HDU 3622 Bomb Game（2-sat）

Bomb Game

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3881    Accepted Submission(s): 1346

Problem Description

Robbie is playing an interesting computer game. The game field is an unbounded 2-dimensional region. There are N rounds in the game. At each round, the computer will give Robbie two places, and Robbie should choose one of them to put a bomb. The explosion area of the bomb is a circle whose center is just the chosen place. Robbie can control the power of the bomb, that is, he can control the radius of each circle. A strange requirement is that there should be no common area for any two circles. The final score is the minimum radius of all the N circles.
Robbie has cracked the game, and he has known all the candidate places of each round before the game starts. Now he wants to know the maximum score he can get with the optimal strategy.

 

Input

The first line of each test case is an integer N (2 <= N <= 100), indicating the number of rounds. Then N lines follow. The i-th line contains four integers x_1i, y_1i, x_2i, y_2i, indicating that the coordinates of the two candidate places of the i-th round are (x_1i, y_1i) and (x_2i, y_2i). All the coordinates are in the range [-10000, 10000].

 

Output

Output one float number for each test case, indicating the best possible score. The result should be rounded to two decimal places.

 

Sample Input

2

1 1 1 -1

-1 -1 -1 1

 

2

1 1 -1 -1

1 -1 -1 1

 

Sample Output

1.41

1.00

 

Source

2010 Asia Regional Tianjin Site —— Online Contest

 

 

距离最少值尽量大，，典型的二分 ， 加上每个bomb有2种放法（只能取任意一种）， 就是一个2-sat.

 

 注意一下精度，这里会卡一下。1e-4 可以过了

 

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <map>
#include <cmath>

using namespace std;

#define eps 1e-5
const int N = 210;
const int M = 400010;

int n , m ;

int st[N] , top ;
bool mark[N];
int eh[N] , et[M] , nxt[M] , tot ;
struct node { double x , y ; }e[N];


void init()
{
    tot = 0 ;
    memset( eh , -1 , sizeof eh );
    memset( mark , false , sizeof mark );
}

void addedge( int u , int v ){
    et[tot] = v , nxt[tot] = eh[u] , eh[u] = tot ++ ;
    et[tot] = u , nxt[tot] = eh[v] , eh[v] = tot ++ ;
}

bool dfs( int u )
{
    if( mark[u] ) return true;
    if( mark[u^1] ) return false;
    mark[u] = true ;
    st[top++] = u ;
    for( int i = eh[u] ; ~i ; i = nxt[i] ){
        int v = et[i];
        if( !dfs(v^1) ) return false;
    }
    return true;
}

bool solve()
{
    for(int i = 0 ; i < 2 * n ; i += 2 ){
        if( !mark[i] && !mark[i+1] ){
            top = 0 ;
            if( !dfs(i) ){
                while( top > 0 ) mark[ st[--top] ] = false;
                if( !dfs(i+1) ) return false ;
            }
        }
    }
    return true;
}

inline double dis( int i , int j )
{
    return sqrt( ( e[i].x - e[j].x ) * ( e[i].x - e[j].x ) + ( e[i].y - e[j].y ) * ( e[i].y - e[j].y ) ) ;
}

bool test( double DIS )
{
    init();
    for( int i = 0 ; i < 2 * n ; i += 2 ){
        for( int j = i + 2 ; j < 2 * n ; j += 2 ){
            if( dis( i , j ) < DIS )addedge( i , j );
            if( dis( i , j^1 ) < DIS )addedge( i , j^1 );
            if( dis( i^1 , j ) < DIS )addedge( i^1 , j ) ;
            if( dis( i^1 , j^1 ) < DIS )addedge( i^1 , j^1 );
        }
    }
    return solve();
}


int main()
{
    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif // LOCAL
    ios::sync_with_stdio(0);

    while( ~scanf("%d",&n) ){
        for( int i = 0 ; i < 2 * n ; ++i ){
            scanf("%lf%lf",&e[i].x,&e[i].y);
        }

        double l = 0.0 , r = sqrt( pow(20000.0,2.0) + pow(20000.0,2.0) );
        while( l + eps <= r ){
            double m = ( l + r ) / 2.0 ;
            
            if( test(m) )
                l = m  ;
            else
                r = m - eps ;
        }
        printf("%.2lf
",l/2);
    }
}