HDOJ 3622 Bomb Game

二分距离2sat

Bomb Game

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3382    Accepted Submission(s): 1161

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

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

const int maxn=222;

struct Edge
{
    int to,next;
}edge[maxn*maxn*3];

int Adj[maxn],Size;

void init()
{
    Size=0; memset(Adj,-1,sizeof(Adj));
}

void Add_Edge(int u,int v)
{
    edge[Size].to=v; edge[Size].next=Adj[u]; Adj[u]=Size++;
}

int Low[maxn],DFN[maxn],Belong[maxn],Instack[maxn],Stack[maxn];
int top,scc,Index;

void tarjan(int u)
{
    Low[u]=DFN[u]=++Index;
    Instack[u]=true; Stack[top++]=u;
    int v;

    for(int i=Adj[u];~i;i=edge[i].next)
    {
        v=edge[i].to;
        if(!DFN[v])
        {
            tarjan(v);
            Low[u]=min(Low[u],Low[v]);
        }
        else if(Instack[v])
        {
            Low[u]=min(Low[u],DFN[v]);
        }
    }

    if(Low[u]==DFN[u])
    {
        scc++;
        do
        {
            v=Stack[--top];
            Belong[v]=scc;
            Instack[v]=false;
        }while(v!=u);
    }
}

bool scc_solve(int n)
{
    memset(DFN,0,sizeof(DFN));
    memset(Instack,0,sizeof(Instack));
    top=scc=Index=0;

    for(int i=0;i<2*n;i++)
        if(!DFN[i]) tarjan(i);

    for(int i=0;i<n;i++)
    {
        if(Belong[i<<1]==Belong[i<<1|1]) return false;
    }
    return true;
}

int n;
int px[maxn][2],py[maxn][2];
int dist(int x,int y)
{
    return x*x+y*y;
}

bool ck(int mid)
{
    init();
    for(int i=0;i<n;i++)
    {
        for(int a=0;a<2;a++)
        {
            for(int j=i+1;j<n;j++)
            {
                for(int b=0;b<2;b++)
                {
                    if(dist(px[i][a]-px[j][b],py[i][a]-py[j][b])<mid)
                    {
                        Add_Edge(i*2+a,j*2+1-b);
                        Add_Edge(j*2+b,i*2+1-a);
                    }
                }
            }
        }
    }
    return scc_solve(n);
}

void solve()
{
    int low=0,mid,high=2000000000,ans=-1;
    while(low+1<high)
    {
        mid=(low+high)/2.;
        if(ck(mid)) ans=mid,low=mid;
        else high=mid;
    }
    printf("%.2lf\n",sqrt((double)ans)/2.);
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        double x1,x2,y1,y2;
        for(int i=0;i<n;i++)
        {
            scanf("%d%d%d%d",&px[i][0],&py[i][0],&px[i][1],&py[i][1]);
        }
        solve();
    }
    return 0;
}