Bomb Game
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3350 Accepted Submission(s): 1149
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.
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 x1i, y1i, x2i, y2i, indicating
that the coordinates of the two candidate places of the i-th round are (x1i, y1i) and (x2i, y2i). 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
给n对炸弹可以放置的位置(每个位置为一个二维平面上的点),
每次放置炸弹是时只能选择这一对中的其中一个点,每个炸弹爆炸
的范围半径都一样,控制爆炸的半径使得所有的爆炸范围都不相
交(可以相切),求解这个最大半径.
首先二分最大半径值,然后2-sat构图判断其可行性,对于每
两队位置(u,uu)和(v,vv),如果u和v之间的距离小于2*id,也就
是说位置u和位置v处不能同时防止炸弹(两范围相交),所以连边(u,vv)
和(v,uu),求解强连通分量判断可行性.
为了确保精度问题，在计算过程当中先不开方。。。。。
程序：
#include"stdio.h"
#include"string.h"
#include"stack"
#include"math.h"
#define M 209
using namespace std;
stack<int>q;
int head[M],dfn[M],low[M],belong[M],use[M],t,n,index,num,cnt[M];
struct st
{
int u,v,next;
}edge[M*200];
void init()
{
t=0;
memset(head,-1,sizeof(head));
}
void add(int u,int v)
{
edge[t].u=u;
edge[t].v=v;
edge[t].next=head[u];
head[u]=t++;
}
void tarjan(int u)
{
dfn[u]=low[u]=++index;
q.push(u);
use[u]=1;
int i;
for(i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
if(!dfn[v])
{
tarjan(v);
low[u]=min(low[u],low[v]);
}
else if(use[v])
{
low[u]=min(low[u],dfn[v]);
}
}
if(dfn[u]==low[u])
{
int vv;
num++;
do
{
vv=q.top();
belong[vv]=num;
q.pop();
use[vv]=0;
}while(u!=vv);
}
}
void solve()
{
index=num=0;
memset(dfn,0,sizeof(dfn));
memset(use,0,sizeof(use));
for(int i=1;i<=n*2;i++)
{
if(!dfn[i])
tarjan(i);
}
}
double len(double x1,double y1,double x2,double y2)
{
return (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
}
int main()
{
int i,j;
double x1[M],y1[M],x2[M],y2[M];
while(scanf("%d",&n)!=-1)
{
for(i=1;i<=n;i++)
scanf("%lf%lf%lf%lf",&x1[i],&y1[i],&x2[i],&y2[i]);
double max1=-1;
for(i=1;i<=n;i++)
{
for(j=i+1;j<=n;j++)
{
double ans=len(x1[i],y1[i],x1[j],y1[j]);
max1=max(max1,ans);
ans=len(x1[i],y1[i],x2[j],y2[j]);
max1=max(max1,ans);
ans=len(x2[i],y2[i],x1[j],y1[j]);
max1=max(max1,ans);
ans=len(x2[i],y2[i],x2[j],y2[j]);
max1=max(max1,ans);
}
}
double left,right,mid;
left=0;
right=max1*2;
while(fabs(right-left)>=0.005)
{
mid=(left+right)/2;
init();
for(i=1;i<=n;i++)
{
for(j=1+i;j<=n;j++)
{
double ans=len(x1[i],y1[i],x1[j],y1[j]);
if(ans<mid)
{
add(i*2-1,j*2);
add(j*2-1,i*2);
}
ans=len(x1[i],y1[i],x2[j],y2[j]);
if(ans<mid)
{
add(2*i-1,2*j-1);
add(2*j,2*i);
}
ans=len(x2[i],y2[i],x1[j],y1[j]);
if(ans<mid)
{
add(2*i,2*j);
add(2*j-1,2*i-1);
}
ans=len(x2[i],y2[i],x2[j],y2[j]);
if(ans<mid)
{
add(2*i,2*j-1);
add(2*j,i*2-1);
}
}
}
solve();
int flag=0;
for(i=1;i<=n;i++)
{
if(belong[i*2-1]==belong[i*2])
{
flag++;
break;
}
}
if(flag)
{
right=mid;
}
else
{
left=mid;
}
}
printf("%.2lf
",sqrt(mid)/2.0);
}
}