poj_1184_BFS（？可以不用吧，待改…

题目描述：

  给一个初始序列，光标停在序列第一个位置。有6个键盘操作分别做左右移动，交换首末和加减操作，求如何用最短的操作组合数来令初始序列达到某个序列。

 

题目思路：

   BFS。

   待改——代码有点不伦不类。

 

代码：

#include <stdio.h>
#include <stdio.h>
#include <math.h>
//#include <time.h>
//#include <windows.h>

#define N 100000
#define MAX 1000000

typedef struct{
   int array[8];
   int c;
   int pos[6];
}QUEUE;

char flag[MAX][6][10];
int src_array[8], des_array[7], min_cnt, src, des;
QUEUE q[N];
int top, num;

int array_to_int(int a[7])
{
   int sum = 0, i;
  for(i=0;i<6;i++)
     sum = sum*10 + a[i];
   return sum;
}

void find_solution()
{
   QUEUE tq;
   inttmp,j,pos,value,f2=0;
  
  while(top<=num){ //取队列top，加入后续队列
  
     if(q[top].c > min_cnt){
        top ++;
        continue;
     }

     tq =q[top];     //判断是否更新min_cnt
     if(array_to_int(tq.array) == des){
        if(tq.c < min_cnt)
            min_cnt = tq.c;
     }
     else{
        j = 0;
        while(j<6){
           f2 = 0;
           if(tq.pos[j]){ //光标到达过的位置
               tq.c += abs(tq.array[j] - des_array[j]);
               if(tq.c > min_cnt)
                  break;
               f2 = 1;
           }
           else{
              if(tq.pos[j]!=des_array[j])
                 break;
              f2 = 1;
           }
           j++; 
        }
        if(f2)
           min_cnt = tq.c;
     }
     
    //继续寻找可能的组合
    pos = q[top].array[6];
    tq = q[top];
    //swap0  --- 1 -- 光标状态不变 ，光标位置不变
    if(pos!=0 &&tq.array[pos]!=tq.array[0]){
         tmp = tq.array[pos];
         tq.array[pos] = tq.array[0];
         tq.array[0] = tmp;
         value = array_to_int(tq.array);
         tq.c++;
         if(tq.c < flag[value][pos][tq.array[7]]){
            //printf("swap0\n");
            flag[value][pos][tq.array[7]] = tq.c;
            num++;
            q[num] = tq;
         }
     }
     //swap1   --- 1 --- 光标状态+6 ，光标位置不变
     tq = q[top];
     if(pos!=5 &&tq.array[pos]!=tq.array[5]){
         tmp = tq.array[pos];
         tq.array[pos] = tq.array[5];
         tq.array[5] = tmp;
         if(tq.array[7]<5)
            tq.array[7]+=5;
         value = array_to_int(tq.array);
         tq.c++;
         if(tq.c < flag[value][pos][tq.array[7]]){
            //printf("swap1\n");
            flag[value][pos][tq.array[7]] = tq.c;
            num++;
            tq.pos[5] = 1;
            q[num] = tq;
         }
     }
         
     //left  --- 3 ---光标状态不变 ，光标位置-1
     tq = q[top];
     if(pos!=0){
        tq.array[6] = pos-1;
        value = array_to_int(tq.array);
        tq.c ++;
        if(tq.c < flag[value][pos-1][tq.array[7]]){
           //printf("left\n");
           flag[value][pos-1][tq.array[7]] = tq.c;
           num++;
           q[num] = tq;
        }
     }
     //right  ---  4 ---光标状态+1ornot,光标位置+1
     tq = q[top];
     if(pos!=5){
        tq.array[6] = pos+1;
        if(tq.pos[pos+1]==0)
           tq.array[7] += 1;
        value = array_to_int(tq.array);
        tq.c ++;
        if(tq.c < flag[value][pos+1][tq.array[7]]){
           //printf("right\n");
           flag[value][pos+1][tq.array[7]] = tq.c;
           num++;
           tq.pos[pos+1] = 1;
           q[num] = tq;
        }
     }
     top ++; //头部指针后移
   }
}

main()
{
   int i, tmp1, tmp2, j, k;
   //int timebegin;
   scanf("%d%d",&src, &des);
  
   min_cnt = 5;
   tmp1 = src;
   tmp2 = des;
   i=0;
  while(i<6){
     src_array[5-i] = tmp1;
     tmp1 = tmp1/10;
     des_array[5-i] = tmp2;
     tmp2 = tmp2/10;
     min_cnt += fabs(src_array[5-i]-des_array[5-i]);
     i++;
   }
  for(i=0;i<MAX;i++)
     for(j=0;j<6;j++)
        for(k=0;k<10;k++)
           flag[i][j][k] = min_cnt;
     
   src_array[6] = 0;
   src_array[7] = 0;
  flag[array_to_int(src_array)][0][0]=1;

   top = 1;
  for(i=0;i<8;i++)
     q[top].array[i] = src_array[i];
  
   q[top].c = 0;
   memset(q[top].pos, 0,sizeof(q[top].pos));
   q[top].pos[0] = 1;
   num = top;
  
  // timebegin = GetTickCount();
   find_solution();
  // printf("time =%d\n",GetTickCount()-timebegin);
   printf("%d\n",min_cnt);
  
   system("pause");
   return 0;
}