Bomb Game
Time Limit: 3000ms
Memory Limit: 32768KB
This problem will be judged on HDU. Original ID: 3622
64-bit integer IO format: %I64d Java class name: Main
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 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
Source
2010 Asia Regional Tianjin Site —— Online Contest
解题：每行两颗弹，择其一而炸之，但是相邻两弹的爆炸范围不能相交，可以相切，所有弹爆炸范围相同，求最大的爆炸范围。
典型的2-SAT。求最大，就二分好了。
注意精度，至少1e-5，否则会WA.
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <cmath>
5 #include <algorithm>
6 #include <climits>
7 #include <vector>
8 #include <queue>
9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 250;
18 const double exps = 1e-5;
19 struct arc{
20 int to,next;
21 arc(int x = 0,int y = -1){
22 to = x;
23 next = y;
24 }
25 };
26 struct node{
27 int x,y;
28 };
29 node p[maxn];
30 arc e[maxn*maxn*10];
31 int head[maxn],tot,dfn[maxn],low[maxn],belong[maxn],scc,cnt,n;
32 bool instack[maxn];
33 stack<int>stk;
34 void add(int u,int v){
35 e[tot] = arc(v,head[u]);
36 head[u] = tot++;
37 }
38 double dis(const node &a,const node &b){
39 double tmp = (a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y);
40 return sqrt(tmp);
41 }
42 void tarjan(int u){
43 dfn[u] = low[u] = ++cnt;
44 stk.push(u);
45 instack[u] = true;
46 for(int i = head[u]; ~i; i = e[i].next){
47 if(!dfn[e[i].to]){
48 tarjan(e[i].to);
49 low[u] = min(low[u],low[e[i].to]);
50 }else if(instack[e[i].to]) low[u] = min(low[u],dfn[e[i].to]);
51 }
52 if(low[u] == dfn[u]){
53 scc++;
54 int v;
55 do{
56 v = stk.top();
57 stk.pop();
58 instack[v] = false;
59 belong[v] = scc;
60 }while(v != u);
61 }
62 }
63 bool solve(){
64 while(!stk.empty()) stk.pop();
65 for(int i = 0; i < maxn; i++){
66 belong[i] = dfn[i] = low[i] = 0;
67 instack[i] = false;
68 }
69 scc = cnt = 0;
70 for(int i = 0; i < n<<1; i++)
71 if(!dfn[i]) tarjan(i);
72 for(int i = 0; i < n; i++)
73 if(belong[i<<1] == belong[i<<1|1]) return false;
74 return true;
75 }
76 int main() {
77 int x,y;
78 double low,high,mid;
79 while(~scanf("%d",&n)){
80 for(int i = 0; i < n; i++)
81 scanf("%d %d %d %d",&p[i<<1].x,&p[i<<1].y,&p[i<<1|1].x,&p[i<<1|1].y);
82 low = 0;
83 high = 2000.0;
84 while(fabs(high - low) > exps){
85 mid = (high + low)/2.0;
86 memset(head,-1,sizeof(head));
87 for(int i = tot = 0; i < (n-1)<<1; i++){
88 for(int j = i&1?i+1:i+2; j < n<<1; j++){
89 if(dis(p[i],p[j]) < 2.0*mid){//i与j不能共存
90 add(i,j^1);
91 add(j,i^1);
92 }
93 }
94 }
95 if(solve()) low = mid;
96 else high = mid;
97 }
98 printf("%.2f\n",high);
99 }
100 return 0;
101 }