http://acm.hdu.edu.cn/showproblem.php?pid=3836 Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others)Total Submission(s): 4802    Accepted Submission(s): 1725 Problem Description To prove two sets A and B are equivalen…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=3836 Equivalent Sets Description To prove two sets A and B are equivalent, we can first prove A is a subset of B, and then prove B is a subset of A, so finally we got that these two sets are equivalent.Y…

主题链接: http://acm.hdu.edu.cn/showproblem.php? pid=3836 Equivalent Sets Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others) Total Submission(s): 2890    Accepted Submission(s): 1006 Problem Description To prove two se…

Equivalent Sets Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others)Total Submission(s): 3568    Accepted Submission(s): 1235 Problem Description To prove two sets A and B are equivalent, we can first prove A is a su…

Equivalent Sets Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others) Problem Description To prove two sets A and B are equivalent, we can first prove A is a subset of B, and then prove B is a subset of A, so finally…

Equivalent Sets Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others) Total Submission(s): 2798    Accepted Submission(s): 962 Problem Description To prove two sets A and B are equivalent, we can first prove A is a su…

Equivalent Sets Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others) Total Submission(s): 4782    Accepted Submission(s): 1721 Problem Description To prove two sets A and B are equivalent, we can first prove A is a s…

Problem Description To prove two sets A and B are equivalent, we can first prove A is a subset of B, and then prove B is a subset of A, so finally we got that these two sets are equivalent. You are to prove N sets are equivalent, using the method abo…

http://acm.hdu.edu.cn/showproblem.php?pid=3836 判断至少需要加几条边才能使图变成强连通 把图缩点之后统计入度为0的点和出度为0的点,然后两者中的最大值就是需要连的边, 例如,假设入度为0的点多,那么每次把出度为0的点连一条边指向入度为0的点,就构成了一个环, 所以构成了一个强连通分量,同理可得出度为0点多的情况. 这代码用g++ re了,c++才能ac. #include <iostream> #include <cstdio> #in…

题目大意:给出N个点,M条边.要求你加入最少的边,使得这个图变成强连通分量 解题思路:先找出全部的强连通分量和桥,将强连通分量缩点.桥作为连线,就形成了DAG了 这题被坑了.用了G++交的,结果一直RE,用C++一发就过了... #include <cstdio> #include <cstring> #define N 20010 #define M 100010 #define min(a,b) ((a) > (b)? (b): (a)) #define max(a,b)…