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  • hdu 1195 Open the Lock

    Open the Lock

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 3679    Accepted Submission(s): 1611


    Problem Description
    Now an emergent task for you is to open a password lock. The password is consisted of four digits. Each digit is numbered from 1 to 9.
    Each time, you can add or minus 1 to any digit. When add 1 to '9', the digit will change to be '1' and when minus 1 to '1', the digit will change to be '9'. You can also exchange the digit with its neighbor. Each action will take one step.

    Now your task is to use minimal steps to open the lock.

    Note: The leftmost digit is not the neighbor of the rightmost digit.
     
    Input
    The input file begins with an integer T, indicating the number of test cases.

    Each test case begins with a four digit N, indicating the initial state of the password lock. Then followed a line with anotther four dight M, indicating the password which can open the lock. There is one blank line after each test case.
     
    Output
    For each test case, print the minimal steps in one line.
     
    Sample Input
    2
    1234
    2144
     
    1111
    9999
     
    Sample Output
    2
    4
     
    Author
    YE, Kai
     
    Source
    Zhejiang University Local Contest 2005
     
     
     
      1 #include<iostream>
  2 #include<stdio.h>
  3 #include<cstring>
  4 #include<cstdlib>
  5 #include<queue>
  6 using namespace std;
  7 
  8 int final[4],start[4];
  9 struct node
 10 {
 11     int a[4];
 12     int time;
 13 };
 14 bool dp[10][10][10][10];
 15 queue<node>Q;
 16 
 17 int bfs()
 18 {
 19     node cur,now;
 20     int i,k,a,b,c,d;
 21     for(i=0;i<4;i++) cur.a[i]=start[i];
 22     cur.time=0;
 23     Q.push(cur);
 24     dp[cur.a[0]][cur.a[1]][cur.a[2]][cur.a[3]]=true;
 25     while(!Q.empty())
 26     {
 27         cur=Q.front();
 28         Q.pop();
 29         for(i=0,k=0;i<4;i++)
 30         {
 31             if(final[i]==cur.a[i])
 32             {
 33                 k++;
 34             }
 35             if(k==4) return cur.time;
 36         }
 37         for(i=0;i<4;i++)//add 1
 38         {
 39             now=cur;
 40             if(now.a[i]==9)
 41                 now.a[i]=1;
 42             else now.a[i]=now.a[i]+1;
 43             a=now.a[0];
 44             b=now.a[1];
 45             c=now.a[2];
 46             d=now.a[3];
 47             if(dp[a][b][c][d]==false)
 48             {
 49                 dp[a][b][c][d]=true;
 50                 now.time++;
 51                 Q.push(now);
 52             }
 53         }
 54         for(i=0;i<4;i++)//del 1
 55         {
 56             now=cur;
 57             if(now.a[i]==1)
 58                 now.a[i]=9;
 59             else now.a[i]=now.a[i]-1;
 60             a=now.a[0];
 61             b=now.a[1];
 62             c=now.a[2];
 63             d=now.a[3];
 64             if(dp[a][b][c][d]==false)
 65             {
 66                 dp[a][b][c][d]=true;
 67                 now.time++;
 68                 Q.push(now);
 69             }
 70         }
 71         for(i=0;i<3;i++)//change
 72         {
 73             now=cur;
 74             k=now.a[i];
 75             now.a[i]=now.a[i+1];
 76             now.a[i+1]=k;
 77 
 78             a=now.a[0];
 79             b=now.a[1];
 80             c=now.a[2];
 81             d=now.a[3];
 82             if(dp[a][b][c][d]==false)
 83             {
 84                 dp[a][b][c][d]=true;
 85                 now.time++;
 86                 Q.push(now);
 87             }
 88         }
 89     }
 90     return -1;
 91 }
 92 int main()
 93 {
 94     int T,cur,i;
 95     char a[10],b[10];
 96     scanf("%d",&T);
 97     getchar();
 98     while(T--)
 99     {
100         scanf("%s",a);
101         scanf("%s",b); 
102         for(i=0;i<4;i++)
103             start[i]=a[i]-'0';
104         for(i=0;i<4;i++)
105             final[i]=b[i]-'0';
106         memset(dp,false,sizeof(dp));
107         while(!Q.empty())
108         {
109             Q.pop();
110         }
111         cur=bfs();
112         printf("%d
",cur);
113     }
114     return 0;
115 }

    双向广搜

      1 #include<iostream>
  2 #include<stdio.h>
  3 #include<cstring>
  4 #include<cstdlib>
  5 #include<queue>
  6 using namespace std;
  7 
  8 char a[6],b[6];
  9 bool flag;
 10 struct node
 11 {
 12     char cur[4];
 13 };
 14 queue<node>Q[2];
 15 bool dp[2][10][10][10][10];
 16 
 17 void bfs(int x)
 18 {
 19     int size,i;
 20     node t,k;
 21     size=Q[x].size();
 22 
 23     while(size--)
 24     {
 25         k=Q[x].front();
 26         Q[x].pop();
 27 
 28         if(dp[x^1][k.cur[0]-'0'][k.cur[1]-'0'][k.cur[2]-'0'][k.cur[3]-'0']==true)
 29         {
 30             flag=true;
 31             return;
 32         }
 33         for(i=0;i<=2;i++)
 34         {
 35             t=k;
 36             swap(t.cur[i],t.cur[i+1]);
 37             if(dp[x][t.cur[0]-'0'][t.cur[1]-'0'][t.cur[2]-'0'][t.cur[3]-'0']==false)
 38             {
 39                 dp[x][t.cur[0]-'0'][t.cur[1]-'0'][t.cur[2]-'0'][t.cur[3]-'0']=true;
 40                 Q[x].push(t);
 41             }
 42         }
 43         for(i=0;i<=3;i++)
 44         {
 45             t=k;
 46             if(t.cur[i]=='9')
 47                 t.cur[i]='1';
 48             else t.cur[i]++;
 49             if(dp[x][t.cur[0]-'0'][t.cur[1]-'0'][t.cur[2]-'0'][t.cur[3]-'0']==false)
 50             {
 51                 dp[x][t.cur[0]-'0'][t.cur[1]-'0'][t.cur[2]-'0'][t.cur[3]-'0']=true;
 52                 Q[x].push(t);
 53             }
 54             t=k;
 55             if(t.cur[i]=='1')
 56                 t.cur[i]='9';
 57             else t.cur[i]--;
 58             if(dp[x][t.cur[0]-'0'][t.cur[1]-'0'][t.cur[2]-'0'][t.cur[3]-'0']==false)
 59             {
 60                 dp[x][t.cur[0]-'0'][t.cur[1]-'0'][t.cur[2]-'0'][t.cur[3]-'0']=true;
 61                 Q[x].push(t);
 62             }
 63         }
 64     }
 65 }
 66 void dbfs()
 67 {
 68     int cnt=0;
 69     node t;
 70     t.cur[0]=a[0];
 71     t.cur[1]=a[1];
 72     t.cur[2]=a[2];
 73     t.cur[3]=a[3];
 74     dp[0][a[0]-'0'][a[1]-'0'][a[2]-'0'][a[3]-'0']=true;
 75     Q[0].push(t);
 76 
 77     t.cur[0]=b[0];
 78     t.cur[1]=b[1];
 79     t.cur[2]=b[2];
 80     t.cur[3]=b[3];
 81     dp[1][b[0]-'0'][b[1]-'0'][b[2]-'0'][b[3]-'0']=true;
 82     Q[1].push(t);
 83     while(true)
 84     {
 85         cnt++;
 86         if(Q[0].size()<Q[1].size())
 87             bfs(0);
 88         else bfs(1);
 89         if(flag==true) break;
 90     }
 91     printf("%d
",cnt-1);
 92 }
 93 
 94 int main()
 95 {
 96     int T;
 97     scanf("%d",&T);
 98     while(T--)
 99     {
100         scanf("%s%s",a,b);
101         if(strcmp(a,b)==0)
102         {
103             printf("0
");
104             continue;
105         }
106         while(!Q[0].empty())
107         {
108             Q[0].pop();
109         }
110         while(!Q[1].empty())
111         {
112             Q[1].pop();
113         }
114         flag=false;
115         memset(dp,false,sizeof(dp));
116         dbfs();
117     }
118     return 0;
119 }
     
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  • 原文地址：https://www.cnblogs.com/tom987690183/p/3637993.html
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