I. Move Between Numbers

题目：http://codeforces.com/gym/101502/problem/I

time limit per test

2.0 s

memory limit per test

256 MB

input

standard input

output

standard output

You are given n magical numbers a₁, a₂, ..., a_n, such that the length of each of these numbers is 20 digits.

You can move from the i^th number to the j^th number, if the number of common digits between a_i and a_j is exactly 17 digits.

The number of common digits between two numbers x and y is computed is follow:

$\text{[math]}$ .

Where countX_i is the frequency of the i^th digit in the number x, and countY_i is the frequency of the i^th digit in the number y.

You are given two integers s and e, your task is to find the minimum numbers of moves you need to do, in order to finish at number a_e starting from number a_s.

Input

The first line contains an integer T (1 ≤ T ≤ 250), where T is the number of test cases.

The first line of each test case contains three integers n, s, and e (1 ≤ n ≤ 250) (1 ≤ s, e ≤ n), where n is the number of magical numbers, s is the index of the number to start from it, and e is the index of the number to finish at it.

Then n lines follow, giving the magical numbers. All numbers consisting of digits, and with length of 20 digits. Leading zeros are allowed.

Output

For each test case, print a single line containing the minimum numbers of moves you need to do, in order to finish at number a_e starting from number a_s. If there is no answer, print -1.

Example

Input

1
5 1 5
11111191111191111911
11181111111111818111
11811171817171181111
11111116161111611181
11751717818314111118

Output

Note

In the first test case, you can move from a₁ to a₂, from a₂ to a₃, and from a₃ to a₅. So, the minimum number of moves is 3 moves.

题意：1个样咧

5串字符从第1串字符开始到第5串字符结束

。。。。（5串字符）

两串字符中有17个数相同就说明两串字符间有一条边联系

求1------》5的最短经过几个字符串

思路： 建立图，，，先暴力把两串间能连起来的边设为1，，其它设为inf

dijkstra 直接求两点最短路复杂度是o（n*n），，如果数据打就用优先队列优化

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<set>
#include<algorithm>
#include<map>
#define maxn 1005
#define inf 1e9+7
using namespace std;
int mp[maxn][maxn];
int dis[maxn];
int pre[maxn];
void chushihua(){
    int i, j;
    for (i = 1; i < maxn; ++i) {
        for (j = 1; j < maxn; ++j) {
            if (i == j) {
                mp[i][j] = 0;
            }
            else {
                mp[i][j] = inf;
            }
        }
    }
}
void dijkstra(int n,int v)
{
    bool vis[maxn];
    for(int i=1;i<=n;i++)
    {
        vis[i]=false;
        dis[i]=mp[v][i];
        if(dis[i]==inf)
        {
            pre[i]=0;
        }
        else pre[i]=v;
    }
    dis[v]=0;
    vis[v]=true;
    for(int i=2;i<=n;i++)
    {
            int u=v;
            int mazz=inf;
            for(int j=1;j<=n;j++){
        if(!vis[j]&&dis[j]<mazz)
        {
                u=j;
                mazz=dis[j];
        }
        }
        vis[u]=true;
        for(int k=1;k<=n;k++)
        {
            if(!vis[k]&&mp[u][k]<inf)
            {
                if(dis[k]>mp[u][k]+dis[u]){
                    dis[k]=mp[u][k]+dis[u];
                    pre[k]=u;
                }
            }
        }

    }
}
int main()
{
    int T,x,y,k,n;
    int num[maxn][12];
    char s[25];
    scanf("%d",&T);
    while(T--)
    {
        chushihua();
        memset(num,0,sizeof(num));
        scanf("%d%d%d",&n,&x,&y);
        for(int i=1;i<=n;i++)
        {
            scanf("%s",&s);
            for(int k=0;k<20;k++)
        num[i][s[k]-'0']++;
        }
        for(int i=1;i<n;i++)
            for(int j=i+1;j<=n;j++)
        {
            int com=0;
            for(int k=0;k<=9;k++)
            {
                com+=min(num[i][k],num[j][k]);
            }
            if(com==17)
            {
                mp[i][j] = 1;
                mp[j][i] = 1;
            }
        }
        dijkstra(n,x);
        if(dis[y]==inf)
           cout<<"-1"<<endl;
        else
            cout<<dis[y]<<endl;
    }
}

简单题。但算是图论上的练练手吧