[BZOJ1654] [Usaco2006 Jan]The Cow Prom 奶牛舞会

2024-01-09 12:59

本文主要是介绍[BZOJ1654] [Usaco2006 Jan]The Cow Prom 奶牛舞会,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!

传送门

http://www.lydsy.com/JudgeOnline/problem.php?id=1654

题目大意

求图中包括点数大于1的强连通分量数

题解

模板测试

varw:array[0..70000,1..2]of longint;low,dfn,p,t:array[0..10005]of longint;i,j,k:longint;n,m,len,a,b,tmp,ans:longint;
procedure init(a,b:longint);
beginw[len,1]:=b;if w[a,2]=0then w[a,2]:=len else w[w[a,1],2]:=len;w[a,1]:=len; inc(len);
end;function min(a,b:longint):longint;
beginif a>b then exit(b) else exit(a);
end;procedure tarjan(a:longint);
var tt,s:longint;
begindfn[a]:=tmp; low[a]:=tmp; inc(tmp);inc(t[0]); t[t[0]]:=a; p[a]:=1;tt:=w[a,2];while tt<>0 dobeginif dfn[w[tt,1]]=0then begin tarjan(w[tt,1]); low[a]:=min(low[a],low[w[tt,1]]); endelse if p[w[tt,1]]=1 then low[a]:=min(low[a],dfn[w[tt,1]]);tt:=w[tt,2];end;s:=t[0];if dfn[a]=low[a] then beginrepeatp[t[t[0]]]:=0;dec(t[0]);until a=t[t[0]+1];if t[0]<>s-1 then inc(ans);end;
end;beginreadln(n,m); len:=n+1;for i:=1 to m dobegin readln(a,b); init(a,b); end;tmp:=1; ans:=0;for i:=1 to n doif dfn[i]=0 then tarjan(i);writeln(ans);
end.

这篇关于[BZOJ1654] [Usaco2006 Jan]The Cow Prom 奶牛舞会的文章就介绍到这儿,希望我们推荐的文章对编程师们有所帮助!



http://www.chinasem.cn/article/587189

相关文章

【POJ】3660 Cow Contest floyd(可以拓扑排序?)

Cow Contest Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6925 Accepted: 3792 Description N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating i

BFS —— POJ 3278 Catch That Cow

对应POJ题目:点击打开链接 Catch That Cow Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 54686 Accepted: 17096 Description Farmer John has been informed of the location of a fugitive cow

USACO Section 2.3 Cow Pedigrees

题意: N个节点  深度为K  的正则二叉树  求  树有几种形态 思路: 一开始以为是数学题…  看了byvoid的题解才知道是dp… 每棵树由根节点、左子树、右子树构成  由此得状态转移  树=左子树*右子树 节点数和深度是影响答案的属性  所以令dp[i][j]表示i个节点深度在j以内的树的形态数 深度在j以内的树又两个深度在j-1以内的树和一个根节点构成 设左子树k个节

Codeforces 283. B. Cow Program记忆化搜索

output standard output Farmer John has just given the cows a program to play with! The program contains two integer variables, x and y, and performs the following operations on a sequence a1, a2, ..

【POJ3270】【Cow Sorting】

Cow Sorting Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 6411 Accepted: 2488 Description Farmer John's N (1 ≤ N ≤ 10,000) cows are lined up to be milked in the evening. Each

【POJ3268】【Silver Cow Party】【反向dij】【sizeof失效】

Silver Cow Party Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 15522 Accepted: 7039 Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going

POJ 1985 Cow Marathon(树的直径)

题目链接 题意:给出一棵树,求出这个树的直径 解答:任选一点进行dfs,会找到一个最远点s,在以这个最远点s进行dfs,会找到一个最远点是t,那么s-t就是树的直径。 //#include<bits/stdc++.h>#include<cstdio>#include<algorithm>#include<vector>#include<cstring>using namespace

POJ 2184 Cow Exhibition (处理负值的01背包)

【题目链接】:click here~~ 【题意】: 题意:给定n头牛的聪明指数S和幸福指数F,如果存在S的和TS>=0与F的和TF>=0同时成立时, 输出TS与TF的和的最大值sum,否则,输出0。 【思路】:      转化问题,求s和为某个固定值时候最大的f和值,然后遍历这些所有的s和以及对应的f和值,求出总和总和最大的那个。      那么这样就是一个0-1背包问题,可以把s值理解

poj3278--Catch That Cow

Catch That Cow Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 43680 Accepted: 13615 Description Farmer John has been informed of the location of a fugitive cow and wants to catch h

poj3267--The Cow Lexicon(dp:字符串组合)

The Cow Lexicon Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 8211 Accepted: 3864 Description Few know that the cows have their own dictionary with W (1 ≤ W ≤ 600) words, each con