[CF475E]Strongly Connected City 2

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[CF475E]Strongly Connected City 2

题目大意:给一张$n(n\leqslant2000)$个点的无向图,给所有边定向,使定向之后存在最多的有序点对$(a,b)$满足从$a$能到$b$ 题解:先把边双缩点,因为这里面的点一定两两可达. 根据网上题解得知,最优解一定长这样:存在一个点$s$,使得对于任意其他点$t$,要么$s$可以到$t$,要么$t$可以到$s$,就把$s$作为根.(出题人的题解也没给出解答,就感性理解) 所以$s$的每一个子树内的边要么都朝向$s$,要么都远离$s$ 然后可以枚举哪个点作为根,记$w_i$第$i$个双…

cf475B Strongly Connected City

B. Strongly Connected City time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Imagine a city with n horizontal streets crossing m vertical streets, forming an (n - 1) × (m - 1) grid. In order…

codeforces B. Strongly Connected City(dfs水过)

题意:有横向和纵向的街道,每个街道只有一个方向,垂直的街道相交会产生一个节点,这样每个节点都有两个方向, 问是否每一个节点都可以由其他的节点到达.... 思路:规律没有想到,直接爆搜!每一个节点dfs一次,记录每个节节点被访问的次数!如果每个节点最终的访问次数和所有节点的数目相同,则输出“YES", 否则输出”NO“ #include <queue> #include <string> #include <cstdio> #include <cstri…

Codeforces 475 B Strongly Connected City【DFS】

题意:给出n行m列的十字路口,<代表从东向西,>从西向东,v从北向南,^从南向北,问在任意一个十字路口是否都能走到其他任意的十字路口四个方向搜,搜完之后,判断每个点能够访问的点的数目是否相同,相同的话则说明可以到达任意点 #include<iostream> #include<cstdio> #include<cstring> #include <cmath> #include<stack> #include<vector&g…

Codeforces-475B Strongly Connected City

仅仅用推断最外层是不是回路假设是则每两个点之间连通 #include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #define INF 200000 using namespace std; int main() { char a[3][22]; int n,m,i,j; scanf("%d%d",&n,&m); getcha…

PTA Strongly Connected Components

Write a program to find the strongly connected components in a digraph. Format of functions: void StronglyConnectedComponents( Graph G, void (*visit)(Vertex V) ); where Graph is defined as the following: typedef struct VNode *PtrToVNode; struct VNode…

algorithm@ Strongly Connected Component

Strongly Connected Components A directed graph is strongly connected if there is a path between all pairs of vertices. A strongly connected component (SCC) of a directed graph is a maximal strongly connected subgraph. For example, there are 3 SCCs in…