Bomb Game
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2294 Accepted Submission(s): 769
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
Recommend lcy 2-SAT问题。 可行性判别。 题意: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. 用到double的时候都要注意精度。
/* HDU 3622 题意: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. 注意精度问题 */ #include<stdio.h> #include<algorithm> #include<string.h> #include<iostream> #include<math.h> using namespace std; const int MAXN=210; const int MAXM=40005;//边的最大数 const double eps=1e-5; struct Edge { int to,next; }edge1[MAXM],edge2[MAXM]; int head1[MAXN]; int head2[MAXN]; int tol1,tol2; bool vis1[MAXN],vis2[MAXN]; int Belong[MAXN];//连通分量标记 int T[MAXN];//dfs结点结束时间 int Bcnt,Tcnt; void add(int a,int b)//原图和逆图都要添加 { edge1[tol1].to=b; edge1[tol1].next=head1[a]; head1[a]=tol1++; edge2[tol2].to=a; edge2[tol2].next=head2[b]; head2[b]=tol2++; } void init()//建图前初始化 { memset(head1,-1,sizeof(head1)); memset(head2,-1,sizeof(head2)); memset(vis1,false,sizeof(vis1)); memset(vis2,false,sizeof(vis2)); tol1=tol2=0; Bcnt=Tcnt=0; } void dfs1(int x)//对原图进行dfs,算出每个结点的结束时间，哪个点开始无所谓 { vis1[x]=true; int j; for(int j=head1[x];j!=-1;j=edge1[j].next) if(!vis1[edge1[j].to]) dfs1(edge1[j].to); T[Tcnt++]=x; } void dfs2(int x) { vis2[x]=true; Belong[x]=Bcnt; int j; for(j=head2[x];j!=-1;j=edge2[j].next) if(!vis2[edge2[j].to]) dfs2(edge2[j].to); } struct Point { int x,y; }s[MAXN]; double dist(Point a,Point b) { return sqrt((double)(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); } bool ok(int n)//判断可行性 { for(int i=0;i<2*n;i++) if(!vis1[i]) dfs1(i); for(int i=Tcnt-1;i>=0;i--) if(!vis2[T[i]])//这个别写错，是vis2[T[i]] { dfs2(T[i]); Bcnt++; } for(int i=0;i<=2*n-2;i+=2) if(Belong[i]==Belong[i+1]) return false; return true; } int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int n; double left,right,mid; while(scanf("%d",&n)!=EOF) { for(int i=0;i<n;i++) scanf("%d%d%d%d",&s[2*i].x,&s[2*i].y,&s[2*i+1].x,&s[2*i+1].y); left=0; right=40000.0; while(right-left>=eps) { mid=(left+right)/2; init(); for(int i=0;i<2*n-2;i++) { int t; if(i%2==0)t=i+2; else t=i+1; for(int j=t;j<2*n;j++) if(dist(s[i],s[j])<2*mid)//冲突了 { add(i,j^1); add(j,i^1);//注意顺序不能变的 } } if(ok(n))left=mid; else right=mid; } printf("%.2lf\n",right); } return 0; }
