【HDU】3622 Bomb Game 2-sat

傳送門：【HDU】3622 Bomb Game

題目分析：倒是很直覺的題目，二分最大半徑，如果在該半徑下有圓沖突則建邊。不過我倒是因為某個地方寫搓wa的飛起。。。

代碼如下：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

#define REP( i , a , b ) for ( int i = ( a ) ; i < ( b ) ; ++ i )
#define FOR( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define REV( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define CLR( a , x ) memset ( a , x , sizeof a )

const int MAXN = 205 ;
const int MAXE = 40000 ;

struct Edge {
	int v ;
	Edge* next ;
} ;

struct Line {
	int x1 , x2 , y1 , y2 ;
	void input () {
		scanf ( "%d%d%d%d" , &x1 , &y1 , &x2 , &y2 ) ;
	}
} ;

Edge E[MAXE] , *H[MAXN] , *cur ;
int dfn[MAXN] , low[MAXN] , scc[MAXN] , scc_cnt ;
int S[MAXN] , top , dfs_clock ;
int n ;

Line l[MAXN] ;

void init () {
	cur = E ;
	top = 0 ;
	scc_cnt = 0 ;
	dfs_clock = 0 ;
	CLR ( H , 0 ) ;
	CLR ( dfn , 0 ) ;
	CLR ( scc , 0 ) ;
}

void addedge ( int u , int v ) {
	cur -> v = v ;
	cur -> next = H[u] ;
	H[u] = cur ++ ;
}

void tarjan ( int u ) {
	dfn[u] = low[u] = ++ dfs_clock ;
	S[top ++] = u ;
	for ( Edge* e = H[u] ; e ; e = e -> next ) {
		int v = e -> v ;
		if ( !dfn[v] ) {
			tarjan ( v ) ;
			low[u] = min ( low[u] , low[v] ) ;
		} else if ( !scc[v] ) low[u] = min ( low[u] , dfn[v] ) ;
	}
	if ( low[u] == dfn[u] ) {
		++ scc_cnt ;
		do {
			scc[S[-- top]] = scc_cnt ;
		} while ( u != S[top] ) ;
	}
}

/*
void scanf ( int& x , char c = 0 ) {
	while ( ( c = getchar () ) < '0' || c > '9' ) ;
	x = c - '0' ;
	while ( ( c = getchar () ) >= '0' && c <= '9' ) x = x * 10 + c - '0' ;
}
*/

int len ( int x , int y ) {
	return x * x + y * y ;
}

bool ok () {
	REP ( i , 0 , n << 1 ) if ( !dfn[i] ) tarjan ( i ) ;
	REP ( i , 0 , n ) if ( scc[i << 1] == scc[i << 1 | 1] ) return 0 ;
	return 1 ;
}

void solve () {
	REP ( i , 0 , n ) l[i].input () ;
	int low = 0 , high = 1e8 ;
	while ( low < high ) {
		int m = ( low + high + 1 ) >> 1 ;
		init () ;
		REP ( i , 0 , n ) REP ( j , i + 1 , n ) {
			if ( len ( l[i].x1 - l[j].x1 , l[i].y1 - l[j].y1 ) < m ) {
				addedge ( i << 1 , j << 1 | 1 ) ;
				addedge ( j << 1 , i << 1 | 1 ) ;
			}
			if ( len ( l[i].x1 - l[j].x2 , l[i].y1 - l[j].y2 ) < m ) {
				addedge ( i << 1 , j << 1 ) ;
				addedge ( j << 1 | 1 , i << 1 | 1 ) ;
			}
			if ( len ( l[i].x2 - l[j].x1 , l[i].y2 - l[j].y1 ) < m ) {
				addedge ( i << 1 | 1 , j << 1 | 1 ) ;
				addedge ( j << 1 , i << 1 ) ;
			}
			if ( len ( l[i].x2 - l[j].x2 , l[i].y2 - l[j].y2 ) < m ) {
				addedge ( i << 1 | 1 , j << 1 ) ;
				addedge ( j << 1 | 1 , i << 1 ) ;
			}
		}
		if ( ok () ) low = m ;
		else high = m - 1 ;
	}
	printf ( "%.2f\n" , sqrt ( ( double ) low ) / 2 ) ;
}

int main () {
	while ( ~scanf ( "%d" , &n ) ) solve () ;
	return 0 ;
}