编程介的小学生 2017-08-31 09:29 采纳率: 20.5%
浏览 826
已采纳

Contest

Lazyman is going to join the programming contest! There are n problems in the contest. Because Lazyman is so lazy, he just tried each problem once and only once. However each problem in the contest may be not independent, it may have some relationship with other problems. So here is a propbility P n*n matrix. The jth element of the ith row is denoted by Pij. Suppose he have tried m problems, a1,a2,...am (No matter the problems are solved or not.) (1 ≤ a1,a2,...am ≤ n) If he tries the rth problems next, (r!=a1,a2,...am) the propbility that he can solve it is the maximum value of the m + 1 values, Prr, Pa1r%, Pa2r%, ... Pamr%.

Help him to decide the order he tries the problems to make the expected number of the solved problems maximum.

Input

There are multiple test cases. The first line of input is an integer T (0 < T ≤ 100), indicating the number of test cases. Then T test cases follow. The first line of each test case is a positive integer n (0 < n ≤ 10). Then comes n lines. The jth element of the ith line is an integer Pij. (0 ≤ Pij ≤100 1 ≤ i,j ≤ n)

Output

For each test case, output 2 lines. The first line is the maxmum expected number of problems, accurated to the nearest 0.01. The second the order he should try each problems to achieve this expected number of problems. If there are several sequeces achieve the same maximum expected number, choose the lexicographically smallest one. Use capital letters to denote each problems please. (Start from A)

Sample Input

2
1
0
2
2 3
5 7
Sample Output

0.00
A
0.12
BA
Hint

For the 2nd sample, if the order is AB:
The probility of the result that he solves problem A is 0.02*(1-0.07)=0.0186
The probility of the result that he solves problem B is (1-0.02)*0.07=0.0686
The probility of the result that he solves problem A and B is 0.02*0.07=0.0014
So the expected number of the problem sovled is 1*0.0186+1*0.0686+2*0.0014=0.09

For the 2nd sample, if the order is BA:
The probility of the result that he solves problem A is (1-0.07)*0.05=0.0465
The probility of the result that he solves problem B is 0.07*(1-0.05)=0.0665
The probility of the result that he solves problem A and B is 0.07*0.05=0.0035
So the expected number of the problem sovled is 1*0.0465+1*0.0665+2*0.0035=0.12

  • 写回答

1条回答 默认 最新

  • threenewbee 2017-09-15 15:46
    关注
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

报告相同问题?

悬赏问题

  • ¥40 复杂的限制性的商函数处理
  • ¥15 程序不包含适用于入口点的静态Main方法
  • ¥15 素材场景中光线烘焙后灯光失效
  • ¥15 请教一下各位,为什么我这个没有实现模拟点击
  • ¥15 执行 virtuoso 命令后,界面没有,cadence 启动不起来
  • ¥50 comfyui下连接animatediff节点生成视频质量非常差的原因
  • ¥20 有关区间dp的问题求解
  • ¥15 多电路系统共用电源的串扰问题
  • ¥15 slam rangenet++配置
  • ¥15 有没有研究水声通信方面的帮我改俩matlab代码