去年google code jam第一轮的B题，我没有想明白，也没有找到题解。没有办法，就去下载了大神的代码来看看。感觉差距好大，别人用十分钟不到就做出来了。题目的描述如下，我就不写翻译了，题意挺明了的。由于墙的原因，我还是把题目贴下面吧！
题目内容
A tree is a connected graph with no cycles.
A rooted tree is a tree in which one special vertex is called the root. If there is an edge between X and Y in a rooted tree, we say that Y is a child of X if X is closer to the root than Y (in other words, the shortest path from the root to X is shorter than the shortest path from the root to Y).
A full binary tree is a rooted tree where every node has either exactly 2 children or 0 children.
You are given a tree G with N nodes (numbered from 1 to N). You are allowed to delete some of the nodes. When a node is deleted, the edges connected to the deleted node are also deleted. Your task is to delete as few nodes as possible so that the remaining nodes form a full binary tree for some choice of the root from the remaining nodes.
Input
The first line of the input gives the number of test cases, T. T test cases follow. The first line of each test case contains a single integer N, the number of nodes in the tree. The following N-1 lines each one will contain two space-separated integers: Xi Yi, indicating that G contains an undirected edge between Xi and Yi.
Output
For each test case, output one line containing “Case #x: y”, where x is the test case number (starting from 1) and y is the minimum number of nodes to delete from G to make a full binary tree.
Limits
1 ≤ T ≤ 100.
1 ≤ Xi, Yi ≤ N
Each test case will form a valid connected tree.
Small dataset
2 ≤ N ≤ 15.
Large dataset
2 ≤ N ≤ 1000.
Sample
Input
Output
3
3
2 1
1 3
7
4 5
4 2
1 2
3 1
6 4
3 7
4
1 2
2 3
3 4
Case #1: 0
Case #2: 2
Case #3: 1
In the first case, G is already a full binary tree (if we consider node 1 as the root), so we don’t need to do anything.
In the second case, we may delete nodes 3 and 7; then 2 can be the root of a full binary tree.
In the third case, we may delete node 1; then 3 will become the root of a full binary tree (we could also have deleted node 4; then we could have made 2 the root).
题意就是给你很多对点，每对点之间有无向边。输入保证这个图是相连的，不是森林。题目要求你去掉一些点，使这个图成为完全二叉树。完全二叉树的是指任意节点的子节点数目为0或2。需要注意的是点对的父子关系不是确定的，这也是题目的难度所在。
我思考这个问题的时候主要是卡在了树的结构上。题目保证了没有孤立的点集，所以思考这个问题的第一步就是忘掉什么二叉树，把这个问题的输入看做是一个无向图，也不要管什么边上的父子关系。一个非常明确是结论是：对于一个连同的无向图，以任意节点为根都可以构造出一棵树，不考虑兄弟节点的顺序的话，这棵树是唯一的。当然这棵树很可能不是二叉树。
然后这个问题的思路就清晰了，求解以每个顶点为根的树形成二叉树的最小代价。然后找到最小值输出即可。求最小代价的原理就看注释吧！这下边是别人的代码，知道逻辑代码还是容易看懂的。
#include<stdio.h>
#include<algorithm>
#include<vector>
using namespace std;
vector<int>E[1010];
int n, Res, D[1010]; //题目的关键点在于无向五环图，题目的图没有孤立点
void DFS(int a, int par){ //任意形式的摆放都可以保证树的结构正确，不过不能保证是二叉树
int i, cnt = 0; //子节点多于两个的就必须选择删减代价最小的两个子树，把其余子树删除掉
D[a] = 1; //字节点少于两个的顶点的子节点全部删除掉
for (i = 0; i < E[a].size(); i++){
if (E[a][i] != par){
DFS(E[a][i], a);
cnt++;
}
}
if (cnt <= 1)return;
int M1 = 0, M2 = 0;
for (i = 0; i < E[a].size(); i++){
if (E[a][i] != par){
if (M1 < D[E[a][i]]){
M2 = M1; M1 = D[E[a][i]]; //
}
else if (M2 < D[E[a][i]]){
M2 = D[E[a][i]];
}
}
}
D[a] = M1 + M2 + 1;
} //
void Do(int a){
int i;
DFS(a, -1);
if (Res > n - D[a]) Res = n - D[a];
}
int main()
{
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
int i, TC, T, a, b;
scanf("%d", &TC);
for (T = 1; T <= TC; T++){
printf("Case #%d: ", T);
scanf("%d", &n);
for (i = 1; i < n; i++){
scanf("%d%d", &a, &b);
E[a].push_back(b);
E[b].push_back(a);
}
Res = n;
for (i = 1; i <= n; i++){
Do(i);
}
printf("%d\n", Res);
for (i = 1; i <= n; i++)E[i].clear();
}
}