POJ1753 Flip Game 枚舉

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

bwwbbbwbbwwbbwww

Sample Output

4題目大意：給一個4×4的棋盤狀態，該棋盤有黑白兩種棋子，你的任務是把棋盤全部翻轉爲白和全部爲黑。翻轉的規則是該棋和它的上下左右（如果存在）。作爲一個acm新手，初次拿到這道題，一絲思路都沒有，感覺很復雜。後來冷靜下來，慢慢想想（慶幸自己有這種精神，而不是不會就馬上找題解），每個棋子要麼就不翻轉，要麼就翻轉一次，決不可能翻轉兩次及以上（翻轉兩次等同沒翻轉，翻轉3次等同翻轉1次，其他類似）。一共就只有16個棋子，所以一個一個枚舉就可以（剛開始理解錯，以爲是16！，其實不然，事實上只有2^16,這是以爲每個棋子只有兩種狀態，一共16個棋子，所以爲2^16種翻轉方案）。
#include <iostream>#include <cstdio>#include <cstdlib>#include <string>#include <cstring>using namespace std;int vis[16];int piece[4][4];int ans;void judge(){    int tag = piece[0][0];    for(int i = 0; i < 4; ++i)        for(int j = 0; j < 4; ++j)        if(tag != piece[i][j]) return;    int sum = 0;    for(int i = 0; i < 16; ++i) if(vis[i]) sum++;    ans = min(ans, sum);}void run(int pos){    int x = pos / 4;    int y = pos % 4;    piece[x][y] = !piece[x][y];    if(y-1 >= 0) piece[x][y-1] = !piece[x][y-1];    if(y+1 <= 3) piece[x][y+1] = !piece[x][y+1];    if(x-1 >= 0) piece[x-1][y] = !piece[x-1][y];    if(x+1 <= 3) piece[x+1][y] = !piece[x+1][y];}void solve(int cur){    if(cur >= 16) return;    judge();    solve(cur+1);    vis[cur] = 1;    run(cur);    judge();    solve(cur+1);    run(cur);    vis[cur] = 0;}int main(){    string ins;    for(int i = 0; i < 4; ++i){        cin >> ins;        for(int j = 0; j < 4; ++j){            if(ins[j] == 'w') piece[i][j] = 1;            else piece[i][j] = 0;        }    }    memset(vis, 0, sizeof(vis));    ans = 1e4;    int cur = 0;    solve(cur);    if(ans == 1e4) printf("Impossible\n");    else printf("%d\n", ans);    return 0;}