[HDU] 2295 Radar 2점 + 중복 적용

3220 단어 HDULinksDancing
라우터: [HDU] 2295 Radar
제목 분석: 2점 중복 덮어쓰기...없어지다
코드는 다음과 같습니다.
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

#define REP( i , a , b ) for ( int i = a ; i < b ; ++ i )
#define REV( i , a , b ) for ( int i = a - 1 ; i >= b ; -- i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define FOV( i , a , b ) for ( int i = a ; i >= b ; -- i )
#define CLR( a , x ) memset ( a , x , sizeof a )
#define REC( i , A , o ) for ( int i = A[o] ; i != o ; i = A[i] )

const int MAXC = 60 ;
const int MAXR = 60 ;
const int MAXN = 60 ;
const int MAXNODE = 4000 ;
const int INF = 0x3f3f3f3f ;
const double eps = 1e-8 ;

struct Node {
	int x , y ;
	void input () {
		scanf ( "%d%d" , &x , &y ) ;
	}
} ;

struct DLX {
	int L[MAXNODE] , R[MAXNODE] , U[MAXNODE] , D[MAXNODE] ;
	int row[MAXNODE] , col[MAXNODE] ;
	int H[MAXR] , S[MAXC] ;
	bool vis[MAXC] ;
	int deep ;
	int size ;
	int nv ;

	Node a[MAXN] , b[MAXN] ;
	double dis[MAXN][MAXN] ;
	int n , m , k ;
	char s[20] ;
	
	void init () {
		CLR ( H , -1 ) ;
		FOR ( i , 0 , nv ) {
			S[i] = 0 ;
			U[i] = D[i] = i ;
			L[i] = i - 1 ;
			R[i] = i + 1 ;
		}
		L[0] = nv ;
		R[nv] = 0 ;
		size = nv ;
		deep = INF ;
	}
	
	void link ( int r , int c ) {
		++ size ;
		++ S[c] ;
		row[size] = r ;
		col[size] = c ;
		U[size] = U[c] ;
		D[size] = c ;
		D[U[c]] = size ;
		U[c] = size ;
		if ( ~H[r] ) {
			L[size] = L[H[r]] ;
			R[size] = H[r] ;
			L[R[size]] = size ;
			R[L[size]] = size ;
		}
		else
			H[r] = L[size] = R[size] = size ;
	}
	
	void remove ( int c ) {
		REC ( i , D , c ) {
			L[R[i]] = L[i] ;
			R[L[i]] = R[i] ;
		}
	}
	
	void resume ( int c ) {
		REC ( i , U , c ) {
			L[R[i]] = i ;
			R[L[i]] = i ;
		}
	}
	
	int h () {
		int cnt = 0 ;
		CLR ( vis , 0 ) ;
		REC ( i , R , 0 )
			if ( !vis[i] ) {
				++ cnt ;
				vis[i] = 1 ;
				REC ( j , D , i )
					REC ( k , R , j )
						vis[col[k]] = 1 ;
			}
		return cnt ;
	}
	
	int dance ( int d ) {
		if ( d + h () > k )
			return 0 ;
		if ( R[0] == 0 )
			return 1 ;
		int c = R[0] ;
		REC ( i , R , 0 )
			if ( S[c] > S[i] )
				c = i ;
		REC ( i , D , c ) {
			remove ( i ) ;
			REC ( j , R , i )
				remove ( j ) ;
			if ( dance ( d + 1 ) )
				return 1 ;
			REC ( j , L , i )
				resume ( j ) ;
			resume ( i ) ;
		}
		return 0 ;
	}
	
	double dist ( int i , int j ) {
		double x = a[i].x - b[j].x ;
		double y = a[i].y - b[j].y ;
		return sqrt ( x * x + y * y ) ;
	}
	
	int f ( double r ) {
		init () ;
		FOR ( j , 1 , m )
			FOR ( i , 1 , n )
				if ( dis[i][j] <= r )
					link ( j , i ) ;
		return dance ( 0 ) ;
	}
	
	void solve () {
		scanf ( "%d%d%d" , &n , &m , &k ) ;
		nv = n ;
		FOR ( i , 1 , n )
			a[i].input () ;
		FOR ( i , 1 , m )
			b[i].input () ;
		FOR ( i , 1 , n )
			FOR ( j , 1 , m )
				dis[i][j] = dist ( i , j ) ;
		double l = 0 , r = 1500 ;
		while ( r - l > eps ) {
			double mid = ( l + r ) / 2 ;
			if ( f ( mid ) )
				r = mid ;
			else
				l = mid ;
		}
		printf ( "%f
" , l ) ; } } dlx ; int main () { int T ; scanf ( "%d" , &T ) ; while ( T -- ) dlx.solve () ; return 0 ; }

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