POJ1573(枚举，动态规划)

Flip Game

Time Limit: 1000 MS Memory Limit: 65536 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

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
********************************************************分割线**********************************************************************************************************************************
学校竟然把它放到枚举的练习题里，我都惊呆了，不过仔细一想的确是一道枚举的题，但我是用动态规划的思想所以我又把它算是动态规划了，

 1 //动态规划的思想，每一个点都有翻不翻两种情况，复杂度为2^n,需注意的是scanf的问题
 2 #include<stdio.h>
 3 #include<string.h>
 4 using namespace std;
 5 bool chess[6][6]={false};
 6 int step;
 7 bool flag;
 8 int r[5]={-1,1,0,0,0};
 9 int c[5]={0,0,-1,1,0};//(0,0)biaoshizijihenhaoyonga
10 bool isOver(){//jianchashifoujieshu
11     int i,j;
12     for(i=1;i<5;i++)
13         for(j=1;j<5;j++)
14             if(chess[i][j]!=chess[1][1])
15                 return false;
16     return true;
17 }
18 void flip(int row,int col){//反转辞典对应可以翻转的所有点
19     for(int i=0;i<5;i++)
20         chess[row+r[i]][col+c[i]]=!chess[row+r[i]][col+c[i]];
21 }
22 void dfs(int row,int col,int deep){
23     if(deep==step){
24       //  printf("HHH%d==%dflag==%d
",deep,step,flag);
25         flag=isOver();
26         return ;
27     }
28 
29     if(flag||row==5) return ;
30     flip(row,col);//翻转
31     if(col<4) dfs(row,col+1,deep+1);
32     else dfs(row+1,1,deep+1);
33     flip(row,col);
34     if(col<4) dfs(row,col+1,deep);
35     else dfs(row+1,1,deep);
36     return ;
37 }
38 int main()
39 {
40     int i,j;
41     char temp;
42    // memset(chess,0,sizeof(chess));
43     for(int i=1;i<5;i++){
44         for(int j=1;j<5;j++){
45             scanf("%c",&temp);//cin不会接受换行符，但scanf会连换行符一起读入
46             if(temp=='b') chess[i][j]=true;
47         }
48         getchar();
49     }
50     /*for(i=1;i<5;i++)
51     for(j=1;j<5;j++)
52     printf("%d ",chess[i][j]);*/
53     for(step=0;step<=16;step++){//最大是16次翻转，因为16次之后的每次翻转都会造成和不翻那个格子相同的情况。
54         dfs(1,1,0);
55        // printf("STEP=%d
",step);
56         if(flag) break;
57     }
58     if(flag) printf("%d
",step);
59     else printf("Impossible
");
60 
61 }

View Code

这里是以翻几个棋子为dp线索的，和子集生成好像啊！每一个都是翻不翻，而子集生成是每一个都取不取以个数为线索寻找结束条件。

我暂时理解为暴力，dp，深搜法(我还是个小白啊,会的少，勿怪，暂时对知识点的总结还不是很成熟)