[POJ1753]Flip Game【枚举】

Flip Game

Time Limit: 1000MS Memory Limit: 65536K

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:
1. Choose any one of the 16 pieces.
2. 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

Source

Northeastern Europe 2000

题解

我们可以发现，重复两次翻同一个地方，等于没翻，所以我们只要每个点DFS一次就可以了。
然后呢，我们会去考虑顺序，其实前一个想通了，这个也就想通了，每个点的状态是这个点被翻的次数，跟先后没关系。
这样就吧复杂度降到O(2^16)
完美解决。

代码如下

#include<cstdio>
#include<algorithm>
using namespace std;
int n=4,mp[10][10],ans=1<<30,que[25];
const int f[5][2]={{1,0},{0,1},{0,0},{-1,0},{0,-1}};
char read(){
    char ch=getchar();
    while(ch^'w'&&ch^'b') ch=getchar();
    return ch=='w';
}
bool chck(int x,int y){
    if(x<0||x>=n||y<0||y>=n) return 0;
    return 1;
}
void check(int t){
    int now[10][10];
    for(int i=0;i^n;i++)
    for(int j=0;j^n;j++) now[i][j]=mp[i][j];
    for(int i=0;i^t;i++)//翻转 
    for(int j=0;j^5;j++)
    if(chck(que[i]/n+f[j][0],que[i]%n+f[j][1])) now[que[i]/n+f[j][0]][que[i]%n+f[j][1]]=1-now[que[i]/n+f[j][0]][que[i]%n+f[j][1]];
    bool tt=1;
    for(int i=0;i^n;i++)//check是否全白 
    for(int j=0;j^n;j++) if(now[i][j]^1){tt=0;break;}
    if(tt){ans=min(ans,t);return;}
    tt=1;
    for(int i=0;i^n;i++)//check是否全黑 
    for(int j=0;j^n;j++) if(now[i][j]^0){tt=0;break;}
    if(tt){ans=min(ans,t);return;}
}
void DFS(int x,int t){
    if(t>ans) return;
    if(x==16){check(t);return;}
    que[t]=0;DFS(x+1,t);
    que[t]=x;DFS(x+1,t+1);
}
int main(){
    for(int i=0;i^n;i++)
    for(int j=0;j^n;j++) mp[i][j]=read();
    DFS(0,0);
    if(ans==1<<30) printf("Impossible\n");else printf("%d\n",ans);
    return 0;
}