Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5970 Accepted Submission(s): 2666
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.
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.
#include<cstdio> #include<cstring> #include<algorithm> #include<math.h> #include<queue> #include<iostream> using namespace std; typedef long long LL; bool vis[10][10][10][10]; struct Node{ int v[4]; int step; }; Node s,t; bool _equal(Node a,Node b){ for(int i=0;i<4;i++){ if(a.v[i]!=b.v[i]) return false; } return true; } Node operate(int x,Node now){ Node next; for(int i=0;i<4;i++){ next.v[i] = now.v[i]; } next.step=now.step+1; if(x<4){ ///+ if(now.v[x]==9) next.v[x]=1; else next.v[x]=now.v[x]+1; }else if(x<8){ ///- if(now.v[x%4]==1) next.v[x%4]=9; else next.v[x%4]=now.v[x%4]-1; }else{ ///exchange int a = now.v[x%4]; int b = now.v[x%4+1]; next.v[x%4]=b; next.v[x%4+1] = a; } return next; } int bfs(Node s){ memset(vis,false,sizeof(vis)); queue <Node>q; vis[s.v[0]][s.v[1]][s.v[2]][s.v[3]]=true; q.push(s); s.step = 0; while(!q.empty()){ Node now = q.front(); q.pop(); if(_equal(now,t)){ return now.step; } for(int i=0;i<11;i++){ ///总共11种操作,[1-4]+ [1-4]- exwchange[1,2][2,3][3,4] Node next=operate(i,now); if(vis[next.v[0]][next.v[1]][next.v[2]][next.v[3]]==false){ vis[next.v[0]][next.v[1]][next.v[2]][next.v[3]]=true; q.push(next); } } } return -1; } int main() { char s1[5],s2[5]; int tcase; scanf("%d",&tcase); while(tcase--){ scanf("%s",s1); scanf("%s",s2); for(int i=0;i<4;i++){ s.v[i]=s1[i]-'0'; t.v[i]=s2[i]-'0'; } int res = bfs(s); printf("%d\n",res); } return 0; }