D. Choosing Capital for Treeland time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output The country Treeland consists of n cities, some pairs of them are connected with unidirectional roads. Over…

D. Choosing Capital for Treeland time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output The country Treeland consists of n cities, some pairs of them are connected with unidirectional roads. Over…

Choosing Capital for Treeland Description The country Treeland consists of n cities, some pairs of them are connected with unidirectional roads. Overall there are n - 1 roads in the country. We know that if we don't take the direction of the roads in…

Choosing Capital for Treeland time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output The country Treeland consists of n cities, some pairs of them are connected with unidirectional roads. Overall…

D. Choosing Capital for Treeland time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output The country Treeland consists of n cities, some pairs of them are connected with unidirectional roads. Over…

传送门 The country Treeland consists of n cities, some pairs of them are connected with unidirectional roads. Overall there are n - 1 roads in the country. We know that if we don't take the direction of the roads into consideration, we can get from any…

D. Choosing Capital for Treeland 链接:http://codeforces.com/problemset/problem/219/D   The country Treeland consists of n cities, some pairs of them are connected with unidirectional roads. Overall there are n - 1 roads in the country. We know that if…

题意翻译 题目描述 Treeland国有n个城市,这n个城市连成了一颗树,有n-1条道路连接了所有城市.每条道路只能单向通行.现在政府需要决定选择哪个城市为首都.假如城市i成为了首都,那么为了使首都能到达任意一个城市,不得不将一些道路翻转方向,记翻转道路的条数为k.你的任务是找到所有满足k最小的首都. 输入输出格式 输入格式 输入包含多个测试点.对于每个测试点,每个测试点的第一行为一个正整数n(2<=n<=2e5).接下来n-1行,每行两个正整数ai,bi,表示城市a到城市b有一条单向通行的道…

题目传送门 /* 题意:求一个点为根节点,使得到其他所有点的距离最短,是有向边,反向的距离+1 树形DP:首先假设1为根节点,自下而上计算dp[1](根节点到其他点的距离),然后再从1开始,自上而下计算dp[v], 此时可以从上个节点的信息递推出来 */ #include <cstdio> #include <cstring> #include <cmath> #include <vector> using namespace std; ; const in…

D. Choosing Capital for Treeland 题意 给出一颗有方向的n个节点的树,现在要选择一个点作为首都. 问最少需要翻转多少条边,使得首都可以到所有其他的城市去,以及相应的首都可以是哪些点. 思路 先忽略掉树中的方向,dp[i]表示i节点到它的子树所有点最少需要翻转的边. 进行第一遍dfs 如果u-v的方向是u-->v,那么dp[u]=dp[u]+dp[v];,否则dp[u]=dp[u]+dp[v]+1;,表示u-v这条边要翻转. 这时根节点的dp值就是根节点作为首都需要…