|
|
back to boardWhy Wa#3? Plz help me~~~ I use dinic to solve this problem. but wa#3. I think maxflow can work it out. I don't know why. could any one give me some test data. Thank you! Edited by author 10.10.2010 09:05 Re: Why Wa#3? Plz help me~~~ Posted by svr 10 Oct 2010 09:14 You are right! This is absolutely maxflow problem. But in my reprizetory I didn't find variant of algo on 300*2000 vertices graph. P.S.AC!!!! Ford_Falkerson! n=102+2000=2012 m=n+2000*n+200~200000 F<=2*n=200 complex=O((n+m)*F)~40 000 000 -normaly Edited by author 10.10.2010 14:06 Re: Why Wa#3? Plz help me~~~ My MAXN=2200; MAXM=2200000; but I still wa#3... I want to ask a question: Does this problem have Special Judge? Edited by author 10.10.2010 19:44 Re: Why Wa#3? Plz help me~~~ And this is my code(C++) [code deleted]. Edited by author 08.11.2010 15:16 Re: Why Wa#3? Plz help me~~~ My node 0 to Max is for time, and S will link to every time node(flow=k). node Max+1 to Max+N is for people, time node will link to the people node which includes this time(flow=1). then people node will link to T(flow=2). S=MAXN-1, T=Max+N+1. Re: Why Wa#3? Plz help me~~~ I use this way to find the answer: if answer is "Yes" then I will regard the vertice which flow is full as answer. Is it right? I think it can give me a feasible answer. but I can't ensure it's a minimal answer... so I wonder is there any special judge. Edited by author 10.10.2010 19:46 Edited by author 11.10.2010 11:41 Re: Why Wa#3? Plz help me~~~ now,I get AC. I have made a stupid mistake. For the input t[i] ans s[i], I think it means the i-th recruit will wait from t[i] to s[i]... but it means he will wait from t[i] to t[i]+s[i]-1... be careful~ Edited by author 08.11.2010 15:20 |
|
|