题目链接:http://lightoj.com/volume_showproblem.php?problem=1003 题意是有m个关系格式是a b:表示想要和b必须喝a,问一个人是否喝醉就看一个人是否能把所有种类的饮料喝完,能输出Yes,不能输出No: 其实就是有向图判断是否存在环,可以用拓扑排序来求 拓扑排序:每次找到一个入度为0的点加入队列,删除与之相连的边,循环操作,得到的序列就是拓扑序(前面的必须出现在后面的前面): #include<stdio.h> #include<str…

One of my friends is always drunk. So, sometimes I get a bit confused whether he is drunk or not. So, one day I was talking to him, about his drinks! He began to describe his way of drinking. So, let me share his ideas a bit. I am expressing in my wo…

Triangle LOVE Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 3566    Accepted Submission(s): 1395 Problem Description Recently, scientists find that there is love between any of two people. For…

题目: 给出几种正方形,每种正方形有无穷多个.在连接的时候正方形可以旋转.翻转. 正方形的每条边上都有一个大写英文字母加‘+’或‘-’.00,当字母相同符号不同时,这两条边可以相连接,00不能和任何边相连. 判断给出的正方形如果能无限连接下去就输出unbounded.不能就输出bounded. 思路: 这个题读完之后,一点思路都没有,看完紫书上的分析知道是用拓扑排序来判断有向环,但具体的构造还是不会…… 1.将每个正方形看作一条边,在正方形每两条边上的字母(不包括00)之间连一条有向边构成一个有…

1003 - Drunk PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB One of my friends is always drunk. So, sometimes I get a bit confused whether he is drunk or not. So, one day I was talking to him, about his drinks! He began to…

hihocoder 1174 [算法]: 计算每一个点的入度值deg[i],这一步需要扫描所有点和边,复杂度O(N+M). 把入度为0的点加入队列Q中,当然有可能存在多个入度为0的点,同时它们之间也不会存在连接关系,所以按照任意顺序加入Q都是可以的. 从Q中取出一个点p.对于每一个未删除且与p相连的点q,deg[q] = deg[q] - 1:如果deg[q]==0,把q加入Q. 不断重复第3步,直到Q为空. 最后剩下的未被删除的点,也就是组成环的点了. [代码]: #include <bits…

//拓扑排序判断是否有环 #include<cstdio> #include<algorithm> #include<string.h> #include<math.h> #include<queue> using namespace std; typedef long long ll; ; int G[maxn][maxn]; int in[maxn]; void init() { memset(G,,sizeof(G)); //图 memse…

One of my friends is always drunk. So, sometimes I get a bit confused whether he is drunk or not. So, one day I was talking to him, about his drinks! He began to describe his way of drinking. So, let me share his ideas a bit. I am expressing in my wo…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3342 题目大意:n个点,m条有向边,让你判断是否有环. 解题思路:裸题,用dfs版的拓扑排序直接套用即可. 代码: #include<iostream> #include<cstdio> #include<cstring> using namespace std; ; int n,m; int vis[N]; bool G[N][N]; bool dfs(int u){ v…

Is It A Tree? Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28399   Accepted: 9684 Description A tree is a well-known data structure that is either empty (null, void, nothing) or is a set of one or more nodes connected by directed edge…