Life Winner Bo

Problem Description
 
Bo is a "Life Winner".He likes playing chessboard games with his girlfriend G.

The size of the chessboard is N×M.The top left corner is numbered(1,1) and the lower right corner is numberd (N,M).

For each game,Bo and G take turns moving a chesspiece(Bo first).At first,the chesspiece is located at (1,1).And the winner is the person who first moves the chesspiece to (N,M).At one point,if the chess can't be moved and it isn't located at (N,M),they end in a draw.

In general,the chesspiece can only be moved right or down.Formally,suppose it is located at (x,y),it can be moved to the next point (x′,y′) only if x′≥x and y′≥y.Also it can't be moved to the outside of chessboard.

Besides,There are four kinds of chess(They have movement rules respectively).

1.king.

2.rook(castle).

3.knight.

4.queen.

(The movement rule is as same as the chess.)

For each type of chess,you should find out that who will win the game if they both play in an optimal strategy.

Print the winner's name("B" or "G") or "D" if nobody wins the game.

 
Input
 
In the first line,there is a number T as a case number.

In the next T lines,there are three numbers type,N and M.

"type" means the kind of the chess.

T≤1000,2≤N,M≤1000,1≤type≤4

 
Output
 
For each question,print the answer.
 
Sample Input
 
4
1 5 5
2 5 5
3 5 5
4 5 5
 
Sample Output
 
G
G
D
B
 

题意:

  4种棋。

  棋子首先在(1,1),你要移动到(n,m);

  谁先移动到终点谁就赢了,注意骑士可能走向平局

题解:

  首先国王可以预处理答案

  皇后是威佐夫博弈

  车是个典型的nim博弈

  骑士呢,你注意到有个平局,就是说,如果当前走下一个状态的时候,面临了必败和平局的情况,你是要走平局的,这个点注意一下就能AC了

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const int N = 3e3+, M = 2e5+, mod = 1e9+, inf = 2e9;
int h[],gw[N][N],qs[N][N];
void init() {
gw[][] = ;
for(int i = ; i <= ; ++i) {
for(int j = ; j <= ; ++j) {
memset(h,,sizeof(h));
if(i>=)h[gw[i-][j]] = ;
if(j>=)h[gw[i][j-]] = ;
if(i>=&&j>=) h[gw[i-][j-]] = ;
if(h[]) gw[i][j]=;
else gw[i][j] = ;
}
}
memset(qs,-,sizeof(qs));
qs[][] = ;int cnt = ;
for(int i = ; i <= ; ++i) {
for(int j = ; j <= ; ++j) {
int ok = ;
memset(h,,sizeof(h));
if(i>=&&j>=&&qs[i-][j-]!=-&&qs[i-][j-]<=)h[qs[i-][j-]] = ,ok = ;
if(j>=&&i>=&&qs[i-][j-]!=-&&qs[i-][j-]<=)h[qs[i-][j-]] = ,ok++;
if(i == && j == ) continue;
if(ok) {
if(h[]) qs[i][j] = ;
else if(!h[] && ok < ) qs[i][j] = -;
else qs[i][j] = ;
}
else qs[i][j] = -; }
}
// cout<<cnt<<endl;
// cout<<qs[1][2]<<endl;
}
int main() {
int T;init();
scanf("%d",&T);
while(T--) {
int type,n,m;
scanf("%d%d%d",&type,&n,&m);
n--,m--;
if(type == ) {//国王
if(gw[n][m])printf("B\n");
else printf("G\n");
} else if(type == ) {//车
if((n ^ m) != )printf("B\n");
else printf("G\n");
} else if(type == ) {//马
if(qs[n][m]==-) printf("D\n");
else if(qs[n][m]) printf("B\n");
else printf("G\n");
} else {//皇后
if(n > m) swap(n,m);
double k = (sqrt()-1.0)/2.0;
int j = n * k;
if(n != (int) (j*(+k))) j++;
if(n + j == m) printf("G\n");
else printf("B\n");
}
}
return ;
}

HDU 5754 Life Winner Bo 组合博弈的更多相关文章

  1. HDU 5754 Life Winner Bo (博弈)

    Life Winner Bo 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5754 Description Bo is a "Life W ...

  2. HDU 5754 Life Winner Bo (各种博弈) 2016杭电多校联合第三场

    题目:传送门 题意:一个国际象棋棋盘,有四种棋子,从(n,m)走到(1,1),走到(1,1)的人赢,先手赢输出B,后手赢输出G,平局输出D. 题解:先把从(n,m)走到(1,1)看做是从(1,1)走到 ...

  3. HDU 5754 Life Winner Bo (找规律and博弈)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5754 给你四种棋子,棋子一开始在(1,1)点,两个人B和G轮流按每种棋子的规则挪动棋子,棋子只能往右下 ...

  4. HDU 5754 Life Winner Bo(各类博弈大杂合)

    http://acm.hdu.edu.cn/showproblem.php?pid=5754 题意: 给一个国际象棋的棋盘,起点为(1,1),终点为(n,m),现在每个棋子只能往右下方走,并且有4种不 ...

  5. 【博弈论】HDU 5754 Life Winner Bo

    题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5754 题目大意: 4种棋子,象棋中的 1王,2车,3马,4后,选其一,B和G轮流走,不能往左上走,一 ...

  6. HDU 5754 Life Winner Bo

    四种棋子实质上都是一样的思路: 如果某位置的棋子,它下一步可以走到的位置中 能找到有后手胜的位置,那么该位置先手必胜. 如果某位置的棋子,它下一步可以走到的位置中 全是先手胜,那么该位置后手必胜. 其 ...

  7. hdu 5754 Life Winner Bo 博弈论

    对于king:我是套了一个表. 如果起点是P的话,则是后手赢,否则前手赢. 车:也是画图推出来的. 马:也是推出来的,情况如图咯. 对于后:比赛时竟然推错了.QAQ最后看了题解:是个威佐夫博奕.(2, ...

  8. HDU5754 Life Winner Bo(博弈)

    题目 Source http://acm.hdu.edu.cn/showproblem.php?pid=5754 Description Bo is a "Life Winner" ...

  9. hdu-5754 Life Winner Bo(博弈)

    题目链接: Life Winner Bo Time Limit: 2000/1000 MS (Java/Others)     Memory Limit: 131072/131072 K (Java/ ...

随机推荐

  1. Mac Brew Install Nginx Summary

    ==> Downloading https://homebrew.bintray.com/bottles/nginx-1.10.1.el_capitan.bot################# ...

  2. webapi集成owin使用Oauth认证时能获取accee_token仍无法登录的解决办法

    HttpConfiguration webapiConfig = new HttpConfiguration(); IIocBuilder iocBuilder = new OwinAutofacIo ...

  3. net.sf.json.JSONException: There is a cycle in the hierarchy!的解决办法

    使用Hibernate manytoone属性关联主表的时候,如果使用JSONArray把pojo对象转换成json对象时,很容易出现循环的异常.解决的办法就是, 在转换json对象时忽略manyto ...

  4. 对日期 ,和 master.dbo.spt_values 表操作

    if object_id('tempdb..#xs') is not null drop table #xs; ), date datetime, sale float) insert into #x ...

  5. 【学习笔记】Struts2 应用开发步骤

    1.在web.xml文件中定义核心Filter来拦截用户请求. <!-- 定义Struts 2的核心Filter --> <filter> <filter-name> ...

  6. [Linux] Linux常用文本操作命令整理

    简单的总结一下常用的一些实用的Linux文本操作命令,包括wc(统计).cut(切分).sort(排序).uniq(去重).grep(查找).sed(替换.插入.删除).awk(文本分析). 1.统计 ...

  7. JS判断字符串长度(中文长度为2,英文长度为1)

    目的:计算字符串长度(英文占1个字符,中文汉字占2个字符) 方法一: String.prototype.gblen = function() { var len = 0; for (var i=0; ...

  8. Android 系统工具类SystemUtils

    包含的功能有: 获取系统中所有APP应用.获取用户安装的APP应用.根据包名和Activity启动类查询应用信息.跳转到WIFI设置.WIFI网络开关.移动网络开关.GPS开关 当前若关则打开 当前若 ...

  9. tomcat 快速部署静态文件

    server.conf配置: <?xml version='1.0' encoding='utf-8'?> <!-- Licensed to the Apache Software ...

  10. MarkdownPad2.5 注册码

    邮箱: Soar360@live.com 授权秘钥: GBPduHjWfJU1mZqcPM3BikjYKF6xKhlKIys3i1MU2eJHqWGImDHzWdD6xhMNLGVpbP2M5SN6b ...