题目描述 The N (2 <= N <= 10,000) 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 D…

题目链接:https://www.luogu.org/problemnew/show/P2863 求强连通分量大小>自己单个点的 #include <stack> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; const int maxn = 100000 + 10; struct edge…

题目传送门 这个题还是个缩点的板子题...... 答案就是size大于1的强连通分量的个数 加一个size来统计就好了 #include <iostream> #include <cstdlib> #include <cstdio> using namespace std; const int N=1e5+5; const int M=5e5+5; struct edge{ int to,next; }e[M]; int n,m,dfn[N],low[N],cnt,he…

P2863 [USACO06JAN]牛的舞会The Cow Prom 求点数$>1$的强连通分量数,裸的Tanjan模板. #include<iostream> #include<cstdio> #include<cstring> using namespace std; int min(int &a,int &b){return a<b?a:b;} #define N 10002 #define M 50002 int n,m,clo,df…

洛谷——P2863 [USACO06JAN]牛的舞会The Cow Prom 题目描述 The N (2 <= N <= 10,000) 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 D…

https://www.luogu.org/problem/show?pid=2863#sub 题目描述 The N (2 <= N <= 10,000) 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 th…

传送门 题目大意:形成一个环的牛可以跳舞,几个环连在一起是个小组,求几个小组. 题解:tarjian缩点后,求缩的点包含的原来的点数大于1的个数. 代码: #include<iostream> #include<cstdio> #include<cstring> #define maxn 10009 using namespace std; int n,m,sumedge,top,sumclr,tim,ans; int Stack[maxn],instack[maxn]…

代码是粘的,庆幸我还能看懂. #include<iostream> #include<cstdio> #include<cmath> #include<algorithm> #include<stack> #include<vector> using namespace std; struct node { int num,par; }p[]; ,x,y,cnt; vector<]; ]; ]; stack<int>…

每日一题 day11 打卡 Analysis 好久没大Tarjan了,练习练习模板. 只要在Tarjan后扫一遍si数组看是否大于1就好了. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #define maxn 10000+10 #define maxm 50000+10 using namespace std; inline int read() { ;…

一道tarjan的模板水题 在这里还是着重解释一下tarjan的代码 #include<iostream> #include<cstdio> #include<algorithm> using namespace std; int n,m; int cnt;//记录强联通分量的个数 int visitnum;//遍历的步数 int dfn[100010];//记录元素第一次被访问的步数 int low[100010];//包含i的强联通分量最早被访问的步数 int nu…