Happy Vertices

中文题意：求一个图中那些    拥有儿子节点数   比他爸爸拥有儿子节点数    更多的节点的总数目，文中称为快乐节点。

You are given a graph with $N$ vertices and $M$ edges. Master parent is the vertex which has no parent but may have 0 or more children. In any connected component of the graph,vertex with the lowest value in that component serves as the master parent. 
A vertex is called happy if it has more children than its parent. Count the number of happy vertices in the given graph.The graph has no cycles or self loops.

Input Format:

First line consists of two space separated integers denoting $N$ and $M$ and the following $M$ lines consist of two space separated integers $X$and $Y$ denoting there is an edge between vertices $X$ and $Y$.

Output Format:

Print the number of happy vertices in the graph.

Constraints:
$1\le N\le 100000$
$0\le M\le N-1$ 
$1\le X,Y\le N$

SAMPLE INPUT

 

4 31 22 32 4

SAMPLE OUTPUT

 

1

Explanation

In this graph, we have vertices 1,2,3 and 4. Since 1 is the lowest among these, so it becomes the master vertex. Now, 1 has only 1 child while 2 has two children.So, 2 is a happy vertex. There are no more happy vertices in the graph.

Time Limit:1.0 sec(s) for each input file.

Memory Limit:256 MB

Source Limit:

1024 KB

#include <iostream>#include <vector>using namespace std;vector <int>adj[100005];bool visited[100005];int counti= 0 ;int children[100005];int T=0;void dfs(int s){visited[s] = true; for (int i = 0; i<adj[s].size(); ++i){if (visited[adj[s][i]] == false){if (T == 0){if (adj[s].size()<adj[adj[s][i]].size()-1)counti++;T = 1;}else{if (adj[s].size()<adj[adj[s][i]].size())counti++;}dfs(adj[s][i]);} }}void initialize(){ for (int i = 0; i < 100005; ++i)visited[i] = false;}int main() {ios_base::sync_with_stdio(false); cin.tie(0);int nodes, edges, count = 0, x, y;cin >> nodes >> edges;for (int i = 0; i < edges; ++i){cin >> x >> y;adj[x].push_back(y);adj[y].push_back(x); }initialize();for (int i = 1; i <= nodes; ++i){if (visited[i] == false)dfs(i);T = 0;}cout << counti<< endl;}