306111: CF1147F. Zigzag Game

Memory Limit:256 MB Time Limit:5 S
Judge Style:Text Compare Creator:
Submit:0 Solved:0

Description

Zigzag Game

题意翻译

**这是一道交互题。** 给出一个两边均有 $n$ 个节点的完全二分图,左部点由 $1$ 到 $n$ 编号,右部点由 $n+1$ 到 $2n$ 编号。输入中会给出一个 $n \times n$ 的矩阵 $a$,其中 $a_{i,j}$ 表示连接 $i$ 和 $j+n$ 的边的边权。 现在有两个玩家 Alice 和 Bob 在这个图上玩游戏。首先 Alice 会选择 "increasing"(增加)或者 "decreasing"(减少)中的一个属性,Bob 则会自动选择另一个。接下来由 Alice 在图上选择一个起始节点 $p$ 放上一枚棋子,Bob 选择一条连接节点 $p$ 的边 $(p,q)$,将棋子移动到 $q$。 接下来由 Alice 先手轮流进行操作。每一次操作时操作方需要选择一个与当前节点 $x$ 相邻的未被经过的节点 $y$。设 $w$ 为 $(x,y)$ 的边权,$w'$ 为上一次移动时的边权,那么如果当前操作方在最开始选择 "increasing",则需要满足 $w > w'$,否则需要满足 $w < w'$。然后当前操作方将棋子移动到 $y$。无法移动的操作方失败。 现在你需要与交互库模拟这样一盘游戏,包括选择先后("Alice" 或者 "Bob")、选择属性("increasing" 或者 "decreasing")、选择初始节点、选择移动节点和游戏过程。你需要保证在与交互库的游戏过程中获胜。 Translation By Itst.

题目描述

You are given a complete bipartite graph with $ 2n $ nodes, with $ n $ nodes on each side of the bipartition. Nodes $ 1 $ through $ n $ are on one side of the bipartition, and nodes $ n+1 $ to $ 2n $ are on the other side. You are also given an $ n \times n $ matrix $ a $ describing the edge weights. $ a_{ij} $ denotes the weight of the edge between nodes $ i $ and $ j+n $ . Each edge has a distinct weight. Alice and Bob are playing a game on this graph. First Alice chooses to play as either "increasing" or "decreasing" for herself, and Bob gets the other choice. Then she places a token on any node of the graph. Bob then moves the token along any edge incident to that node. They now take turns playing the following game, with Alice going first. The current player must move the token from the current vertex to some adjacent unvisited vertex. Let $ w $ be the last weight of the last edge that was traversed. The edge that is traversed must be strictly greater than $ w $ if the player is playing as "increasing", otherwise, it must be strictly less. The first player unable to make a move loses. You are given $ n $ and the edge weights of the graph. You can choose to play as either Alice or Bob, and you will play against the judge. You must win all the games for your answer to be judged correct.

输入输出格式

输入格式


输出格式


Each test contains multiple test cases. The first line contains the number of test cases $ t $ ( $ 1 \le t \le 50 $ ). Description of the test cases follows. Each test starts with an integer $ n $ ( $ 1 \leq n \leq 50 $ ) — the number of nodes on each side of the bipartition. The next $ n $ lines contain $ n $ integers $ a_{ij} $ ( $ 1 \leq a_{ij} \leq n^2 $ ). $ a_{ij} $ denotes the weight of the edge between node $ i $ and node $ j+n $ . All $ a_{ij} $ will be distinct. You first print a string "A" or "B", denoting which player you want to play ("A" for Alice and "B" for Bob). If playing as Alice, first print either "I" or "D" (denoting whether you choose "increasing" or "decreasing"). Then, print the node that you wish to start at. If playing as Bob, read in a character "I" or "D" (denoting whether the judge chose "increasing" or "decreasing"), then the node that the judge chooses to start at. To make a move, print the index of the node that you wish to go to. To read what move the judge made, read an integer from standard in. If the judge responds with $ -1 $ , then that means the judge has determined it has no legal moves (or it may have just given up) and that you won the case. Stop processing this case immediately and start processing the next case. If the judge responds with $ -2 $ , that means that the judge has determined that you made an invalid move, so you should exit immediately to avoid getting other verdicts. If you are unable to make a move or give up, print $ -1 $ and then exit immediately. This will give you a Wrong answer verdict. After printing a move do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use: - fflush(stdout) or cout.flush() in C++; - System.out.flush() in Java; - flush(output) in Pascal; - stdout.flush() in Python; - see documentation for other languages. Hack Format To hack, use the following format. Note that you can only hack with one test case. The first line should contain a single integer $ t $ ( $ t=1 $ ). The second line should contain an integer $ n $ ( $ 1 \leq n \leq 50 $ ). The next $ n $ lines should contain $ n $ integers each, the $ a_{ij} $ ( $ 1 \leq a_{ij} \leq n^2 $ ). $ a_{ij} $ denotes the weight of the edge between node $ i $ and node $ j+n $ . All $ a_{ij} $ must be distinct. The judge will play a uniformly random legal move with this case versus the hacked program. If it has no legal moves left, it will give up and declare the player as the winner.

输入输出样例

输入样例 #1

2
3
3 1 9
2 5 7
6 4 8
6
-1
1
1
I 1
-1

输出样例 #1

A
D 3
2
B
2

说明

The first example has two test cases. In the first test case, the graph looks like the following. ![](https://cdn.luogu.com.cn/upload/vjudge_pic/CF1147F/75cfe5fcf597027e3291b6485fcfe46a40219dfc.png)In the sample output, the player decides to play as Alice and chooses "decreasing" and starting at node $ 3 $ . The judge responds by moving to node $ 6 $ . After, the player moves to node $ 2 $ . At this point, the judge has no more moves (since the weight must "increase"), so it gives up and prints $ -1 $ . In the next case, we have two nodes connected by an edge of weight $ 1 $ . The player decides to play as Bob. No matter what the judge chooses, the player can move the token to the other node and the judge has no moves so will lose.

Input

题意翻译

**这是一道交互题。** 给出一个两边均有 $n$ 个节点的完全二分图,左部点由 $1$ 到 $n$ 编号,右部点由 $n+1$ 到 $2n$ 编号。输入中会给出一个 $n \times n$ 的矩阵 $a$,其中 $a_{i,j}$ 表示连接 $i$ 和 $j+n$ 的边的边权。 现在有两个玩家 Alice 和 Bob 在这个图上玩游戏。首先 Alice 会选择 "increasing"(增加)或者 "decreasing"(减少)中的一个属性,Bob 则会自动选择另一个。接下来由 Alice 在图上选择一个起始节点 $p$ 放上一枚棋子,Bob 选择一条连接节点 $p$ 的边 $(p,q)$,将棋子移动到 $q$。 接下来由 Alice 先手轮流进行操作。每一次操作时操作方需要选择一个与当前节点 $x$ 相邻的未被经过的节点 $y$。设 $w$ 为 $(x,y)$ 的边权,$w'$ 为上一次移动时的边权,那么如果当前操作方在最开始选择 "increasing",则需要满足 $w > w'$,否则需要满足 $w < w'$。然后当前操作方将棋子移动到 $y$。无法移动的操作方失败。 现在你需要与交互库模拟这样一盘游戏,包括选择先后("Alice" 或者 "Bob")、选择属性("increasing" 或者 "decreasing")、选择初始节点、选择移动节点和游戏过程。你需要保证在与交互库的游戏过程中获胜。 Translation By Itst.

加入题单

上一题 下一题 算法标签: