poj 1753 -- Flip Game

最新推荐文章于 2019-08-27 16:39:20 发布

weixin_34179968

最新推荐文章于 2019-08-27 16:39:20 发布

阅读量78

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/ubuntu-kevin/p/3727098.html

本文介绍了一款名为FlipGame的游戏，该游戏在一个4x4的棋盘上进行，玩家的目标是通过翻转棋子使所有棋子颜色统一。文章详细解释了游戏规则，并提供了一种使用位运算进行搜索的方法来找出达到目标所需的最少翻转次数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Flip Game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 29059		Accepted: 12568

Description

Flip game is played on a rectangular 4x4 field with two-sided pieces placed on each of its 16 squares. One side of each piece is white and the other one is black and each piece is lying either it's black or white side up. Each round you flip 3 to 5 pieces, thus changing the color of their upper side from black to white and vice versa. The pieces to be flipped are chosen every round according to the following rules:

Choose any one of the 16 pieces.
Flip the chosen piece and also all adjacent pieces to the left, to the right, to the top, and to the bottom of the chosen piece (if there are any).

Consider the following position as an example:

bwbw
wwww
bbwb
bwwb
Here "b" denotes pieces lying their black side up and "w" denotes pieces lying their white side up. If we choose to flip the 1st piece from the 3rd row (this choice is shown at the picture), then the field will become:

bwbw
bwww
wwwb
wwwb
The goal of the game is to flip either all pieces white side up or all pieces black side up. You are to write a program that will search for the minimum number of rounds needed to achieve this goal.

Input

The input consists of 4 lines with 4 characters "w" or "b" each that denote game field position.

Output

Write to the output file a single integer number - the minimum number of rounds needed to achieve the goal of the game from the given position. If the goal is initially achieved, then write 0. If it's impossible to achieve the goal, then write the word "Impossible" (without quotes).

Sample Input

bwwb
bbwb
bwwb
bwww

Sample Output

4


这个题也是用位运算搜索（枚举）

　　0xc800,0xe400,0x7200,0x3100,
　　0x8c80,0x4e40,0x2720,0x1310,
　　0x08c8,0x04e4,0x0272,0x0131,
　　0x008c,0x004e,0x0027,0x0013,

用十六进制表示比较方便，这就是十六种状态，可以枚举每种即可，用异或操作，可以改变每种状态。

 1 /*======================================================================
 2  *           Author :   kevin
 3  *         Filename :   FlipGame.cpp
 4  *       Creat time :   2014-05-12 20:25
 5  *      Description :
 6  ========================================================================*/
 7 #include <iostream>
 8 #include <algorithm>
 9 #include <cstdio>
10 #include <cstring>
11 #include <queue>
12 #include <cmath>
13 #define M 6
14 #define Max 0xffff
15 using namespace std;
16 int cnt[Max+100];
17 int dir[M][M]={
18     0x0000,0x0000,0x0000,0x0000,0x0000,0x0000,
19     0x0000,0xc800,0xe400,0x7200,0x3100,0x0000,
20     0x0000,0x8c80,0x4e40,0x2720,0x1310,0x0000,
21     0x0000,0x08c8,0x04e4,0x0272,0x0131,0x0000,
22     0x0000,0x008c,0x004e,0x0027,0x0013,0x0000,
23     0x0000,0x0000,0x0000,0x0000,0x0000,0x0000
24 };
25 int BFS(int t)
26 {
27     queue<int>pq;
28     cnt[t] = 1;
29     pq.push(t);
30     while(!pq.empty()){
31         int num = pq.front();
32         pq.pop();
33         for(int i = 1; i <= 4; i++){
34             for(int j = 1; j <= 4; j++){
35                 int d = num^dir[i][j];
36                 if(d == 0x0000 || d == 0xffff){
37                     cnt[d] = cnt[num];
38                     return cnt[d];
39                 }
40                 if(!cnt[d]){
41                     cnt[d] = cnt[num]+1;
42                     pq.push(d);
43                 }
44             }
45         }
46     }
47     return -1;
48 }
49 void Init()
50 {
51     char str[M];
52     int number = 0,steps;
53     while(cin>>str){
54         number = 0;
55         memset(cnt,0,sizeof(cnt));
56         for(int j = 0; j < 4; j++){
57             number = number<<1;
58             if(str[j] == 'w'){
59                 number = number|1;
60             }
61         }
62         for(int i = 0; i < 3; i++){
63             cin>>str;
64             for(int j = 0; j < 4; j++){
65                 number = number<<1;
66                 if(str[j] == 'w'){
67                     number = number|1;
68                 }
69             }
70         }
71         if(number == 0x0000 || number == 0xffff){
72             printf("0\n");
73             continue;
74         }
75         else{
76             steps = BFS(number);
77         }
78         if(steps == -1){
79             printf("Impossible\n");
80         }
81         else{
82             printf("%d\n",steps);
83         }
84     }
85 }
86 int main(int argc,char *argv[])
87 {
88     Init();
89     return 0;
90 }