首页 > 代码库 > HDU 4714 Tree2cycle (树形DP)
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;}
HDU 4714 Tree2cycle (树形DP)
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。