给你一棵有向树,需要选定一个点为capital,满足翻转边数最小 思路:先求出1为capital 的答案,然后向下更新孩子节点 dp[i]=dp[i-1]+judge(i); #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<queue> #include<stack> #in…

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…

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…

题目是给一张边有向的树形图.要选出首都的点,首都要都能走到其他点,因此要反转一些边的方向.问可以选哪几个点作为首都,使它们所需反转边的数量最少. 这题挺好想的,因为做过HDU2196. 首先就不妨设正向边权值为0,反向边权值为1,那样就是各个点出发到其他点经过边所需的最少权值和. 然后对于每个点,分两个部分考虑:以这个点为根的子树.这个点往上走的部分: dp[0][u]表示以u点作为首都且以u点为根的子树部分所需反转边的数量,容易知道就等于子树内边权和 dp[1][u]表示以u点作为首都且u点向…