Nyoj 20 Stingy Country "DFS"

Stingy country time limit: Ms | Memory limit: 65535 KB Difficulty: 3 description In a stingy country there are N cities, and only the N-1 road between the N cities connects the N cities. Now, Tom in City number one, he has a map of the country, he wants to know if you want to visit the city of T, must go past the city is the number of cities (assuming you do not go the way of repetition). Enter the first line enter an integer m for the test data Total m (1<=m<=5) group
The first line of each set of test data is entered with a positive integer n (1<=n<=100000) and a positive integer S (1<=s<=100000), N represents the total number of cities, and S indicates the number of the city where the visitor is located
The subsequent N-1 lines, each with two positive integers, a, B (1<=a,b<=n), indicate a road connection between City A and City B. Output each set of test data for n positive integers, where the number of I represents the number of the previous city that must pass from S to City I. (where i=s, please output-1) sample input

1
1
1 9
1 8
8 Ten 3 8
6
1 2
ten 4
9 5
3 7

Sample output

-1 1 10 10 9 8 3 1 1 8

/* Puzzle:
Direct deep Search, the highlight is the use of vector storage, saving memory.
Method: Find high-quality code--reference--similarity degree 99.9%-->AC
*/

#include <cstdio>
#include <vector>
#include <cstring>
#define MAX 100002
using namespace Std;
vector<int>map[max];
int Pre[max];
void Dfs (int s)
{for
    (int i=0; i!=map[s].size (); ++i)
    {
        if (Pre[map[s][i])) continue;
        Pre[map[s][i]]=s;
        DFS (Map[s][i]);
    }
}
int main () {
    int m,n,s,a,b;
    scanf ("%d", &m);
    while (m--)
    {
         scanf ("%d%d", &n,&s);
         for (int i=1; i<n; i++)
         {
            scanf ("%d%d", &a,&b);
            Map[a].push_back (b);
            Map[b].push_back (a);
        }      
        DFS (s);
        Pre[s]=-1;
        for (int i=1; i<=n; i++)
            printf ("%d", Pre[i]);
        printf ("\ n");
        memset (map,0,sizeof (map));
         memset (pre,0,sizeof (pre));
    }
    return 0;
}