{"id":327,"date":"2010-07-29T17:50:00","date_gmt":"2010-07-29T09:50:00","guid":{"rendered":"http:\/\/localhost\/?p=327"},"modified":"2010-07-29T17:50:00","modified_gmt":"2010-07-29T09:50:00","slug":"codeforces19_e","status":"publish","type":"post","link":"https:\/\/www.shuizilong.com\/wjmzbmr\/?p=327","title":{"rendered":"CodeForces#19 E"},"content":{"rendered":"

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\u7136\u540e\u5bf9\u6bcf\u4e2a\u4e24\u7aef\u989c\u8272\u4e0d\u4e00\u6837\u7684\u975e\u6811\u8fb9\u7684\u4e24\u70b9\u4e4b\u95f4\u7684\u8def\u5f84+1\uff0c\u663e\u7136\u8981\u5220\u6389\u6811\u8fb9\u4e0d\u80fd\u5728\u88ab\u4e24\u7aef\u989c\u8272\u4e00\u6837\u7684\u975e\u6811\u8fb9\u7684\u4e24\u70b9\u95f4\u7684\u8def\u5f84\u8de8\u8d8a\u7684\u60c5\u51b5\u4e0b\u518d\u88ab\u8fd9\u4e2a\u4e0d\u4e00\u6837\u7684\u975e\u6811\u8fb9\u7684\u4e24\u70b9\u4e4b\u95f4\u7684\u8def\u5f84\u8de8\u8d8a\u3002\u3002
\u7136\u540e\u7ef4\u62a4\u4e24\u4e2a\u6811\u94fe\u5256\u5206\u3002\u3002\u5206\u522b\u641e\u6811\u8fb9\u548c\u975e\u6811\u8fb9\u3002\u3002\u5c31\u6ca1\u4ec0\u4e48\u95ee\u9898\u4e86\u3002\u3002
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\u6211\u53d1\u73b0\u8fd8\u662f\u6211\u5728codeforces\u4e0a\u82f1\u6587\u7684\u8bf4\u660e\u66f4\u6e05\u695a\u4e00\u70b9\u554a\u56e7\u3002\u3002\u3002
It’s a interesting problem.If you for every edge, 
try to remove it and check if it is a <\/strong>bipartite 
graph..I think it will get TLE..so let’s analysis 
the property of bipartite graph..
It should never contain a cycle of odd length…
and it can be 2-colored..
so first build a spanning forest for the graph..
and do the 2-color on it(Tree can be 2-colored).
for convenience.
Let TreeEdge={all edge in forest}
NotTreeEdge={All edge}\/TreeEdge
ErrorEdge={all edge that two endpoint have the same color..}
NotErorEdge=NotTreeEdge\/ErroEdge..
First,consider a edge form NotTreeEdge,remove it 
can’t change any node’s color..so..if |ErrorEdge|=0 of course we can remove all NotTreeEdgeif =1 we just can remove the ErrorEdgeif >1 we can’t remove any from NotTreeEdge
Now,Let consider a Edge e from TreeEdge..
Let Path(Edge e)=the path in forest between e’s two endpoints..
if there is a Edge e’ from ErrorEdge that Path(e’) 
didn’t  go through e..it will destroy the bipartite 
graph..
if there is a Edge e’ from ErrorEdge that Path(e’) go through e and there is a Edge e” from NotErrorEdge that Path(e”) go through e..it will also destroy the bipartite graph..
so now we need to know for every edge,how many such path go through it..it require a data structure…
one way is to use heavy-light decomposition then we can update every path in O(LogN^2)…
another way is to use Link-Cut Tree..It can do the same in O(LogN)….
Code\uff1a
http:\/\/www.ideone.com\/dPS5N<\/a>
<\/a> <\/p>\n","protected":false},"excerpt":{"rendered":"

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