1647: [Usaco2007 Open]Fliptile 翻格子游戏
1647: [Usaco2007 Open]Fliptile 翻格子游戏
Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 423 Solved: 173
Description
Farmer John knows that an intellectually satisfied cow is a happy cow who will give more milk. He has arranged a brainy activity for cows in which they manipulate an M x N grid (1 <= M <= 15; 1 <= N <= 15) of square tiles, each of which is colored black on one side and white on the other side. As one would guess, when a single white tile is flipped, it changes to black; when a single black tile is flipped, it changes to white. The cows are rewarded when they flip the tiles so that each tile has the white side face up. However, the cows have rather large hooves and when they try to flip a certain tile, they also flip all the adjacent tiles (tiles that share a full edge with the flipped tile). Since the flips are tiring, the cows want to minimize the number of flips they have to make. Help the cows determine the minimum number of flips required, and the locations to flip to achieve that minimum. If there are multiple ways to achieve the task with the minimum amount of flips, return the one with the least lexicographical ordering in the output when considered as a string. If the task is impossible, print one line with the word "IMPOSSIBLE".
约翰知道,那些高智力又快乐的奶牛产奶量特别高.所以他做了一个翻瓦片的益智游戏来娱乐奶牛.在一个M×N(1≤M,N≤15)的骨架上,每一个格子里都有一个可以翻转的瓦片.瓦片的一面是黑色的,而另一面是白色的.对一个瓦片进行翻转,可以使黑变白,也可以使白变黑.然而,奶牛们的蹄子是如此的巨大而且笨拙,所以她们翻转一个瓦片的时候,与之有公共边的相邻瓦片也都被翻转了.那么,这些奶牛们最少需要多少次翻转,使所有的瓦片都变成白面向上呢?如杲可以做到,输出字典序最小的结果(将结果当成字符串处理).如果不能做到,输出“IMPOSSIBLE”.
Input
* Line 1: Two space-separated integers: M and N
* Lines 2..M+1: Line i+1 describes the colors (left to right) of row i of the grid with N space-separated integers which are 1 for black and 0 for white
第1行输入M和N,之后M行N列,输入游戏开始时的瓦片状态.0表示白面向上,1表示黑面向上.
Output
* Lines 1..M: Each line contains N space-separated integers, each specifying how many times to flip that particular location.
输出M行,每行N个用空格隔开的整数,表示对应的格子进行了多少次翻转.
Sample Input
4 4 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1
Sample Output
0 0 0 0 1 0 0 1 1 0 0 1 0 0 0 0 OUTPUT DETAILS: After flipping at row 2 column 1, the board will look like: 0 0 0 1 1 0 1 0 1 1 1 0 1 0 0 1 After flipping at row 2 column 4, the board will look like: 0 0 0 0 1 0 0 1 1 1 1 1 1 0 0 1 After flipping at row 3 column 1, the board will look like: 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 1 After flipping at row 3 column 4, the board will look like: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Another solution might be: 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 but this solution is lexicographically higher than the solution above.
HINT
Source
题解:没记错的话,这个是当年囧神(囧神=JSZKC,省选前夕orz一下攒攒RP)出的NOIP模拟题里面的(T_3),当时的我只知道 (O({2}^{NM})) 的纯暴力枚举,但事实上不用这样——
其实还是枚举,但实际上只要 (O({2}^{N} )) 的枚举即可,看到 (N) ,显然就是枚举第一行啦,事实上第一行的决策将直接决定下一行的决策,然后下一行影响下一行,也就是说实际上第一行的决定决定了全局的决策,然后根据第一行二进制穷举出来的进行模拟,然后判断此方案是否可行,然后打擂台记录下来
有人可能会在比对过程中再弄个字典序比较,因为题目中说要字典序最小,但是还有一个细节你似乎忽略了——我们二进制穷举从小数字到大数字本身就符合字典序上升,对于一个第一行决策情况,最多只有一种合法解,所以完全不必比较,直接“先入为主”即可
1 /**************************************************************
2 Problem: 1647
3 User: HansBug
4 Language: Pascal
5 Result: Accepted
6 Time:500 ms
7 Memory:232 kb
8 ****************************************************************/
9
10 var
11 i,j,k,l,m,n,ans:longint;
12 a,c,e,f:array[0..20,0..20] of longint;
13 b:array[0..20] of longint;
14 begin
15 readln(n,m);
16 for i:=1 to n do
17 begin
18 for j:=1 to m do read(a[i,j]);
19 readln;
20 end;
21 fillchar(b,sizeof(b),0);ans:=maxlongint;
22 while b[0]=0 do
23 begin
24 for i:=1 to m do c[0,i]:=b[i];
25 for i:=1 to n do
26 for j:=1 to m do c[i,j]:=a[i,j];
27 fillchar(e,sizeof(e),0);k:=0;
28 for i:=1 to n do
29 begin
30 for j:=1 to m do
31 if c[i-1,j]=1 then
32 begin
33 e[i,j]:=1;inc(k);
34 c[i,j]:=1-c[i,j];
35 c[i,j-1]:=1-c[i,j-1];
36 c[i,j+1]:=1-c[i,j+1];
37 c[i-1,j]:=1-c[i-1,j];
38 c[i+1,j]:=1-c[i+1,j];
39 end;
40 end;
41 l:=0;
42 for i:=1 to m do inc(l,c[n,i]);
43 if l=0 then
44 begin
45 if k<ans then
46 begin
47 ans:=k;
48 for i:=1 to n do
49 for j:=1 to m do
50 f[i,j]:=e[i,j];
51 end;
52 end;
53 i:=m;
54 while b[i]=1 do
55 begin
56 b[i]:=0;
57 dec(i);
58 end;
59 b[i]:=1;
60 end;
61 if ans=maxlongint then
62 begin
63 writeln('IMPOSSIBLE');
64 halt;
65 end;
66 for i:=1 to n do
67 for j:=1 to m do
68 if j<m then write(f[i,j],' ') else writeln(f[i,j]);
69 readln;
70 end.
- JavaScript 教程
- JavaScript 编辑工具
- JavaScript 与HTML
- JavaScript 与Java
- JavaScript 数据结构
- JavaScript 基本数据类型
- JavaScript 特殊数据类型
- JavaScript 运算符
- JavaScript typeof 运算符
- JavaScript 表达式
- JavaScript 类型转换
- JavaScript 基本语法
- JavaScript 注释
- Javascript 基本处理流程
- Javascript 选择结构
- Javascript if 语句
- Javascript if 语句的嵌套
- Javascript switch 语句
- Javascript 循环结构
- Javascript 循环结构实例
- Javascript 跳转语句
- Javascript 控制语句总结
- Javascript 函数介绍
- Javascript 函数的定义
- Javascript 函数调用
- Javascript 几种特殊的函数
- JavaScript 内置函数简介
- Javascript eval() 函数
- Javascript isFinite() 函数
- Javascript isNaN() 函数
- parseInt() 与 parseFloat()
- escape() 与 unescape()
- Javascript 字符串介绍
- Javascript length属性
- javascript 字符串函数
- Javascript 日期对象简介
- Javascript 日期对象用途
- Date 对象属性和方法
- Javascript 数组是什么
- Javascript 创建数组
- Javascript 数组赋值与取值
- Javascript 数组属性和方法
- PHP yield关键字功能与用法分析
- php PDO属性设置与操作方法分析
- PHP时间日期增减操作示例【date strtotime实现加一天、加一月等操作】
- CI框架实现创建自定义类库的方法
- php如何计算两坐标点之间的距离
- Python调用C语言程序方法解析
- php workerman定时任务的实现代码
- Yii2.0 RESTful API 基础配置教程详解
- opencv 形态学变换(开运算,闭运算,梯度运算)
- 使用darknet框架的imagenet数据分类预训练操作
- php两点地理坐标距离的计算方法
- tp5(thinkPHP5)框架连接数据库的方法示例
- CI框架附属类用法分析
- ThinkPHP5.0框架结合Swoole开发实现WebSocket在线聊天案例详解
- PHP封装的page分页类定义与用法完整示例