POJ 3180 Tarjan】的更多相关文章

题意:找强连通中点数大于2的强连通分量个数 思路:Tarjan // By SiriusRen #include <cstdio> #include <algorithm> using namespace std; int n,m,ans=0,t=0,cnt=0,tot=1,top=0,dfn[50050],low[50050],p[10050],s[10050]; int xx,yy,jy,first[10050],next[50050],v[50050]; bool vis[5…
Description The N ( <= N <= ,) cows are so excited: it's prom night! They are dressed in their finest gowns, complete with corsages and new shoes. They know that tonight they will each try to perform the Round Dance. Only cows can perform the Round…
题目用google翻译实在看不懂 其实题目意思如下 给一个有向图,求点个数大于1的强联通分量个数 #include<cstdio> #include<algorithm> #include<cstring> #include<vector> #include<stack> #define M 50010 #define N 10010 using namespace std; ,ans,sz[N],belong[N],dfn[N],low[N],…
http://poj.org/problem?id=3180 The Cow Prom Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 1428   Accepted: 903 Description The N (2 <= N <= 10,000) cows are so excited: it's prom night! They are dressed in their finest gowns, complete…
Going from u to v or from v to u? Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 15494   Accepted: 4100 Description In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors…
Going from u to v or from v to u? Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 17383   Accepted: 4660 Description In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors…
[题目链接] http://poj.org/problem?id=3180 [题目大意] N头牛,M条有向绳子,能组成几个歌舞团?要求顺时针逆时针都能带动舞团内所有牛. [题解] 等价于求点数大于1的SCC数量. [代码] #include <cstdio> #include <algorithm> #include <vector> #include <cstring> using namespace std; const int MAX_V=10000;…
Network Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 7298   Accepted: 2651 Description A network administrator manages a large network. The network consists of N computers and M links between pairs of computers. Any pair of computers…
蕾姐讲过的例题..玩了两天后才想起来做 貌似省赛之后确实变得好懒了...再努力两天就可以去北京玩了! 顺便借这个题记录一下求强连通分量的算法 1 只需要一次dfs 依靠stack来实现的tarjan算法 每次走到一个点 马上把它压入栈中 每次对与这个点相连的点处理完毕 判断是否low[u]==dfn[u] 若是 开始退栈 直到栈顶元素等于u才退出(当栈顶元素等于u也需要pop) 每次一起退栈的点属于同一个强连通分量 储存图可以用链式前向星也可以用邻接矩阵更可以用vector 蕾姐说不会超时 我信…
题目大意: 约翰的N(2≤N≤10000)只奶牛非常兴奋,因为这是舞会之夜!她们穿上礼服和新鞋子,别上鲜花,她们要表演圆舞.           只有奶牛才能表演这种圆舞.圆舞需要一些绳索和一个圆形的水池.奶牛们围在池边站好,顺时针顺序由1到N编号.每只奶牛都面对水池,这样她就能看到其他的每一只奶牛.为了跳这种圆舞,她们找了M(2≤M≤50000)条绳索.若干只奶牛的蹄上握着绳索的一端,绳索沿顺时针方绕过水池,另一端则捆在另一些奶牛身上.这样,一些奶牛就可以牵引另一些奶牛.有的奶牛可能握有很多绳…