pii breach

Do not write the Normal template, or need priority queue optimization of the Ang1#include //basic need for header files2#include string.h>3#include 4#include 5#include 6 using namespacestd;7typedef pairint,int>PII;8 Const intinf=0x3f3f3f3f;9 Ten One structcmp{//Change priority queue to small Gan A BOOL operator() (PII a,pii b) { - returnA.first>B.fir

Package android.zhh.com Import android.app.Activity import android.os.Bundle import android.util.Log Import Java.math.BigInteger class Main2activity:activity () {Override fun OnCreate (Savedinstancestate:bundle?) {super.oncreate (savedinstancestate) Setcontentview (r.layout.activity_main2)/** * Function General notation *///line1; /** * When the content of the LINE1 function is only one line, the function above can be written in this way *///line2 (on); /** * When the con

Topic Links:Codeforces 332BMain topic:Give a sequence to find the two length k segments that do not cover each other, ask for the weight and maximum of the two segments, sort by the left end of the left segment, and the left end of the right segment as the second keyword to get the first answer.Topic Analysis: Very water data structure of the topic, we only need to first use the prefix and preprocessing all the length of the total weight of the section K. The RMQ is then used to maintai

the priority queue of data too much caused by the memory exceeded the limit, so hurriedly added a limit. Sure enough, it's over.　　 　　 Here is a Boolean array trainofpath[v] recorded to the V point of this side is not a railroad. If the road was originally a railroad, it is now the road that is updated, only to be updated; If you want to get a MLE code, if you are working on a highway that is currently being updated, some per

Topic Radio Wave: poj--3268 Silver Cow PartyRun two times, Dijkstra. Mentally retarded problem#include #include#include#include#includeusing namespacestd;#defineMAXN 10001#defineINF 0x3f3f3f#definePII pairstructac{intTo,va; AC () {} ac (intAintb) { to=a,va=b; }};vectorQ[MAXN],W[MAXN];intDIS[MAXN],DIST[MAXN];BOOLFA[MAXN];voidDijkstraints) {memset (FA,0,sizeof(FA)); memset (Dis,inf,sizeof(DIS)); Priority_queuePQ; Dis[s]=0; Pq.push (PII (dis[s],s)); w

New Week/w \Everyone is reading in the library. I sleep in the library ...5.31Do CF encounter a problem of BFS, think of this question has not yet mendedHiho 1233 BoxesLast year a God taught me ... But it's not going to happen again.Or is the representation of the State not to be clear, is the weight of each block multiplied by the position of the number representing a state1#include 2#include 3#include 4#include 5#include 6 using namespacestd;7 8typedef pairint,int>

people who are not 0, but it is a priority to manually kill the less durable ones.Conclusion: After killing the first person with a minimum endurance of 0, I manually kill from small to large with durability.It is also important to note that, after killing people in turn, if the rest of the people can be killed automatically, I do not need to manually kill the person.In the end, there are two cases where the most value is taken.Code:#include #include#include#include#include#includeSet>#include#

Test instructionsGive you n personal scores and change the values, ask you to arrange a change in order to make n individual rank change and maximumIdeas:N^2 enumerates the effects on the results of the change in the rank of each of the two people./************************************************author:d evil*********************************************** * */#include#include#include#include#include#include#includeSet>#include#include#include#includestring>#include#include#include#defineLL Long

Portal: http://codeforces.com/contest/765A: Give your home name, and n the beginning and end of the ticket, Jinotega start at home, you have to judge according to the beginning and end of these tickets Jinotega is not at home. Direct record the number of times the beginning and end of the home appear, if the equal description Jinotega at home, otherwise not at home.#include #include#include#include#include#include#include#includestring>#include#include#includeSet>#include#defineX First#defineY S

Have to say, Taobao has a lot of bad sellers, encounter bad sellers, we would like to complain about them. But Taobao complains that sellers still have time and conditions to limit. Let's follow the small weave to see. 　　Taobao buyers to initiate complaints against the rights of all have a time, to meet the following time conditions: 1, the transaction status of "transaction close" can initiate the following types of complaints: breach of

sufficient number of clicks or does not produce enough advertising effective display times (in any case, based on Google) and without notice. Parts 3rd, 6th to 10th and 14th to 17th remain in force for any reason to terminate the participation of any resources in the scheme or to terminate this agreement. Confidential . You agree not to disclose Google confidential information without prior written permission from Google. Google Confidential information includes but is not limited to:

Label:Oracle version: 11.2.0.2 Connection mode: Spring JdbcTemplate connects to the database and loads the configuration file via Classpathxmlapplicationcontext ("Vodbosscontext.xml"). Do the following error in the unit test package: Org.springframework.jdbc.CannotGetJdbcConnectionException:Could not get JDBC Connection; Nested exception is Org.apache.commons.dbcp.SQLNestedException:Cannot create poolableconnectionfactory (breach of agreement) At Org.

sample. There is 6 triples: (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1). Because N = 3, the cost of needed to build the network are always D(1, 2) + D(2, 3) + D (3, 1) for all the triples. So, the expected cost equals to D(1, 2) + D(2, 3) + D(3, 1). It's a pity that I didn't make it in this game (or yellow). Test instructions is given a tree with weights.Ask you to choose any of the 3 points in the tree C1, C2, C3 D (C1,C2) + D (C2,C3) + D (C1, C3) expectations.Because

, integrated Nic, etc.) are all controlled by the nanqiao chip. The nanqiao chip is usually exposed next to the PCI slot, and the block size is relatively large. A channel is required for data transmission between the North and South bridges at any time. This channel is the north and south bridge bus. The wider the north-south bridge bus, the more convenient data transmission. In the Motherboard chipset of each manufacturer, the North-South Bridge bus is named respectively ., For example, Intel'

It's only been a week, Qaq.2.21CF 629 E famil Door and RoadsIn fact, when Tarjan to engage in, O (n+m) of Acridine.1#include 2#include 3#include 4 using namespacestd;5typedefLong LongLL;6typedef pairint,int>PII;7 Const intMAXN = 1e5 +Ten;8VectorQ[MAXN];9 DoubleANS[MAXN];Ten One //Tree A intcnt, H[MAXN]; - structEdge - { the intto, pre; -} e[maxn1]; - - voidAddint from,intto ) + { -cnt++; +E[cnt].pre = h[ from]; AE[cnt].to =to ; ath[ from] =CNT;

subsequence is.Sample Input2 5 35 14 23 12 43 1 5 45 14 23 12 43 1Sample output44 Test instructions is given two sequences a, B. Then asked to use the most m operation (Swap (AI,BI)), so that the longest ascending sequence of the length of a sequence of the longest is not difficult to think of a DP is, dp[i][j][k] means that the last element of the oldest sequence is I, with a J operation, K indicates that element I has not exchanged (0 means no, 1 means there is )。 Then the transfer is dp[i][

Sweep with priority queue to get the minimum value of the sum of the values greater than and less than the median, and sweep again to get the optimal solution#include #include#include#include#include#include#defineINF 0X3FFFFFFFusing namespaceStd;typedefLong Longll;intN,c,f;typedef pairint,int>PII;Const intmaxv=1e5+ -;p Airint,int> COW[MAXV];////score,aidll LOW[MAXV],UPP[MAXV];BOOLCMP (PII a,

;intdata[4][MAXN];intvec[maxn*MAXN];int_lower_bound (int*a,intLintRintv) { intm; while(lR) {m= (l+r) >>1; if(A[M]GT;=V) R =m; ElseL = m+1; } returnL;}int_upper_bound (int*a,intLintRintv) { intm; while(lR) {m= (l+r) >>1; if(A[M]GT;V) R =m; ElseL = m+1; } returnL;}intMain () {intT scanf"%d",T); int*a = data[0], *b = data[1], *c = data[2],*d = data[3]; while(t--){ intN scanf"%d",N); for(inti =0; I ) {scanf ("%d%d%d%d", a+i,b+i,c+i,d+i); } intSZ =0; for(inti =0; I )

represents the composition of the starry sky.OutputLine: Indicates the number of stars in the current constellation.Sample Input10 5.. *.....**.**.. *****.*...*...... ****.***.. ****.***15 8**.**......*.. *.. *.**.*...*...*.*.**.*****.**...***.****.**....**.. *.*.....*****.. *****.. *....**...*.. *..*.*...*.*.*.***Sample Output27SourceProblem solving: Searching1#include 2 #definePII pair3 using namespacestd;4 intw,h;5 Chartable[1010][ -];6 Const intdir[8][2] = {7-1,0,1,0,0,-1,0,1,8-1,-1,1,1,-1,

. No. 0, at a speed of.Outputonly one row, sequentially outputting the city from 0 to D. Ensure that the quickest route is a single line.Sample Input6 15 10 1 25 680 2 30 500 5 0 1011 2 70 771 3 35 422 0 0 222 1 40 862 3 0 232 4 45 403 1 64 143 5 0 234 1 95 85 1 0 845 2 90 645 3 36 40Sample Output0 5 2) 3 1HINT"Data Range"30% n100% 2SourceSpfaIdea: Actually very obviously, here not only to record to this point of time, but also to this point of the speed, so record a two-dimensional dist[x][y] t