Flip Game [bfs经典的二进制处理]【非原创】

Flip Game

Time Limit: 1000MS  Memory Limit: 65536K

Total Submissions: 3162  Accepted: 1310

 

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

bwwwSample Output

 

4

Source

 

Northeastern Europe 2000

 

第一次写位运算的BFS，原来好有趣呀-_-!

用一个int型的低16位代表状态，黑为1白为0,如果是全白（全为0,即10进制为0)或全黑（即为二进制的1111111111111111,即十进位的65535)则输出.二维数组change存的是进行异或运算的东东，化为相应二进位即可明白.

 

#include #include #define SIZE 10000000 const int change[][4] = {​{51200, 58368, 29184, 12544}, {35968, 20032, 10016, 4880}, {2248, 1252, 626, 305}, {140, 78, 39, 19}}; struct check { int dis; int visited; }d[67000]; struct Queue { int q[SIZE]; int head, tail; void clear() {head = tail = 0;} int empty() {return (head - tail) == 0 ? 1 : 0;} void push(int n) { if(tail == SIZE) tail = 1; q[tail] = n; tail++; } int pop() { int x; if(head == SIZE) head = 1; x = q[head]; head++; return x; } }Q; int bfs(int result) { int front, tmp, i, j; memset(d, 0, sizeof(d)); Q.clear(); Q.push(result); d[result].visited = 1; while(!Q.empty()) { front = Q.pop(); for(i = 0; i < 4; i++) for(j = 0; j < 4; j++) { tmp = front ^ change[i][j]; if(!d[tmp].visited) { Q.push(tmp); d[tmp].visited = 1; d[tmp].dis = d[front].dis + 1; if(tmp == 0 || tmp == 65535) return d[tmp].dis; } } } return -1; } int main() { char str; int result, num, i, count; while(scanf("%c", &str) != EOF) { result = 0; num = 1 << 15; if(str == 'b') result += num; num >>= 1; for(i = 2; i <= 16; i++) { scanf("%c", &str); if(str == 'b') result += num; num >>= 1; if(i % 4 == 0) getchar(); } if(result == 0 || result == 65535) {printf("0/n"); continue;} count = bfs(result); if(count == -1) puts("Impossible"); else printf("%d/n", count); } return 0; }   

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