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 ", ans); } return 0; }