scanjet 4370

) {Priority_queueint>que; intT; for(intI=1; ii) lowcost[i]=INF; Que.push (start); while(!Que.empty ()) {T=Que.top (); Que.pop (); for(intI=1; ii)if(i!=t)if(Lowcost[t]==inf | | lowcost[i]>lowcost[t]+Cost[t][i]) {Lowcost[i]= (Lowcost[t]==inf?)0: lowcost[t]) +Cost[t][i]; Que.push (i); } }}intMAP1[MAXN][MAXN];intANS[MAXN];intMain () {//freopen ("In.txt", "R", stdin); //freopen ("OUT.txt", "w", stdout); intN; intt1,t2; while(~SCANF ("%d",N)) { for(intI=1; ii) for(in

#include The topic requires 01 of the MatrixThe first line, except A11, can only have 1 1The last column except Ann can only have 1 1In addition to the four edges of the matrix, the points inside require that the number of rows 1 of the point be equalFinally we can find that if the topic gives a complete graph of the Benquan matrix, then the 01 matrix shows a path on this graphCleverly modified SPFA to find the smallest ring on OKusing namespacestd;#defineLL Long LongConst intmaxn=3e2+Ten;intDIS

adequate, but not necessary. (Sorry for the team that caused the problem) B: Starting from 1, taking a ring (at least one point, that is, not a self-ring) and returning to 1; Starting from N, taking a ring (likewise) and returning to n. It is easy to verify, which meets the requirements of the question. And a | B is a sufficient condition for this question. Because the edge weight is not negative, the two rings correspond to two simple rings. Therefore, we can start from 1 and find a m

; Q.push (ST); while(!Q.empty ()) { intu=Q.front (); Q.pop (); Inq[u]=0; for(intI=1; i) { if(d[u]+w[u][i]D[i]) {D[i]=d[u]+W[u][i]; if(!inq[i]) {inq[i]=1; Q.push (i);} } } if(ORZ) d[st]=inf,orz=0; } returnD[en];}intMain () {//freopen ("In.txt", "R", stdin); while(SCANF ("%d", n)! =EOF) {memset (W,0,sizeof(w)); for(intI=1; i) for(intj=1; j) scanf ("%d",W[i][j]); intANS=SPFA (1, N); intRES1=SPFA (1,1); intRes2=SPFA (n

1001. http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4370 A topic similar to God, which is modeled using graph theory ~ N has n points in total. X12 + x13 +... x1n = 12. x1n + x2n +... xN-1n = 1 It is understood that the outbound degree of point 1 is 1, and the inbound degree of point N is 1; The matrix is interpreted as the weight on the edge, so it can be understood: 1. The shortest path from 1 to n 2.1 and N respectively. Mark wrote a dijikstra an

The main idea: to give a nxn matrix, to find a nxn containing only 0 and 1 of the Matrix, so that two matrix position corresponding to the elements of the multiplication and the smallest. Ideas: Ingenious ideas, the three conditions given by the

Tomcat abnormal stop, process is killedThe corresponding Tomcat logs are as follows:/application/tomcat-service-8080/bin/catalina.sh:line 386:4370 killed "/usr/local/jdk/bin/java" "- Djava.util.logging.config.file=/application/tomcat-service-8080/conf/loggIng.properties "-DJAVA.UTIL.LOGGING.MANAGER=ORG.APACHE.JULI.CLASSLOADERLOGMANAGER-SERVER-XMS4096M-XMX4096M-XX: Permsize=128m-xx:maxnewsize=512m-xx:maxpermsize=256m-djava.awt.headless=true-djava.endor

( photovieweractivity.java:125) atcom.android.mediascape.activity.photovieweractivity$33$ 1.run (photovieweractivity.java:4558) atandroid.os.handler.handlecallback (Handler.java : 587) atandroid.os.handler.dispatchmessage (handler.java:92) Atandroid.os.Looper.loop (looper.java:1) atandroid.app.activitythread.main (activitythread.java:4370) Atjava.lang.reflect.Method.invokeNative (Nativemethod) Atjava.lang.reflect.Method.invoke (method.java:521) at

(bufferedinputstream.java:157) Atjava.io.BufferedInputStream.read ( bufferedinputstream.java:346) Atandroid.graphics.BitmapFactory.nativeDecodeStream (Native Method) Atandroid.graphics.BitmapFactory.decodeStream (bitmapfactory.java:459) Atcom.android.mediascape.activity.PhotoViewerActivity.getPreviewImage (photovieweractivity.java:4465) Atcom.android.mediascape.activity.PhotoViewerActivity.dispPreview (photovieweractivity.java:4406) atcom.android.mediascape.activity.photoviewerac

) Atandroid.graphics.BitmapFactory.nativeDecodeStream (Native Method) Atandroid.graphics.BitmapFactory.decodeStream (bitmapfactory.java:459) Atcom.android.mediascape.activity.PhotoViewerActivity.getPreviewImage (photovieweractivity.java:4465) Atcom.android.mediascape.activity.PhotoViewerActivity.dispPreview (photovieweractivity.java:4406) atcom.android.mediascape.activity.photovieweractivity.access$6500 (photovieweractivity.java:125) Atcom.android.mediascape.activity.photovieweractivity$33$1.run

:459) atcom.android.mediascape.activity.PhotoViewerActivity.getPreviewImage(PhotoViewerActivity.java:4465) atcom.android.mediascape.activity.PhotoViewerActivity.dispPreview(PhotoViewerActivity.java:4406) atcom.android.mediascape.activity.PhotoViewerActivity.access$6500(PhotoViewerActivity.java:125) atcom.android.mediascape.activity.PhotoViewerActivity$33$1.run(PhotoViewerActivity.java:4558) atandroid.os.Handler.handleCallback(Handler.java:587) atandroid.os.Handler.dispatchMessa

)atorg.apache.harmony.luni.net.SocketInputStream.read(SocketInputStream.java:87)atorg.apache.harmony.luni.internal.net.www.protocol.http.HttpURLConnection$LimitedInputStream.read(HttpURLConnection.java:303)atjava.io.InputStream.read(InputStream.java:133)atjava.io.BufferedInputStream.fillbuf(BufferedInputStream.java:157)atjava.io.BufferedInputStream.read(BufferedInputStream.java:346)atandroid.graphics.BitmapFactory.nativeDecodeStream(Native Method)atandroid.graphics.BitmapFactory.decodeStream(Bit

windows GreekCp1254 windows TurkishCp1255 windows HebrewCp1256 windows ArabicCp1257 windows BalticCp1258 windows VietnameseCp1381 ibm OS/2, DOS People's Republic of China (PRC)Cp1383 ibm aix People's Republic of China (PRC)Cp273 IBM Austria, GermanyCp277 IBM Denmark, NorwayCp278 IBM Finland, SwedenCp280 IBM ItalyCp284 IBM Catalan/Spain, Spanish Latin AmericaCp285 IBM United Kingdom, IrelandCp297 IBM FranceCp33722 IBM-eucjp-Japan (superset of 5050)Cp420 IBM ArabicCp424 IBM HebrewMS-DOS United St

11 36 3:2 11 1 6 20:2 18:2 2d 2f 33:2 2f:2 11 36 3:2 11 16 18 20:2 18:2 11:2 2 92:9 1 8,340:2 1:5 2:2 3:2 1:7 5:5 6:5 7 : 5 8:5 9:5 a:b c:11 d e:2 d:5 e:2 d:5 e:2 d ef:2 d:5 f:2 d:5 f:2 d f 10:2 d:5 10:3 d:3 11:2 b:7 1 1fb4:3 0 1:3 0 2:a 0 7e 1:7 05: 2 0 3 4:3 0 3:7 0 54:3 0 5:3 04:3 0 7 7e 2 A:2 0 B 4:3 0 7:2 0 7 d f:6 0 07c 0 6:6 0f 0 d9:3 0 1 4:7 017 0 7c0 8:6 0 9:3 0 19:7 0 1c1a 0 7c 0a:6 0 c80 one 9:3 01e:7 0 1f0 7c 0 b:6 0 9:3 0 23:7 0 07c 0 c:6 0e:3 0 9:3 028:7 0 2b 290 7c 0 d:6 0 f:3 0

22249896-4a9e-403a-ade0-70fbffeeb934 TxServiceLastModified 9c8a5443-5e09-4bdf-8bb9-63a4d58d6896 TxStartDate 32f590a3-4b46-433a-9061-49b31d91b22a TxEffectiveCreated f1962adb-b231-42ac-9c48-60dbe899280b TxEffectiveLastModified 74bec003-cac3-4e53-a3d8-60d6c7ddc226 TxEffectiveLocationGroupOrder 150bb05a-af1e-4e06-bafd-9b3ccd2d2074 TxEffectiveOrder 26eeefb6-05bc-459b-9959-65ac996a2afb TIxLocationId 09068dab-

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line parameters are used internally, for example:Erl ... -kernel inet_dist_listen_min 4370 inet_dist_listen_max 4375Third, Erlang cookiesAn Erlang cookie is a string of numbers and letters. An Erlang node reads a cookie from the command-line parameter-setcookie at startup. If not specified, the cookie file is read from the $HOME/.erlang.cookie. If the file does not exist, it will be created immediately with a random cookie. Two Erlang nodes communica

Graph Coloring Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 4370 Accepted: 1980 Special Judge DescriptionYou is to write a program this tries to find a optimal coloring for a given graph. Colors is applied to the nodes of the graph and the only available Colors is black and white. The coloring of the graph is called optimal if a maximum of nodes are black. The col

set of test data.The first behavior of each set of test data is two integers n, M, meaning as described earlier.Second behavior for each set of test data n integers, where the I integer VI represents the score of the node labeled IEach group of test data of the line 3~n+1, each line describes a stick, wherein the I+1 Act Two integer Ai,bi, which represents the number of two balls connected by the root of the I stick.For 100% of data, meet NLittle hi Tip: Those with array storage tree edge remem

ID Origin Title 8/22 Problem A HDU 4358 Boring counting 35/52 Problem B HDU 4359 Easy Tree DP? 31/79 Problem C HDU 4362 Dragon Ball 1/2 Problem D HDU 4363 Draw and Paint 18/56 Problem E HDU 4365 Palindrome graph 3/17 Problem F HDU 4367 The War of Virtual world