(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): 7603    Accepted Submission(s): 3397

 

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

 

題目分析：

問從初始的密碼到目的的密碼的最少操作次數，每位上的數有三種操作，加一，減一，交換。沒什麼問題，就是將三種情況算一下，然後入隊列，出隊的時候判斷一下情況就行了。已得到的密碼記得標記一下。

代碼：

#include<iostream>
#include<queue>
#include<string>
using namespace std;
struct Node {
	string k;
	int step;
	Node(string kk = "", int s = 0) {
		k = kk; step = s;
	}
};
string key;
string target;
queue<Node> q;
int vis[100005];
string add(string t, int index) {
	int a = t[index] - '0';
	if(a!=8)
	t[index] = (a + 1) % 9 + '0';
	else t[index] = '9';
	return t;
}
string Minus(string t, int index) {
	int a = t[index] - '0';
	if(a!=1)
	t[index] = (a - 1) % 9+'0';
	else t[index] = '9';
	return t;
}
string exchange(string t, int index1, int index2) {
	char temp = t[index1];
	t[index1] = t[index2];
	t[index2] = temp;
	return t;
}
int BFS() {
	Node temp;
	while (!q.empty())q.pop();
	q.push(Node(key,0));
	while (!q.empty())
	{
		temp = q.front();
		q.pop();
		
		if (temp.k == target) { return temp.step; }
		int st = temp.step+1;
		for (int i = 0; i < 4; i++)
		{
			//加一操作
			string p = add(temp.k, i);
			if (!vis[(p[0] - '0') * 1000 + (p[1] - '0') * 100 + (p[2] - '0') * 10 + (p[3] - '0')]) {
				vis[(p[0] - '0') * 1000 + (p[1] - '0') * 100 + (p[2] - '0') * 10 + (p[3] - '0')] = 1;
				q.push(Node(p, st));
			}
			//減一操作
			p = Minus(temp.k, i);
			if (!vis[(p[0] - '0') * 1000 + (p[1] - '0') * 100 + (p[2] - '0') * 10 + (p[3] - '0')]) {
				vis[(p[0] - '0') * 1000 + (p[1] - '0') * 100 + (p[2] - '0') * 10 + (p[3] - '0')] = 1;
				q.push(Node(p, st));
			}
			//交換操作
			if (i != 3) {
				if (temp.k[i] != temp.k[i + 1]) {
					p = exchange(temp.k, i, i + 1);
					if (!vis[(p[0] - '0') * 1000 + (p[1] - '0') * 100 + (p[2] - '0') * 10 + (p[3] - '0')]) {
						vis[(p[0] - '0') * 1000 + (p[1] - '0') * 100 + (p[2] - '0') * 10 + (p[3] - '0')] = 1;
						q.push(Node(p, st));
					}
				}
			}
		}
	}
}
int main() {
	int T;
	cin >> T;
	while (T--)
	{
		memset(vis, 0, sizeof(vis));
		cin >> key>>target;
		cout << BFS() << endl;
	}
	return 0;
}