2010 Liaoning Province Nbut 1218 "DFS implementation tree traversal and update"

• [1218] You is my brother
• Time limit: Ms Memory limit: 131072 K
• Links: nbut 1218
• Description of the problem
• Input
• There is multiple test cases.
  The first line have a single integer, N (n<=1000). The next n lines has integers a and B (1<=a,b<=2000) each, indicating B is the father of a. One person has exactly one father, of course. Little A is numbered 1 and Little B is numbered 2.
  Proceed to the end of file.
• Output
• For each test case, if Little B is Little A's younger, print "You are my younger". Otherwise, if Little B is Little A's elder, print "You Are my Elder". Otherwise, print "You Are my brother". The output for each test case is occupied exactly one line.
• Sample input

• 51 32 43 54 65 661 32 43 54 65 76 7

• Sample output

• You are my elderyou is my brother

Test Instructions:

Given N -times input, each time a row a,b, indicating b is a father. Finally , the relationship between node 1 and node 2 "is the comparison depth".

N <=1000 , 1<=a,b<=2000

Analysis:

Test instructions very clear, I use a value for each nodeRanksaves the node's depth, so each time the nodeamerging the tree of the root node into a nodebas the root node of the tree above, I'm going to put the slave nodeaand NodeaThe following nodes are allRankvalues are added to the nodebof theRankvalues. In this way, a struct holds the information for each node, each time it merges two trees.DFSupdating these nodes isOKup.

Code implementation:

#include <cmath> #include <cstdio> #include <string> #include <cstring> #include <iostream > #include <algorithm>using namespace std; #define FIN freopen ("Input.txt", "R", stdin) struct node{in    T Rank; Vector<int> Son; nodes[2000 + 5];void dfs (int pos, const int& val) {Nodes[pos].    Rank + = val; int num = Nodes[pos].    Son.size ();    if (num = = 0) return; for (int i = 0; i < num; i++) {int &np = Nodes[pos].        Son[i];    DFS (NP, Val); }}void Init () {for (int i = 1; i <=; i++) {nodes[i].        Rank = 1; Nodes[i].    Son.clear ();    }}int Main () {//FIN;    int n, a, B;        while (~SCANF ("%d", &n)) {init ();            for (int i = 1; I <= n; i++) {scanf ("%d%d", &a, &b); NODES[B].            Son.push_back (a); DFS (A, nodes[b].        Rank); } if (Nodes[1]. Rank = = Nodes[2].        Rank) printf ("You Are my brother\n"); else if (nodes[1]. Rank > Nodes[2].        Rank) printf ("You Are my elder\n");    else printf ("You Are my younger\n"); } return 0;}

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2010 Liaoning Province Nbut 1218 "DFS implementation tree traversal and update"