UVA - 12633 Super Rooks on Chessboard FFT

Posted 2020-09-26 Meek

题目链接：点这里

题意:

　　给你一个矩阵R*C，n个点，与给的的n个点同一列同一行，同一条主对角线上的点都被染黑

　　问你最后有多少个点没有被染黑

题解：

　　把每一列每一行没有被染黑的x,y找出来，其任意组合起来是没这种情况下的答案（同一条主对角线上的点都被染黑）

　　对于 x - y，我们可以拿来判断两个点是不是相同的一条主对角线上

　　那么对x、-y进行任意组合，FFT加速

　　总的答案就是，没有被染过行数*没有被染过的列数 - （找不到相同的x- y）

　　具体看代码吧

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double pi = acos(-1.0);
const int N = 1e5+10, M = 1e3+20,inf = 2e9,mod = 1e9+7;


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 , t , k ; i < n ; ++ i ) {
        for ( j = 0 , t = i , k = n >> 1 ; 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 = Complex ( 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 ;
}

Complex s[N*4],t[N*4];
int R,C,n,T,col[N],row[N],d[N*2];
int main() {
    int cas = 1;
    scanf("%d",&T);
    while(T--) {
        scanf("%d%d%d",&R,&C,&n);
        for(int i = 1; i <= R; ++i) col[i] = 1;
        for(int i = 1; i <= C; ++i) row[50000 - i] = 1;
        memset(d,0,sizeof(d));
        for(int i = 1; i <= n; ++i) {
            int x,y;
            scanf("%d%d",&x,&y);
            d[x - y + 50000] = 1;
            col[x] = 0;
            row[50000 - y] = 0;
        }
        int n1 = 1;
        while(n1 < 50000*2) n1<<=1;
        
        for(int i = 0; i <= R; ++i)s[i] = Complex(col[i],0);
        for(int i = R+1; i < n1; ++i)s[i] = Complex(0,0);
        
        for(int i = 0; i < 50000 - C; ++i)t[i] = Complex(0,0);
        for(int i = 1; i <= C; ++i) t[50000 - i] = Complex(row[50000 - i],0);
        for(int i = 50000; i < n1; ++i) t[i] = Complex(0,0);
        
        FFT(s,n1,1);FFT(t,n1,1);
        for(int i = 0; i < n1; ++i) s[i] = s[i] * t[i];
        FFT(s,n1,-1);
        LL A = 0, B = 0;
        for(int i = 1; i <= R; ++i) if(col[i]) A++;
        for(int i = 1; i <= C; ++i) if(row[50000 - i]) B++;
        LL ans = A*B;
        for(int i = 1+50000 - C; i <= 50000*2 - 1; ++i) {
            int x = (int ) (s[i].r + 0.5);
            if(d[i]) ans -= x;
        }
        printf("Case %d: %lld\n",cas++,ans);
    }
    return 0;
}