = 1 ; i <= r ; ++ i ) x1[i - 1] = Complex ( visx[i] == 0 , 0 ) ; for ( int i = r ; i < n ; ++ i ) x1[i] = Complex ( 0 , 0 ) ; for ( int i = 1 ; i <= c ; ++ i ) x2[c - i] = Complex ( visy[i] == 0 , 0 ) ; for ( int i = c ; i < n ; ++ i ) x2[i] = Complex ( 0 , 0 ) ; FFT ( x1 , n , 1 ) ; FFT ( x2 , n , 1 ) ; for ( int i = 0 ; i < n ; ++ i ) x1[i] = x1[i] * x2[i] ; FFT ( x1 , n , -1 ) ; LL ans = ( LL ) numx * numy ; for ( int i = 0 ; i < n ; ++ i ) ans -= ( LL ) ( x1[i].r + 0.5 ) * visd[i + 1] ; printf ( "Case %d: %lld\n" , T , ans ) ;}int main () { int T ; scanf ( "%d" , &T ) ; for ( int i = 1 ; i <= T ; ++ i ) solve ( i ) ; return 0 ;}
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