HDU1195:Open the Lock(BFS)
2014-03-27 10:43
387 查看
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
Sample Output
普通的BFS,将已经访问过的状态标记下即可
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
普通的BFS,将已经访问过的状态标记下即可
#include <stdio.h> #include <algorithm> #include <string.h> #include <queue> using namespace std; struct node { int num[4],step; } first,last; int vis[11][11][11][11]; void bfs() { int i; node a,next; queue<node> q; a = first; a.step = 0; q.push(a); vis[a.num[0]][a.num[1]][a.num[2]][a.num[3]] = 1; while(!q.empty()) { a = q.front(); q.pop(); if(a.num[0] == last.num[0] && a.num[1] == last.num[1] && a.num[2] == last.num[2] && a.num[3] == last.num[3]) { printf("%d\n",a.step); return ; } for(i = 0; i<4; i++) //+1 { next = a; next.num[i]++; if(next.num[i]==10) next.num[i] = 1; if(!vis[next.num[0]][next.num[1]][next.num[2]][next.num[3]]) { vis[next.num[0]][next.num[1]][next.num[2]][next.num[3]] = 1; next.step++; q.push(next); } } for(i = 0; i<4; i++) //-1 { next = a; next.num[i]--; if(next.num[i]==0) next.num[i] = 9; if(!vis[next.num[0]][next.num[1]][next.num[2]][next.num[3]]) { vis[next.num[0]][next.num[1]][next.num[2]][next.num[3]] = 1; next.step++; q.push(next); } } for(i = 0; i<3; i++) //交换 { next = a; next.num[i] = a.num[i+1]; next.num[i+1] = a.num[i]; if(!vis[next.num[0]][next.num[1]][next.num[2]][next.num[3]]) { vis[next.num[0]][next.num[1]][next.num[2]][next.num[3]] = 1; next.step++; q.push(next); } } } } int main() { int i,j,t; char s1[10],s2[10]; scanf("%d",&t); while(t--) { scanf("%s%s",s1,s2); for(i = 0; i<4; i++) { first.num[i] = s1[i]-'0'; last.num[i] = s2[i]-'0'; } memset(vis,0,sizeof(vis)); bfs(); } return 0; }
相关文章推荐
- hdu1195-Open the Lock-双向&&单向bfs
- hdu1195 Open the Lock(bfs水题)
- hdu1195 Open the Lock--单向BFS & 双向BFS
- hdu1195 Open the Lock (bfs)
- HDU1195:Open the Lock(BFS)
- hdu1195 Open the Lock(BFS)
- hdu1195 Open the Lock (BFS)
- hdu 1195 Open the Lock(BFS)
- HDOJ 1195 Open the Lock bfs 双向bfs
- hdu1195 Open the Lock
- HDU1195-Open the Lock
- Bfs++ open the lock
- hdu 1195 Open the Lock (bfs)
- hdu1195 Open the Lock 广搜BFS 四维数组标记
- hdu 1195:Open the Lock(暴力BFS广搜)
- HDU-1195-Open the Lock(BFS)
- 【更新】HDOJ 1195 Open the Lock (双向BFS)
- 双向BFS-->hdu 1195 Open the Lock
- 【非图BFS+STL】电脑修好了—Open the Lock_HDU 1195
- hdu 1195 Open the Lock(双向bfs)