题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2295
Radar
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4106 Accepted Submission(s): 1576
Problem Description
N cities of the Java Kingdom need to be covered by radars for being in a state of war. Since the kingdom has M radar stations but only K operators, we can at most operate K radars. All radars have the same circular coverage with a radius of R. Our goal is to minimize R while covering the entire city with no more than K radars.
Input
The input consists of several test cases. The first line of the input consists of an integer T, indicating the number of test cases. The first line of each test case consists of 3 integers: N, M, K, representing the number of cities, the number of radar stations and the number of operators. Each of the following N lines consists of the coordinate of a city.
Each of the last M lines consists of the coordinate of a radar station.
All coordinates are separated by one space.
Technical Specification
1. 1 ≤ T ≤ 20
2. 1 ≤ N, M ≤ 50
3. 1 ≤ K ≤ M
4. 0 ≤ X, Y ≤ 1000
Output
For each test case, output the radius on a single line, rounded to six fractional digits.
Sample Input
1 3 3 2 3 4 3 1 5 4 1 1 2 2 3 3
Sample Output
Source
The 4th Baidu Cup final
题解：
超时方法：
1.对于DLX的矩阵：行代表着雷达与城市的距离， 列代表着城市。矩阵大小250*50。
2.Dancing跳起来，当R[0]==0时， 取当前所选行中，距离的最大值dis（这样才能覆盖掉所有城市），然后再更新答案ans，ans = min（ans， dis）。
3.结果矩阵有点大， 超时了。
4.错误思想分析：把雷达与城市的距离作为行，实际上是太明智的。因为题目说明了每个雷达的接收半径是相同的，而以上方法选出来的每个雷达的接收半径是相异的，然后又再取最大值，那为何不每次都取最大值（相同值）呢？ 如果取相同值，那么行就是雷达，列就是城市，矩阵的大小就减少了。但是又怎么确定雷达的接收半径呢？如下：
正确方法：
1.雷达作为行， 城市作为列。
2.二分雷达的接收范围，每次二分都：根据接收半径更新矩阵中所含的元素，然后再进行一次Dance()，如果能覆盖掉所有城市，则缩小半径，否则扩大半径。
超时方法：
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正确方法：
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