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。

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstdlib>

 4 #include<cstring>

 5 #include<string>

 6 #include<queue>

 7 #include<algorithm>

 8 #include<map>

 9 #include<iomanip>

10 #include<climits>

11 #include<string.h>

12 #include<cmath>

13 #include<stdlib.h>

14 #include<vector>

15 #include<stack>

16 #include<set>

17 using namespace std;

18 #define INF 1000000007

19 #define MAXN 4010

20 #define Mod 1000007

21 #define N 100010

22 #define NN 30

23 #define sigma_size 3

24 const int MAX = 1000100;

25 const int maxn = 6e5 + 10;

26 using namespace std;

27 typedef long long LL;

28 

29 int T;

30 int n, s;

31 int u, v;

32 vector<int> G[100010];

33 int flag[100010];

34 

35 void dfs(int u)

36 {

37     for (int i = 0; i < G[u].size(); ++i) {

38         int v = G[u][i];

39         if (!flag[v]) {

40             flag[v] = u;

41             dfs(v);

42         }

43     }

44 }

45 

46 void run()

47 {

48     cin >> n >> s;

49     for (int i = 0; i <= n; ++i) {

50         G[i].clear();

51         flag[i] = 0;

52     }

53     for (int i = 1; i < n; ++i) {

54         scanf("%d%d",&u,&v);

55         G[u].push_back(v);

56         G[v].push_back(u);

57     }

58     flag[s] = -1;

59     dfs(s);

60     printf("%d", flag[1]);

61     for (int i = 2; i <= n; ++i) {

62         printf(" %d", flag[i]);

63     }

64     puts("");

65 }

66 

67 int main()

68 {

69     cin >> T;

70     while (T--)

71         run();

72     //system("pause");

73     return 0;

74 }