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. CSS 高级布局技巧

    随着 IE8 逐渐退出舞台,很多高级的 CSS 特性都已被浏览器原生支持,再不学下就要过时了. 用 :empty 区分空元素 兼容性:不支持 IE8 /*假如我们有以上列表:*/ <div cl ...

  2. [Storm] 并发度的理解

    Tasks & executors relation Q1. However I'm a bit confused by the concept of "task". Is ...

  3. [C++]for同时遍历两个数组

    C++11同时遍历两个数组 #define for2array(x,y,xArray,yArray) \ for(auto x=std::begin(xArray), x##_end=std::end ...

  4. MarkdownPad 2 在win10下出错:HTML 渲染错误(This view has crashed) 的解决办法 + MarkdownPad2.5 注册码

    首先附上MarkdownPad2.5的注册码. 邮箱:Soar360@live.com 授权密钥: GBPduHjWfJU1mZqcPM3BikjYKF6xKhlKIys3i1MU2eJHqWGImD ...

  5. 汤姆大叔 javascript 系列 第20课 最后的5到javascript题目

    博客链接:http://www.cnblogs.com/TomXu/archive/2012/02/10/2342098.html 原题: 大叔注:这些题目也是来自出这5个题目的人,当然如果你能答对4 ...

  6. angularjs 中的setTimeout(),setInterval() / $interval 和 $timeout

    $interval window.setInterval的Angular包装形式.Fn是每次延迟时间后被执行的函数. 间隔函数的返回值是一个承诺.这个承诺将在每个间隔刻度被通知,并且到达规定迭代次数后 ...

  7. ArrayList和HashSet的Contains()方法(转)

    来源: ArrayList和HashSet的Contains()方法 笔试题: package com.champion.test.exam; import java.util.ArrayList; ...

  8. Sublime text 3如何编辑less并转(编译)成css文件

    今天开始学习使用less这个强大方便的前端工具,本来是考虑用koala(专门编辑less的软件)来使用less的,但是发现sublime编辑器也可以实现对less的编译及高亮显示代码,这样既能少用一个 ...

  9. Mosquitto pub/sub服务实现代码浅析-主体框架

    Mosquitto 是一个IBM 开源pub/sub订阅发布协议 MQTT 的一个单机版实现(目前也只有单机版),MQTT主打轻便,比较适用于移动设备等上面,花费流量少,解析代价低.相对于XMPP等来 ...

  10. 360随身wifi在win10中连不上网络

    找到服务"Wired AutoConfig"和"WLAN AutoConfig"项,点击"启动"按钮,确保使其正常启动. 讲本地网卡共享到移 ...