Problem:
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Breadth First Search

Write a program which reads an directed graph $G = (V, E)$, and finds the shortest distance from vertex $1$ to each vertex (the number of edges in the shortest path). Vertices are identified by IDs $1, 2, ... n$.

Input

In the first line, an integer $n$ denoting the number of vertices, is given. In the next $n$ lines, adjacent lists of vertex $u$ are given in the following format:

$u$ $k$ $v_1$ $v_2$ ... $v_k$

$u$ is ID of the vertex and $k$ denotes its degree.$v_i$ are IDs of vertices adjacent to $u$.

Constraints

$1 \leq n \leq 100$

Output

For each vertex $u$, print $id$ and $d$ in a line. $id$ is ID of vertex $u$ and $d$ is the distance from vertex $1$ to vertex $u$. If there are no path from vertex $1$ to vertex $u$, print -1 as the shortest distance. Print in order of IDs.

Sample Input 1

4
1 2 2 4
2 1 4
3 0
4 1 3

Sample Output 1

1 0
2 1
3 2
4 1

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in C (judge verdict: Time Limit Exceeded)

#include <stdio.h>
void yuusen(int a,int b);

int connect[100][100],connecter[100];
int thisNO[100];

int main (void){
    int n;
    int bangou,yazirusi;
    int loop,loop_in_loop;//???????????°?????????,????????????????????????????????????
    
    scanf("%d",&n);//????????°????????????
    for(loop=0;loop<n;loop++){//????????¨?????????????????\???????????????loop+1??§??????
        scanf("%d %d",&bangou,&yazirusi);
        //??\?¶????????????????
        for(loop_in_loop=0;loop_in_loop<yazirusi;loop_in_loop++){
            scanf("%d",&connect[loop][loop_in_loop]);//???????????????????????????
        }
        connecter[loop]=yazirusi;//??????????????¢??°?????°???connecter[loop]??§loop+1?????????????????¢??°?????°??????????????????
    }
    //?????????
    for(loop=0;loop<100;loop++){
        thisNO[loop]=10000;
    }
    
    //????????¢?´¢????§?
    yuusen(1,0);
    
    //????????????????????£?????´??????-1???
    for(loop=0;loop<n;loop++){
        if(thisNO[loop]==10000)thisNO[loop]=-1;
    }
    
    //??????
    for(loop=0;loop<n;loop++){
        printf("%d %d\n",loop+1,thisNO[loop]);
    }
    
    return 0;
}

void yuusen(int no,int bangou){//???????????????????????????
    int loop;

    if(connecter[no-1]==0){//?????????????????????????????´???
    }else{//??????????????´???
        for(loop=0;loop<connecter[no-1];loop++){
            if(thisNO[connect[no-1][loop]-1]>bangou){//??\????????¨???????????´?????????
                yuusen(connect[no-1][loop],bangou+1);
            }
        }
    }
    //????????????????????¢????¨????
    if(thisNO[no-1]>bangou){
        thisNO[no-1]=bangou;
    }

    return;
}