HDU 4714 Tree2cycle (树形DP)

题意：给定一棵树，断开一条边或者接上一条边都要花费 1，问你花费最少把这棵树就成一个环。

析：树形DP，想一想，要想把一棵树变成一个环，那么就要把一些枝枝叶叶都换掉，对于一个分叉是大于等于2的我们一定要把它从父结点上剪下来是最优的，

因为如果这样剪下来再粘上花费是2（先不管另一端），如果分别剪下来再拼起来，肯定是多花了，因为多了拼起来这一步，知道这，就好做了。先到叶子结点，

然后再回来计算到底要花多少。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 5;
const LL mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
struct Edge{
    int next, to;
};
Edge edge[maxn<<1];
int cnt, ans;
int head[maxn];void add(int u, int v){
    edge[cnt].to =v;
    edge[cnt].next = head[u];
    head[u] = cnt++;
}int dfs(int u, int fa){
    int tmp = 0;
    for(int i = head[u]; ~i; i = edge[i].next){
        int v = edge[i].to;
        if(v == fa)  continue;
        tmp += dfs(v, u);
    }    if(tmp >= 2){
        if(1 == u)  ans += 2 * (tmp - 2);
        else ans += 2 * (tmp - 1);
        return 0;
    }
    return 1;
}int main(){
    int T;  cin >> T;
    while(T--){
        scanf("%d", &n);
        int u, v;
        cnt = 0;
        memset(head, -1, sizeof head);
        for(int i = 1; i < n; ++i){
            scanf("%d %d", &u, &v);
            add(u, v);
            add(v, u);
        }
        ans = 1;
        dfs(1, -1);
        printf("%d\n", ans);
    }
    return 0;
}