HDU1195 BFS或者DP

Open the Lock

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5936    Accepted Submission(s): 2651

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

21234214411119999

 

Sample Output

24

每个点由四种状态，加1减1，和左边交换或者和右边交换。可以用搜索的来查询每种操作。
要求代价最小可以近似的理解为寻找最短路，那么很自然的选择用BFS来做。

#include<iostream>#include<queue>#include<cstdio>using namespace std;int s1,s2,s3,s4;int e1,e2,e3,e4;int steps;bool vis[10000];struct Q{    int x[4];    int steps;};void bfs(){    queue<Q>q;    Q temp,now;    int i,j;    int x[4];    int bit[4]={1000,100,10,1};    while(!q.empty())q.pop();    temp.x[0]=s1; temp.x[1]=s2;    temp.x[2]=s3; temp.x[3]=s4;    temp.steps=0;    q.push(temp);    steps=0xfffffff;    memset(vis,0,sizeof(vis));    vis[s1*1000+s2*100+s3*10+s4]=1;    while(!q.empty())    {        now=q.front();        q.pop();        if(now.x[0]==e1&&now.x[1]==e2&&now.x[2]==e3&&now.x[3]==e4)        {            if(steps>now.steps)            {                steps=now.steps;            }        }        for(i=0;i<4;i++)        {            for(j=0;j<4;j++)            {                x[0]=now.x[0];x[1]=now.x[1];                x[2]=now.x[2];x[3]=now.x[3];                temp=now;                if(j==0)                {                    x[i]++;                    if(x[i]>9)x[i]=1;                }                else                if(j==1)                {                    x[i]--;                    if(x[i]<1)x[i]=9;                }                else                if(j==2)                {                    if(i<=0)continue;                    int t=x[i];                    x[i]=x[i-1];                    x[i-1]=t;                }                else                {                    if(i>=3)continue;                    int t=x[i];                    x[i]=x[i+1];                    x[i+1]=t;                }                //printf("%d %d %d %d\n",x[0],x[1],x[2],x[3]);                if(vis[x[0]*bit[0]+x[1]*bit[1]+x[2]*bit[2]+x[3]*bit[3]])continue;                vis[x[0]*bit[0]+x[1]*bit[1]+x[2]*bit[2]+x[3]*bit[3]]=1;                temp.x[0]=x[0];temp.x[1]=x[1];temp.x[2]=x[2];temp.x[3]=x[3];                temp.steps++;                q.push(temp);            }        }    }    return;}int main(){    int t;    char str[5];    scanf("%d",&t);    while(t--)    {        scanf("%s",str);        s1=str[0]-'0'; s2=str[1]-'0';        s3=str[2]-'0'; s4=str[3]-'0';        scanf("%s",str);        e1=str[0]-'0'; e2=str[1]-'0';        e3=str[2]-'0'; e4=str[3]-'0';        bfs();        printf("%d\n",steps);    }    return 0;}