Problem - D - Codeforces  Fix a Tree 看完第一名的代码,顿然醒悟... 我可以把所有单独的点全部当成线,那么只有线和环. 如果全是线的话,直接线的条数-1,便是操作数. 如果有环和线,环被打开的同时,接入到线上.那就是线和环的总数-1. 如果只有环的话,把所有的环打开,互相接入,共需n次操作. #include <cstdio> #include <algorithm> using namespace std; ; int cur[maxn];…

Fix a Tree time limit per test2 seconds A tree is an undirected connected graph without cycles. Let's consider a rooted undirected tree with n vertices, numbered 1 through n. There are many ways to represent such a tree. One way is to create an array…

题目链接:http://codeforces.com/contest/699/problem/D D. Fix a Tree time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A tree is an undirected connected graph without cycles. Let's consider a root…

D. Fix a Tree time limit per test 2 seconds memory limit per test 256 megabytes     A tree is an undirected connected graph without cycles. Let's consider a rooted undirected tree with n vertices, numbered 1 through n. There are many ways to represen…

D. Fix a Tree time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A tree is an undirected connected graph without cycles. Let's consider a rooted undirected tree with n vertices, numbered 1 th…

Fix a Tree A tree is an undirected connected graph without cycles. Let's consider a rooted undirected tree with n vertices, numbered 1 through n. There are many ways to represent such a tree. One way is to create an array with n integers p1, p2, ...,…

原题:http://codeforces.com/contest/699/problem/D 题目中所描述的从属关系,可以看作是一个一个块,可以用并查集来维护这个森林.这些从属关系中会有两种环,第一种是一个点从自身出发到自己,这说明该点是一棵子树的根:第二种是从一点出发到另外一个点.这两种情况在并查集合并的时候都会失败,因为合并时他们都已经属于一个子树,我们现在需要做的就是将这些子树合并,这时我们要优先对生成第二种环的子树进行合并,因为这些从属关系一定是需要修改的,第一种情况有一个点可以不需要修…

题目链接: http://codeforces.com/problemset/problem/698/B http://codeforces.com/problemset/problem/699/D 题目大意: 通过给定当前节点的父亲给你一棵有错的树,可能有多个根和环,输出改成正确的一棵树至少要修改几个节点的父亲和修改后所有点的父亲值 题目思路: [并查集][模拟] 用并查集把成环的归在一起(类似强连通分量),然后统计分量数并修改. 第一个出现的当作根,其余的每一块连通分量都去掉一条边改为连接到…

题目链接: http://codeforces.com/problemset/problem/698/B 题解: 还是比较简单的.因为每个节点只有一个父亲,可以直接建反图,保证出现的环中只有一条路径. 然后发现,有多少个环,就需要改多少条边.然后设出现连向自己的节点为x,这些也要改边,对答案的贡献应为:$max(x-1,0)$.对于最后的根节点,有自环选一个,没自环在其他环上任选一个点就行. #include<cstdio> inline int min(int a,int b){return…

题目链接:http://codeforces.com/problemset/problem/698/B题意:告诉你n个节点当前的父节点,修改最少的点的父节点使之变成一棵有根树.思路:拆环.题解:http://codeforces.com/blog/entry/46148代码: #include <cstdio> ; int f[maxn], vis[maxn], n, s, cnt, idx; int Find(int x) { vis[x] = ++ idx; while (!vis[ f[…