ZOJ 2610 Puzzle 模拟
大模拟:枚举6个方向。检查每一个0是否能移动
Puzzle
Time Limit: 2 Seconds Memory Limit: 65536 KB
Little Georgie likes puzzles very much. Recently he has found a wooden triangle in the box with old toys. The side of the triangle is n inches long. The triangle is divided into n2unit
triangles with lines drawn on his surface.
The interesting fact about that triangle is that it is not solid - it consists of two parts. Each of the parts is a connected set of unit triangles. Georgie has put his triangle onto
the table and now wonders whether he can separate the parts. He wants to separate them without taking any part of the triangle off the table, just moving the parts by the table surface. The triangle has a small but non-zero thickness, so while being moved
the parts must not intersect.
For example, if the triangle is divided into parts as it is shown on the top picture below, Georgie can separate the parts the way he wants. However in the case displayed on the bottom
picture, he cannot separate the parts without lifting one of them.
Help Georgie to determine whether he can separate the parts moving them by the surface of the table.
| <img src="http://acm.zju.edu.cn/onlinejudge/showImage.do?name=0000%2F2610%2Fg.gif" <="" img=""> | <img src="http://acm.zju.edu.cn/onlinejudge/showImage.do?name=0000%2F2610%2Fg2.gif" <="" img=""> |
Two puzzles corresponding the samples
Input
Input file contains one or more testcases. The first line of each testcase contains n (2 <= n <= 50). Next n lines contain the description of the triangle, i-th of these lines contains
2i - 1 characters, describing unit triangles in the i-th row, from left to right. Character '0' means that the triangle belongs to the first part of the main triangle, '1' means that it belongs to the second one.
Testcase with n = 0 designates the end of the test data, this testcase must not be processed. There is no blank line in the input file.
Output
For each puzzle output the line with its number followed by the line that states whether the parts can be separated. Do not output any blank lines.
Sample Input
6
0
001
00011
0000011
000111111
00111111111
6
0
001
00111
0011011
000000111
00111111111
0
Sample Output
Puzzle 1
Parts can be separated
Puzzle 2
Parts cannot be separated
Author: Andrew Stankevich
Source: Andrew Stankevich's Contest #7
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm> using namespace std; int n;
int tu[55][200]; bool inmap(int x,int y)
{
if((x>=1&&x<=n)&&(y>=1&&y<=2*x-1)) return true;
return false;
} bool check_down(int x,int y,int c,int kind)
{
if(kind==1)
{
/// x+1 --- y,y+1,y+2
/// x+2 --- y+1,y+2,y+3
for(int i=0;i<3;i++)
{
int nx=x+1,ny=y+i;
if(inmap(nx,ny))
{
if(tu[nx][ny]!=c) return false;
}
}
for(int i=1;i<4;i++)
{
int nx=x+2,ny=y+i;
if(inmap(nx,ny))
{
if(tu[nx][ny]!=c) return false;
}
}
}
else if(kind==0)
{
/// x --- y-1 y+1
/// x+1 --- y+1 y+2 y+3
/// x+2 -- y+2 for(int i=-1;i<=1;i++)
{
int nx=x,ny=y+i;
if(inmap(nx,ny))
{
if(tu[nx][ny]!=c) return false;
}
} for(int i=1;i<=3;i++)
{
int nx=x+1,ny=y+i;
if(inmap(nx,ny))
{
if(tu[nx][ny]!=c) return false;
}
} if(inmap(x+2,y+2))
{
if(tu[x+2][y+2]!=c) return false;
}
} return true;
} bool check_left(int x,int y,int c,int kind)
{
if(kind==1)
{
/// x -- y-1 y-2 y-3
/// x-1 y-2,y-3,y-4 for(int i=1;i<=3;i++)
{
int nx=x,ny=y-i;
if(inmap(nx,ny))
{
if(tu[nx][ny]!=c) return false;
}
} for(int i=2;i<=4;i++)
{
int nx=x-1,ny=y-i;
if(inmap(nx,ny))
{
if(tu[nx][ny]!=c) return false;
}
}
}
else if(kind==0)
{
/// x -- y-1 y-2
/// x-1 -- y-1 y-2 y-3 y-4 for(int i=1;i<=2;i++)
{
int nx=x,ny=y-i;
if(inmap(nx,ny))
{
if(tu[nx][ny]!=c) return false;
}
} for(int i=1;i<=4;i++)
{
int nx=x-1,ny=y-i;
if(inmap(nx,ny))
{
if(tu[nx][ny]!=c) return false;
}
}
} return true;
} bool check_right(int x,int y,int c,int kind)
{
if(kind==1)
{
/// x -- y+1 y+2 y+3
/// x-1 y y+1 y+2 y+3 for(int i=1;i<=3;i++)
{
int nx=x,ny=y+i;
if(inmap(nx,ny))
{
if(tu[nx][ny]!=c) return false;
}
} for(int i=0;i<=3;i++)
{
int nx=x-1,ny=y+i;
if(inmap(nx,ny))
{
if(tu[nx][ny]!=c) return false;
}
} }
else if(kind==0)
{
/// x y+1 y+2
/// x-1 y-1 ... y+3
for(int i=1;i<=2;i++)
{
int nx=x,ny=y+i;
if(inmap(nx,ny))
{
if(tu[nx][ny]!=c) return false;
}
}
for(int i=-1;i<=3;i++)
{
int nx=x-1,ny=y+i;
if(inmap(nx,ny))
{
if(tu[nx][ny]!=c) return false;
}
}
}
return true;
} bool check_heng()
{
bool flag=true;
for(int i=1;i<=n&&flag;i++)
{
for(int j=1;j<=2*i-1&&flag;j++)
{
if(tu[i][j]==0)
{
if(inmap(i,j-1)==false) continue;
if(tu[i][j-1]==0) continue;
else flag=false;
}
}
}
if(flag) return true;
flag=true;
for(int i=1;i<=n&&flag;i++)
{
for(int j=1;j<=2*i-1&&flag;j++)
{
if(tu[i][j]==0)
{
if(inmap(i,j+1)==false) continue;
if(tu[i][j+1]==0) continue;
else flag=false;
}
}
}
if(flag) return true;
return false;
} bool check_shu()
{
bool flag=true;
for(int i=1;i<=n&&flag;i++)
{
for(int j=1;j<=2*i-1&&flag;j++)
{
if(tu[i][j]==0)
{
if(j%2==1)
{
int nx=i, ny=j+1;
if(inmap(nx,ny)==false) continue;
if(tu[nx][ny]==0) ;
else flag=false;
}
else if(j%2==0)
{
int nx=i-1, ny=j-1;
if(inmap(nx,ny)==false) continue;
if(tu[nx][ny]==0) ;
else flag=false;
}
}
}
}
if(flag) return true;
flag=true;
for(int i=1;i<=n&&flag;i++)
{
for(int j=1;j<=2*i-1&&flag;j++)
{
if(tu[i][j]==0)
{
if(j%2==1)
{
int nx=i+1, ny=j+1;
if(inmap(nx,ny)==false) continue;
if(tu[nx][ny]==0) ;
else flag=false;
}
else if(j%2==0)
{
int nx=i, ny=j-1;
if(inmap(nx,ny)==false) continue;
if(tu[nx][ny]==0) ;
else flag=false;
}
}
}
}
if(flag) return true;
return false;
} bool check_xie()
{
bool flag=true;
for(int i=1;i<=n&&flag;i++)
{
for(int j=1;j<=2*i-1&&flag;j++)
{
if(tu[i][j]==0)
{
if(j%2==1)
{
int nx=i, ny=j-1;
if(inmap(nx,ny)==false) continue;
if(tu[nx][ny]==0) ;
else
{
flag=false;
}
}
else if(j%2==0)
{
int nx=i-1, ny=j-1;
if(inmap(nx,ny)==false) continue;
if(tu[nx][ny]==0) ;
else
{
flag=false;
}
}
}
}
}
if(flag) return true;
flag=true;
for(int i=1;i<=n&&flag;i++)
{
for(int j=1;j<=2*i-1&&flag;j++)
{
if(tu[i][j]==0)
{
if(j%2==1)
{
int nx=i+1, ny=j+1;
if(inmap(nx,ny)==false) continue;
if(tu[nx][ny]==0) ;
else
{
flag=false;
}
}
else if(j%2==0)
{
int nx=i, ny=j+1;
if(inmap(nx,ny)==false) continue;
if(tu[nx][ny]==0) ;
else
{
flag=false;
}
}
}
}
}
if(flag) return true;
return false;
} char hang[5000]; int main()
{
int cas=1;
while(scanf("%d",&n)!=EOF&&n)
{
printf("Puzzle %d\n",cas++);
int zero=0,one=0;
for(int i=1;i<=n;i++)
{
scanf("%s",hang+1);
for(int j=1;j<=2*i-1;j++)
{
tu[i][j]=hang[j]-'0';
if(tu[i][j]==0) zero++;
else one++;
}
} if(one==0||zero==0)
{
puts("Parts cannot be separated"); continue;
} bool flag=true; if(check_heng()||check_shu()||check_xie()){puts("Parts can be separated"); continue;} for(int i=1;i<=n&&flag;i++)
{
for(int j=1;j<=2*i-1&&flag;j++)
{
if(tu[i][j]!=0) continue;
if(check_down(i,j,0,j%2)) continue;
else flag=false;
}
} if(flag==true)
{
puts("Parts can be separated"); continue;
} flag=true;
for(int i=1;i<=n&&flag;i++)
{
for(int j=1;j<=2*i-1&&flag;j++)
{
if(tu[i][j]!=0) continue;
if(check_left(i,j,0,j%2)) continue;
else flag=false;
}
} if(flag==true)
{
puts("Parts can be separated"); continue;
} flag=true;
for(int i=1;i<=n&&flag;i++)
{
for(int j=1;j<=2*i-1&&flag;j++)
{
if(tu[i][j]!=0) continue;
if(check_right(i,j,0,j%2)) continue;
else flag=false;
}
} if(flag==true)
{
puts("Parts can be separated"); continue;
}
else
{
puts("Parts cannot be separated");
} }
return 0;
}
ZOJ 2610 Puzzle 模拟的更多相关文章
- A - Jugs ZOJ - 1005 (模拟)
题目链接:https://cn.vjudge.net/contest/281037#problem/A 题目大意:给你a,b,n.a代表第一个杯子的容量,b代表第二个杯子的容量,然后一共有6种操作.让 ...
- Capture the Flag ZOJ - 3879(模拟题)
In computer security, Capture the Flag (CTF) is a computer security competition. CTF contests are us ...
- ZOJ 3019 Puzzle
解题思路:给出两个数列an,bn,其中an,bn中元素的顺序可以任意改变,求an,bn的LCS 因为数列中的元素可以按任意顺序排列,所以只需要求出an,bn中的元素有多少个是相同的即可. 反思:一开始 ...
- ZOJ 3705 Applications 模拟
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include< ...
- ZOJ 3652 Maze 模拟,bfs,读题 难度:2
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4842 要注意题目中两点: 1.在踏入妖怪控制的区域那一刹那,先减行动力,然后才 ...
- [ZOJ 1009] Enigma (模拟)
题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1009 题目大意:给你三个转换轮,只有当第一个转换轮转动一圈后第二 ...
- ZOJ 1122 Clock(模拟)
Clock Time Limit: 2 Seconds Memory Limit: 65536 KB You are given a standard 12-hour clock with ...
- 2018 German Collegiate Programming Contest (GCPC 18)
2018 German Collegiate Programming Contest (GCPC 18) Attack on Alpha-Zet 建树,求lca 代码: #include <al ...
- BFS+模拟 ZOJ 3865 Superbot
题目传送门 /* BFS+模拟:dp[i][j][p] 表示走到i,j,方向为p的步数为多少: BFS分4种情况入队,最后在终点4个方向寻找最小值:) */ #include <cstdio&g ...
随机推荐
- Visual Studio 2013 新增web项目IIS Express的64位版
使用Visual Studio 2012开发SharePoint的应该都遇到过下面的错误“SharePoint 在32位进程中不受支持”,而怎么修改目标平台都不好使,因为VS 2012所配备的IIS ...
- 类似于GROUP BY SUM() 用于字符串连接的语句
CREATE TABLE T ( [f1] VarCHAR(100), [f2] VarCHAR(100))goINSERT INTO T VALUES ('a','abc')INSERT INT ...
- poj 3264 Balanced Lineup 题解
Balanced Lineup Time Limit: 5000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u Subm ...
- Java性能监控工具:VisualVM
VisualVM是JDK自带的一款全能型性能监控和故障分析工具,包括对CPU使用.JVM堆内存消耗.线程.类加载的实时监控,内存dump文件分析,垃圾回收运行情况的可视化分析等,对故障排查和性能调优很 ...
- WEB漏洞挖掘技术总结
漏洞挖掘技术一直是网络攻击者最感兴趣的问题,漏洞挖掘的范围也在随着技术的提升而有所变化.在前期针对缓冲区溢出.格式化字符串.堆溢出.lib库溢出等技术都是针对ELF文件(Linux可执行文件)或者PE ...
- 在css加载完毕后执行后续代码
最近在写项目的framework,写个JQueryMessageBox的类,以使用jquery ui中的dialog()来显示消息框,为了使方法方便调用,便加入了自动判断页面是否加入了ui.js和ui ...
- spring+mybatis 多数据源切换
摘自: http://www.oschina.net/code/snippet_347813_12525 1. 代码: DbContextHolder public class DbContextHo ...
- 【深夜福利】Caffe 添加自己定义 Layer 及其 ProtoBuffer 參数
在飞驰的列车上,无法入眠.外面阴雨绵绵,思绪被拉扯到天边. 翻看之前聊天,想起还欠一个读者一篇博客. 于是花了点时间整理一下之前学习 Caffe 时添加自己定义 Layer 及自己定义 ProtoBu ...
- hdu 4287Intelligent IME(简单hash)
Intelligent IME Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) ...
- 总结一些Android好用的开源库
1.android-viewFlow https://github.com/pakerfeldt/android-viewflow 2. android-viewbadger https://gith ...