nyoj 吝啬的国度 图的遍历

题意：n个点，n-1条路，求出从点s出发到每一个点必须经过的前一个点，到自身为-1

思路：从s出发遍历一遍图就可以了，开flag数组记录下来每一个点的必经的点。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#include<climits>
#include<string.h>
#include<cmath>
#include<stdlib.h>
#include<vector>
#include<stack>
#include<set>
using namespace std;
#define INF 1000000007
#define MAXN 4010
#define Mod 1000007
#define N 100010
#define NN 30
#define sigma_size 3
const int MAX = 1000100;
const int maxn = 6e5 + 10;
using namespace std;
typedef long long LL;

int T;
int n, s;
int u, v;
vector<int> G[100010];
int flag[100010];

void dfs(int u)
{
    for (int i = 0; i < G[u].size(); ++i) {
        int v = G[u][i];
        if (!flag[v]) {
            flag[v] = u;
            dfs(v);
        }
    }
}

void run()
{
    cin >> n >> s;
    for (int i = 0; i <= n; ++i) {
        G[i].clear();
        flag[i] = 0;
    }
    for (int i = 1; i < n; ++i) {
        scanf("%d%d",&u,&v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    flag[s] = -1;
    dfs(s);
    printf("%d", flag[1]);
    for (int i = 2; i <= n; ++i) {
        printf(" %d", flag[i]);
    }
    puts("");
}

int main()
{
    cin >> T;
    while (T--)
        run();
    //system("pause");
    return 0;
}