缩点练习 洛谷 P3387 [模板]缩点 缩点 解题思路: 都说是模板了...先缩点把有环图转换成DAG 然后拓扑排序即可 #include <bits/stdc++.h> using namespace std; /* freopen("k.in", "r", stdin); freopen("k.out", "w", stdout); */ //clock_t c1 = clock(); //std::cerr…

http://poj.org/problem?id=2186 Popular Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 20191   Accepted: 8193 Description Every cow's dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows, you…

题目地址:http://poj.org/problem?id=2186 Popular Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 27496   Accepted: 11059 Description Every cow's dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows…

Popular Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 27698   Accepted: 11148 Description Every cow's dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows, you are given up to M (1 <= M &…

传送门 Popular Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 31808   Accepted: 12921 Description Every cow's dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows, you are given up to M (1 <=…

Popular Cows Description Every cow's dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows, you are given up to M (1 <= M <= 50,000) ordered pairs of the form (A, B) that tell you that cow A thinks that co…

题意:给出一个有向图代表牛和牛喜欢的关系,且喜欢关系具有传递性,求出能被所有牛喜欢的牛的总数(除了它自己以外的牛,或者它很自恋). 思路:这个的难处在于这是一个有环的图,对此我们可以使用tarjan算法求出强连通分量,把强连通分量压缩成一个点,构成一个新的图,这个图一定是没有环的,如果有环就跟强连通分量的矛盾了.压缩成无环图以后这个图里面的点是不具有方向的,我们通过遍历每个节点所能连到的点,如果两点的id值即所在的强连通分量区域不同时,我们就把这个节点的出度加1.最后去找那些出度等于0的点,如果…

题目链接:http://poj.org/problem?id=2186 题目意思大概是:给定N(N<=10000)个点和M(M<=50000)条有向边,求有多少个“受欢迎的点”.所谓的“受欢迎的点”当且仅当任何一个点出发都能到达它. 原来的图是无序且可能有环,用tarjan缩点,变成一个DAG.受欢迎点的出度一定为0,所以缩点后求有多少个连通分量的出度是0,要是大于1则没有受欢迎的点,等于1的话求出这个出度为0的连通分量有多少个点. 强联通分量tarjan算法里邻接表用vector一般好理解,…

题意: 有一群牛, a会认为b很帅, 且这种认为是传递的. 问有多少头牛被其他所有牛认为很帅~ 思路: 关键就是分析出缩点之后的有向树只能有一个叶子节点(出度为0). 做法就是Tarjan之后缩点统计出度. #include <cstdio> #include <cstring> #include <stack> #include <algorithm> using namespace std; const int MAXN = 10005; const i…

/** problem: http://poj.org/problem?id=2186 当出度为0的点(可能是缩点后的点)只有一个时就存在被所有牛崇拜的牛 因为如果存在有两个及以上出度为0的点的话,他们不会互相崇拜所以不存在被所有牛崇拜的牛 牛的个数即该出度为0点(由环缩点而来)有多少牛 **/ #include<stdio.h> #include<stack> #include<algorithm> using namespace std; class Graphic…