1654: [Usaco2006 Jan]The Cow Prom 奶牛舞会 Time Limit: 5 Sec Memory Limit: 64 MB Description 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 ton…
Description 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…
几乎是板子,求有几个size>1的scc 直接tarjan即可 #include<iostream> #include<cstdio> #include<cstring> using namespace std; const int N=10005; int n,m,h[N],cnt,ans,tmp,dfn[N],low[N],s[N],top; bool v[N]; struct qwe { int ne,to; }e[N*10]; int read() { i…
Description 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…
Description 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…
裸的强连通 ; type node=record f,t:longint; end; var n,m,dgr,i,u,v,num,ans:longint; bfsdgr,low,head,f:array[..maxe] of longint; b:array[..maxe] of node; p:array[..maxe] of boolean; procedure insert(u,v:longint); begin with b[i] do begin f:=head[u]; t:=v; e…
http://www.lydsy.com/JudgeOnline/problem.php?id=1720 Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 177 Solved: 90[Submit][Status][Discuss] Description Farmer John wishes to build a corral for his cows. Being finicky beasts, they demand that the corr…
[BZOJ1720][Usaco2006 Jan]Corral the Cows 奶牛围栏 Description Farmer John wishes to build a corral for his cows. Being finicky beasts, they demand that the corral be square and that the corral contain at least C (1 <= C <= 500) clover fields for afterno…
题目 1635: [Usaco2007 Jan]Tallest Cow 最高的牛 Time Limit: 5 Sec Memory Limit: 64 MB Description FJ's N (1 <= N <= 10,000) cows conveniently indexed 1..N are standing in a line. Each cow has a positive integer height (which is a bit of secret). You are t…
本来以为是一道数学题,一顿XJBT导式子,结果就是个幼儿园都会的模拟. Code: #include<bits/stdc++.h> #define ll long long using namespace std; int main(){ ll n; scanf("%lld",&n); int ans=0; for(;n>1;++ans){ if(n%2==0) n>>=1; else n=n*3+1; } printf("%d"…
Description FJ's N (1 <= N <= 10,000) cows conveniently indexed 1..N are standing in a line. Each cow has a positive integer height (which is a bit of secret). You are told only the height H (1 <= H <= 1,000,000) of the tallest cow along with…
Description The pasture contains a small, contiguous grove of trees that has no 'holes' in the middle of the it. Bessie wonders: how far is it to walk around that grove and get back to my starting position? She's just sure there is a way to do it by…
首先来分析一下,这是一张无向图,要求没有两条路联通的点对个数 有两条路连通,无向图,也就是说,问题转化为不在一个点双连通分量里的点对个数 tarjan即可,和求scc还不太一样-- #include<iostream> #include<cstdio> using namespace std; const int N=5005; int n,m,h[N],cnt=1,x,y,dfn[N],low[N],tmp,s[N],top,bl[N],col,d[N],ans; bool v[…
s[i]为差分后的"i这头牛前有几头比它高",计算答案的时候加成前缀和,假设第一头最高减一下即可 用map记录一下被加过的区间,避免重复 #include<iostream> #include<cstdio> #include<map> #include<algorithm> using namespace std; const int N=10005; int n,q,h,r,s[N],f[N],a,b; map<int,bool…
题目简述:一个有向图,求出这个图点数>1的强连通分量的个数. 那么就是tarjan求强联通分量的模板了. 记得要用一个数组标记节点是否在栈中. 1 #include<bits/stdc++.h> 2 using namespace std; 3 const int N=1e5+10; 4 int head[N],nxt[N<<1],to[N<<1],tot; 5 int dfn[N],low[N],st[N],top,idx,cnt,sze[N]; 6 int n…
