POj 1753 -- Flip Game -- bitwise operation + BFS

POj 1753 -- Flip Game -- bitwise operation + BFS

Flip Game Time Limit:1000 MS Memory Limit:65536 K Total Submissions:30669 Accepted:13345 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 bwwbbbwbbwwbbwww Sample Output 4 Source I have to say that bitwise computation is a powerful thing that has never been used before. I did not expect such a question that has no clue. bitwise computation is optimized into a bare one-dimensional BFS, this question is intended to be a flop game with black and white sides. When all the cards are black or white, the game is in the correct status and outputs the minimum number of operation steps. You can flip the cards at each position. However, if you flip the cards at the current position, the cards at the adjacent positions will be turned over. For all states, there are 2 ^ 16 (4*4) For the flop operation, if the current status is s and you want to flip the card at position I, as long as s ^ = 1 < # Include # Include # Include Using namespace std; bool vis [70000]; char ma [20]; typedef struct node {int state, step ;}; void bfs (int s) {queue Q; node t; t. state = s; t. step = 0; vis [s] = 1; Q. push (t); while (! Q. empty () {node v = Q. front (); Q. pop (); if (v. state = 0 | v. state = 65535) {cout < = 0) tem ^ = 1 <(I-4); // upper if (I <12) tem ^ = 1 <(I + 4 ); // lower if (I % 4! = 0) tem ^ = 1 <(I-1); // left if (I + 1) % 4! = 0) tem ^ = 1 <(I + 1); // right if (! Vis [tem]) {vis [tem] = 1; t. state = tem; t. step = v. step + 1; Q. push (t) ;}} puts ("Impossible");} int main () {int I = 0, s = 0; memset (vis, 0, sizeof (vis); while (I <16) {cin> ma [I]; if (ma [I] = 'B') s + = 1 <