hdu1195 Open the Lock （bfs）

hdu1195 Open the Lock （bfs）

Open the Lock

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6618    Accepted Submission(s): 2979

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​​

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广搜解决~

对于每个密码进行加、减、交换操作即可。

#include #include #include using namespace std;struct node{ int cost; char c[10]; friend bool operator<(node a,node b) { return a.cost>b.cost; }};char x_str[10];char y_str[10];int bfs(){ node node1; char ch[10]; priority_queues; bool vis[18][18][18][18]; memset(vis,false,sizeof(vis)); strcpy(node1.c,x_str); node1.cost=0; s.push(node1); while(!s.empty()) { node1=s.top();s.pop(); if(vis[node1.c[0]-'0'][node1.c[1]-'0'][node1.c[2]-'0'][node1.c[3]-'0']) continue; vis[node1.c[0]-'0'][node1.c[1]-'0'][node1.c[2]-'0'][node1.c[3]-'0']=true; if(strcmp(node1.c,y_str)==0) return node1.cost; for(int i=0;i<4;i++) { node node2; node2.cost=node1.cost+1; strcpy(ch,node1.c); if(ch[i]=='1') { //减操作 ch[i]='9'; if(!vis[ch[0]-'0'][ch[1]-'0'][ch[2]-'0'][ch[3]-'0']) { strcpy(node2.c,ch); s.push(node2); } //加操作 ch[i]='2'; if(!vis[ch[0]-'0'][ch[1]-'0'][ch[2]-'0'][ch[3]-'0']) { strcpy(node2.c,ch); s.push(node2); } } else if(ch[i]=='9') { //减操作 ch[i]='8'; if(!vis[ch[0]-'0'][ch[1]-'0'][ch[2]-'0'][ch[3]-'0']) { strcpy(node2.c,ch); s.push(node2); } //加操作 ch[i]='1'; if(!vis[ch[0]-'0'][ch[1]-'0'][ch[2]-'0'][ch[3]-'0']) { strcpy(node2.c,ch); s.push(node2); } } else { char CH=ch[i]; //减操作 ch[i]=CH-1; if(!vis[ch[0]-'0'][ch[1]-'0'][ch[2]-'0'][ch[3]-'0']) { strcpy(node2.c,ch); s.push(node2); } //加操作 ch[i]=CH+1; if(!vis[ch[0]-'0'][ch[1]-'0'][ch[2]-'0'][ch[3]-'0']) { strcpy(node2.c,ch); s.push(node2); } } //i i+1 交换 if(i<3) { strcpy(ch,node1.c); char CH=ch[i]; ch[i]=ch[i+1];ch[i+1]=CH; if(!vis[ch[0]-'0'][ch[1]-'0'][ch[2]-'0'][ch[3]-'0']) { strcpy(node2.c,ch); s.push(node2); } } } }}int main(){ int t; scanf("%d",&t); while(t--) { memset(x_str,0,sizeof(x_str)); memset(y_str,0,sizeof(y_str)); scanf("%s %s",x_str,y_str); printf("%d\n",bfs()); } return 0;}

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