HDU 4714 Tree2cycle

Tree2cycle

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 2174    Accepted Submission(s): 516

 

Problem Description

 

　　A tree with N nodes and N-1 edges is given. To connect or disconnect one edge, we need 1 unit of cost respectively. The nodes are labeled from 1 to N. Your job is to transform the tree to a cycle(without superfluous edges) using minimal cost.

　　A cycle of n nodes is defined as follows: (1)a graph with n nodes and n edges (2)the degree of every node is 2 (3) each node can reach every other node with these N edges.

 

Input

 

　　The first line contains the number of test cases T( T<=10 ). Following lines are the scenarios of each test case.
In the first line of each test case, there is a single integer N( 3<=N<=1000000 ) - the number of nodes in the tree. The following N-1 lines describe the N-1 edges of the tree. Each line has a pair of integer U, V ( 1<=U,V<=N ), describing a bidirectional edge (U, V).

 

Output

 

　　For each test case, please output one integer representing minimal cost to transform the tree to a cycle.

 

Sample Input

 

1

4

1 2

2 3

2 4

 

Sample Output

 

3

 

题意：

　　给你一颗树，问用多少步才能使这棵树成为一个圈。

思路：

　　对于人一个节点，只能有一个儿子，除了树根，其他的统统剪掉

AC代码：

 

# include <bits/stdc++.h>
using namespace std;
const int MAX = 1000010;
struct node
{
    int to;
    int next;
}tree[MAX * 2];
int head[MAX];
int tol;
int sum = 0;
void add(int a, int b)
{
    tree[tol].to = b;
    tree[tol].next = head[a];
    head[a] = tol++;
}

int dfs(int root, int f)
{
    int tmp = 0;
    for(int i = head[root]; i != -1; i = tree[i].next)
    {
        int son = tree[i].to;
        if(son == f)
            continue;
        tmp += dfs(son, root);
        
    }
    if(tmp >= 2) 
    {
        if(root == 1)
            sum += 2 * (tmp - 2);
        else 
            sum += 2 * (tmp - 1);
        return 0; // 被剪断了
    }
    return 1;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        tol = 0;
        sum = 0;
        memset(head, -1, sizeof(head));
        
        int n;
        scanf("%d", &n);
        int a, b;
        for(int i = 1; i < n; i++)
        {
            scanf("%d%d", &a, &b);
            add(a, b);
            add(b, a);
        }
        dfs(1, -1);
        printf("%d\n", sum + 1);
    }
    return 0;
}