【POJ】1753【DFS】【Filp game】【又、全熄灯全亮灯问题】-优快云博客

本文深入探讨了游戏开发领域的关键技术，包括游戏引擎、编程语言、硬件优化等，并重点阐述了AI音视频处理的应用场景和实现方法，如语义识别、物体检测、语音变声等。通过实例分析，揭示了这些技术如何相互融合，提升游戏体验和互动性。

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Description

Flip game is played on arectangular 4x4 field with two-sided pieces placed on each of its16 squares. One side of each piece is white and the other one isblack 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 theirupper side from black to white and vice versa. The pieces to beflipped are chosen every round according to the followingrules:

Choose any one of the 16 pieces.
Flip the chosen piece and also all adjacent pieces tothe left, to the right, to the top, and to the bottom ofthe chosen piece (if there are any).（四周包括自己颜色全变）
Consider the following positionas an example:

bwbw
wwww
bbwb
bwwb
Here "b" denotes pieces lying their black side up and "w" denotespieces lying their white side up. If we choose to flip the 1stpiece from the 3rd row (this choice is shown at the picture), thenthe field will become:

bwbw
bwww
wwwb
wwwb
The goal of the game is to flip either all pieces white side up orall pieces black side up. You are to write a program that willsearch for the minimum number of rounds needed to achieve thisgoal.

Input

The input consists of 4 lineswith 4 characters "w" or "b" each that denote game fieldposition.

Output

Write to the output file asingle integer number - the minimum number of rounds needed toachieve the goal of the game from the given position. If the goalis initially achieved, then write 0. If it's impossible to achievethe goal, then write the word "Impossible" (without quotes).

Sample Input

bwwb
bbwb
bwwb
bwww

Sample Output

//看到这个图我就明白了，再次不解释，呵呵~用到位运算，把每一种状态转换为一个int数据

#include <stdio.h>

#include <stdlib.h>

#define MAX 65540

int flag[MAX],step[MAX];//flag[]作为标记，step[]作为记录次数

//用链表很可恶的RE，所以改为顺序队列

#define int_ int  

typedef struct queue{

 int_ *front;

 int_ *rear;

}Queue;

void InitQueue(Queue *Q){

 Q->front=Q->rear=malloc(sizeof(int_)*65536*2);

}

void EnQueue(Queue *Q,int e){

 *(Q->rear++)=e;

}

void GetPop(Queue *Q,int *e){

 *e=*(Q->front++);

}

int Empty(Queue *Q){

 if(Q->front==Q->rear)

  return 1;

 return 0;

}

int main(){

 Queue s;

 int i,j,n=0,m;

 char c;

 InitQueue(&s);

 for(i=0;i<4;i++){

  for(j=0;j<4;j++){

   scanf("%c",&c);

   n<<=1;

   if(c=='b')

    n++;

  }

  getchar();

 }

 if(n==0 || n==65535){

  printf("0\n");

  return 0;

 }

 EnQueue(&s,n);

 flag[n]=1;

 while(!(Empty(&s))){

  GetPop(&s,&n);//对头元素

  for(i=0;i<4;i++)

   for(j=0;j<4;j++){

    m=n;

    if(i==0)//只翻转下一行的同列元素

     m^=1<<(11-4*i-j);//15-4*(i+1)-j

    else if(i==3)//只翻转上一行的同列元素

     m^=1<<(19-4*i-j);//15-4*(i-1)-j

    else{//上下两行

     m^=1<<(11-4*i-j);

     m^=1<<(19-4*i-j);

    }

    if(j==0)//只翻转同行的下一列元素和自身,因为同一列的11即3的2进制

     m^=3<<(14-4*i-j);//15-4*i-j、15-4*i-(j+1)

    else if(j==3)

     m^=3<<(15-4*i-j);//15-4*i-(j+1)、15-4*i-j

    else

     m^=7<<(14-4*i-j);//111即为7的2进制

    if(m==0 || m==65535){

     printf("%d\n",step[n]+1);

     return 0;

    }

    if(!flag[m]){

     EnQueue(&s,m);

     flag[m]=1;

     step[m]=step[n]+1;

    }

   }

 }

 printf("Impossible\n");

 return 0;

}