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The problem is to give you the number of n, and then give you M T and r of the sequence T value is 1 or 2, when the 1 is the number of n the first number of R is arranged in ascending order, when the number of the first n is 2 is a descending order, ask you this m operation sequence is how much? The first thing to be sure is that if there is an operation R greater than the previous R then the previous operation can be ignored, according to this property we can change the m operation to follow th

Set D (I, j) to the minimum length of the tree connecting the point I to the J point, there is a state transition equation:D (i, j) = min{D (i, K) + D (k + 1, j) + P[k].y-p[j].y + p[k+1].x-p[i].x}Then we use the quadrilateral inequality to optimize it.1#include 2#include 3#include 4#include 5#include 6 #defineMP Make_pair7 #defineX First8 #defineY Second9 using namespacestd;Ten Onetypedef pairint,int>PII; A - Const intMAXN = ++Ten; - Const intINF =0

, A, B, C, root=1, Par[lg_n_max][n_max],//Par[m][u] Indicates the (2 ^ m) th ancestor of UDep[n_max], Disrt[n_max];Charinstr[8];vectorint>G[n_max];typedef pairint,int>Pii;mapint>Len;voidinit () {clr (PAR); CLR (DEP); CLR (DISRT); for(inti =1; I i) g[i].clear (); Len.clear (); Lg_n=0; int_n =1; while(_n N) {_n1; ++Lg_n; } ++Lg_n;}voidAddedge (intAintBintc) {G[a].push_back (b); G[b].push_back (a); Len[make_pair (A, b)]=C; Len[make_pa

double quotes). Otherwise, the output satisfies the nature of the number of squares.Sample Input5 5--+-+..| #...| ##S-+-t####.Sample Output2Sample DescriptionIf you mark an "X" on the map with a square of the satisfying nature , the map looks like this:--+-+..| #X..| ##S-+-t# # #XProblem solving: Direct search ...1#include 2#include 3#include 4#include 5 #definePII pair6 using namespacestd;7 Const intMAXN = -;8 CharMP[MAXN][MAXN];9 BOOLva[maxn][maxn],vb[maxn][maxn],e[maxn*maxn][maxn*MAXN];Ten i

A very challenging topic of direct violence to build the words of No doubt O (n^2) will be tle each layer of virtual a point and will let no point of the layer can also connect the pastRefer to the method of Kuangbin chicory two virtual points per layer n+i*2-1 is the entrance n+i*2 is the exit and then the one-way side.VA once, because MAXN should be twice times bigger than the data. Accidentally ignored as far as MAXM directly to 1e7#include #includestring>#include#include#include#include#incl

Problem Descriptionjohn has several lines. The lines is covered on the X axis. Let's a point which are covered by the most lines. John wants to know how many lines cover A.Inputthe first line contains a single integerT(1≤t≤) (The data forN> Less than cases), indicating the number of test cases.Each test case begins with an integern(1≤n≤5) , indicating the number of lines.Next N lines contains and integersXi andyi(1≤Xi≤Yi≤9) , describing a line. Outputfor each case, output an in

number of squares that can be coated with K is a+n-b, The number of squares that can only be painted in K-1 is x*n-(a+n-b); the coloring scheme for x lines is temp=k^ (a+n-b) * (K-1) ^ (x*n-(A+n-b)) and if temp=r, temp is the answerAgain consider the case of line x+1, if the X Act is not a color of the lattice, then the next row of its adjacent lattice coloring scheme is K, otherwise K-1, the number of K-1 can be painted C, then temp=temp*k^c* (K-1) ^ (n-c), if temp=r, then temp is the answerTh

http://www.lydsy.com/JudgeOnline/problem.php?id=2534Given the string s, the number of substrings in the form of ABA, where the length of B is L.Consider the two starting position x, Y (x1.y>x+l2.LCP (x, y) ≥y-x-lWe press the height from the large to the small enumeration, so that LCP becomes the current height, the two sets of statistical answers and merge. The statistical answer is to enumerate the points within the smaller collection of size as x or Y. Can be maintained with a balance tree or

Topic Links: http://acm.hdu.edu.cn/showproblem.php?pid=1874Thinking Analysis: The problem is given a graph, the starting point and the endpoint, the minimum distance from the starting point to the endpoint is required to find out;The shortest-circuit length from the starting point to all other points is calculated using the Dijkstra algorithm, if the shortest-circuit length is int_max, indicating that no path is connected from the starting point to the point;The code is as follows:#include #incl

-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,- fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,- fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,ofast,inline,- Fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2 ", 3)#include#include#include#include#include#include#include#include#include#includestring>#include#include#include#include#include#include#includeSet>#i

(that is, e[pre[ E[t].to]].length > E[t].length), the edge recorded by the Pity Dorado (pre[e[t].to] = t) is updated.Code: #include #include #include #include #include #define INF 0X7FFFFFFFFFFFFFFF #define PII Pair using namespace Std; Long Long s,n,m,pre[300002],tag[600002],u[600002],v[600002],next[600002],first[300002]; Long Long sum,d[300002],w[600002]; priority_queue BOOL vis[300002]; void input () { Long Long

During the course of learning network technology, we often see that Linux and Linux operating systems are increasingly popular with computer users, you may encounter hardware problems related to the installation and recognition of the Linux operating system. Here we will introduce how to install the Linux operating system and how to understand the Supported Hardware, I will share it with you here. The CPU level must be PII, PIII, P4, K7, or K8 or abov

operations so that the variables in the parent process can be accessed in a local process. Therefore, it is necessary to remove the local process and then use parameters to pass the required variables. Process Parameters In Delphi, the default call convention is register. In this mode, eax, ECx, and EDX can be used to pass parameters. Therefore, there are generally no more than three parameters in the process. In object type methods, we recommend that you set no more than two parameters bec

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4417 Your sister's CIN cout scanf printf. It's far behind. Cin cout will wa But scanf printf will be AC .. I learned about the magic operation of eaual_range .. It is a line segment tree, and the vector is very powerful .. View code # Include # Include # Include # Include # Define Lson L, M, root # Define Rson m + 1, R, root Using Namespace STD; Const Int Maxn = 100001 ; Typedef Vector Int > VI; typedef pair

used 100, 64, and feet)Htqfp: Heat Sink quard flatpacks with flat encapsulation around heat sinkCqfp: ceramic quad flatpack ceramic four-sided flat encapsulation (ceramic carrier, and GeneralQFP is different in that its pin is split from the inside to the peripheral parallel to the chip edge, shapeIt looks like four "horns ")LCC: leadless chip carrier no-lead carrierBGA: Ball Grid Array EncapsulationTCP: tape carrier package on-load encapsulation (such as the encapsulation of black glue protect

Weighted and check Set#include #include#include#include#defineMAXN 20000+10#definePII pairusing namespacestd;intFA[MAXN];intD[MAXN];intN;intAbs (intx) { return(x>0? x:-x);} PII Find (intx) { if(fa[x]!=x) {PII T=find (fa[x]); FA[X]=T.first; D[X]+=T.second; } returnMake_pair (fa[x],d[x]);}voidsolve () {scanf ("%d",N); for(intI=1; i) {Fa[i]=i,d[i]=0; } Charch[9]={0}; while(SCANF ("%s", ch) ch[0]!

with set)Then use a tree-like array to maintain how many ordinary houses are affected.#include #include#include#includeSet>#defineFi first#defineSe Secondusing namespaceStd;typedef pairint,int>PII;Const intMAXN = 1e5 + -;intc[maxn*8], f[maxn][2];intDEEP[MAXN], p[maxn][ -], Col[maxn];vectorint>G[MAXN];Set2];intn, x, tot, q;Charstr[Ten]; PII ans;voidModify (intXints) { for(; x 2*n; x + = x (-X)) c[x] + =

Spit Bubble Topic Link:https://www.nowcoder.com/acm/contest/74/ATopic:Ideas:This topic was stuck to me for a long time, today morning training encountered a problem, one eye on the idea of a stack simulation, and then back to the problem. The problem is very simple, look at the code basically can understand, do not explain.The code is implemented as follows:1#include Set>2#include 3#include 4#include 5#include 6#include 7#include 8#include string>9#include Ten#include One#include A#include -#

and take defensive measures, it will be mistaken, encounter a variety of security risks, such as confidential data leakage, spyware intrusion, etc. Therefore, it is urgent for enterprises to identify these harmful mails effectively. Exchange 2013 Technology Highlights: Data Loss Prevention (DLP) and legal compliance Exchange 2013 provides a range of accurate, automated, and centrally managed data loss prevention capabilities that make it easier for you to monitor and protect sensitive and co

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=4417Title: give you the number of N, subscript 0 to N-1,m query, ask the query interval [l,r] is less than or equal to the number of x is how many.It's funny when I write. or write too little.We query by x from small to large, then find the point on the interval, if it is less or equal, insert it, and see how many points are inserted in the interval.The points can also be sorted.See the code:1#include 2#include 3#include 4#include 5#include 6