uva timecard

=Q.front (); Q.pop (); the if(!u->vis)return false;//**????? AboutAns.push_back (u->v); the if(U->left! =NULL) theQ.push (u->Left ); the if(U->right! =NULL) +Q.push (u->Right ); - } the return true;Bayi } the voidShow () the { - intt=0; - for(intI=0; I) the { the if(i) printf (" "); theprintf"%d", Ans[i]); the } -printf"\ n"); the ans.clear (); the } the intMain ()94 { the while(Input ()) the { the BFS (ans);98 if(error|

(); - if(OK) { -Rep (I,0, N-1){ the if(A[i] PO) { the if(solver.mark[i*2]) printf ("b\n"); the Elseprintf"c\n"); the } - Else { the if(solver.mark[i*2]) printf ("a\n"); the Elseprintf"c\n"); the }94 } the } the returnOK; the }98 About intMain () - {101 while(SCANF ("%d%d", n,m) = =2 N m) {102 intsum =0;103Rep (I,0, N-1) {104A[i] =read (); theSum + =A[i];1

10140-prime DistanceTopic linksTest instructions: Find the nearest and farthest prime pairs in the [l,r] interval.Train of thought: The prime number hits the table, hits to sqrt (Max) to be possible, then uses the big table to sift the prime numbers, because [L, R] is up to 100W, therefore can go through again, finds the answer. Note that 1 of the situation, at the beginning did not infer 1, the result WACode:#include UVA 10140-prime Distance (number

no effect)Input m, there are m electrical appliances, the following m line of two words per line is the name of the appliance and the plug type (the same 24 letter word inside no space, two words separated by space)Input k, there are K converters, the following K lines per line of two words, respectively, the converter's entry type and plug typeThe number of each converter is infinite, and the converter itself can be connected to the converterTo ask you, so that the largest number of appliances

coordinates. The second point, for example, has the X coordinate 2, and the Y coordinate 3. The result for each data set was the tour length, (6.47 for the first data set in the given example).Sample Input31 12 33 141 12 33 14 2Sample Output6.477.89It's a classic question. The coordinates of n points on a given plane (in ascending order of X)Your task is to design a route that departs from the far left, to the far right and back to the left, and every point happens once. Ask for the minimum cos

The main idea: to place the Queen on a chess plate, so that the target is all occupied, the minimum number of Queens.Topic Analysis: Iterate to deepen the search, otherwise time out.Tips: Use Vis[0][r], vis[1][c], vis[2][r+c], Vis[c-r+n] respectively flag (R,C) position corresponding to the row, column, main, sub-diagonal is not occupied (see "Introduction Classic (2nd edition)" P193), where n represents any number that is larger (greater than or equal) than the number of rows and columns.The co

]*1+ weight[7]*1; - returnsqrt (sum_mv*sum_mv+sum_mh*sum_mh); - } the - intMain () - { - + while(1) - { + BOOLOK =false; A for(inti =0; I 8; i++) at { -scanf"%d", weight[i]); - if(Weight[i]) OK =true; - } - if(!ok) Break; - inSort (weight, weight+8); - DoubleAns =INF; to while(Next_permutation (weight, weight+8)) + { -Ans =min (ans, solve ()); the } *printf"%.3lf\n", ans); $ }Panax Notoginse

UVA-147Dollars Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld %llu Submit StatusDescriptionNew Zealand currency consists of $, $, $ $, $ $, $ notes and $ $, $50c, 20c, 10c and 5c coins. Write a program that would determine, for any given amount, in how many ways that amount may is made up. Changing the order of listing does not increase the count. Thus 20c May is made up in 4 ways:1 20c,

Problem A:expanding RodsWhen a thin rod of length l was heated n degrees, it expands to a new length l ' = (1+n*c) *l, where C is the coefficient of heat expansion.When a thin rod was mounted on both solid walls and then heated, it expands and takes the shape of a circular segment, the O Riginal Rod being the chord of the segment.Your task is to compute the distance by which the center of the rod is displaced.The input contains multiple lines. Each line of input contains three non-negative numbe

Title: give you some data structure operations, to determine whether the data structure is a stack, queue, or priority queue.Analysis: Primary DS, simulation. Construct three structures, simulate directly and then judge by the results.Description: The priority queue is implemented with the maximum heap.#include UVa 11995-i Can Guess the Data structure!

-called error vector (12, 13, 14 lines of code), and the size of the cross product is proportional to the gyroscope's integral error, Using this error vector to correct the next gyroscope data can achieve the effect of correcting gyro data error with accelerometer.Then is the px4firmware source in the accelerometer data correction gyroscope data, the source code is as follows.1.//Accelerometer correction 2.//Project ' k ' unit vector of Earth frame to body frame 3.//vectorThe,vectorthe total a

)! =EOF) + { - intflag=1; +memset (D,0,sizeof(d)); Amemset (PRE,0,sizeof(pre)); at for(i=0; i10010; i++) -pre[i]=i; - for(i=1; i) - { -scanf"%d%d",u,v); -d[u]++; ind[v]++; - Unionroot (u,v); to } + - intRoot=find (0); the * if(m==0|| n==0) flag=0;//n=0 can not form a circuit, m=0 time can not $ Panax Notoginseng for(i=0; i){ - if(d[i]!=0){//determine if this point is an isola

starting and ending node of the edge (nodes is labeled from 0 to N -1).The test case was terminated by n more integerss0,..., sn-1 (one per line), WHERE sI represents the number of stones that is initially placed on node i ( 0si10 XX). Each test case was followed by a blank line, and input was terminated by a line containing '0 0' which should not Be processed.OutputFor each test case output a single line with either the word ' First' If the first player would win, or the wo

DescriptionProblem AExpression bracketinginput: standard inputoutput: standard outputTime Limit: 1 secondMemory Limit: MBInthis problem you'll have to find on how many waysn letters can being bracketed so that's the bracketing is Non-bin Arybracketing. For example4 lettershave possible bracketing:xxxx, (xx) xx, x (xx) x, XX (xx), (XXX) x, x (XXX), ((XX) x) x, (x (XX)) x, (XX) (XX), X ((XX) x), X (x (XX)) . Of these the first sixbracketing is not binary. Given the number of letters you'll have a

Topic Test Instructions:There are n different numbers, you can ask A, b which is big, will get the answer, and then ask at least a few times to ensure that the largest and the second biggest number can be selected.Analysis:N number, first in Daleitai way, 22 compared to the largest, n-1 times, and then by the largest PK down the number, compared to the largest, there are log (n) number, need to log (n)-1 times, pay attention to rounding up.#include #include #include using namespace Std;int main

1584-circular SequenceSome DNA sequences exist in circular forms as in the following figure, which shows a circular sequence ' cgagtcagct ', that Was, the last symbol ' T ' in ' CGAGTCAGCT ' was connected to the first symbol ' C '. We always read a circular sequence in the clockwise direction.Since It is isn't easy to store a circular sequence in a computer as it's, we decided to store it as a linear sequence. However, there can be many linear sequences that is obtained from a circular sequence

The problem can be transformed into a lawn where the boundaries are completely covered. Such a circle is replaced by a line segment.Greedy, from the select as few line segments to cover the interval, according to the line to the left end of the order, record a current coverage of the position cur,From the left end of the endpoint is less than or equal to cur select one of the largest right endpoint as the selection of the interval, if there is no qualifying, the description is not possible to co

Ask for a minimum bandwidth arrangement and minimum bandwidth. Bandwidth is the maximum distance from a letter to its adjacent letter.Topic Analysis: In the process of recursive generation of the whole arrangement of pruning, pruning scheme is still two. First, the current solution is inferior to the optimal optimization, minus; second, the prediction of the ideal solution does not need to optimize the optimization, minus. The ideal solution is to take all the letters that are adjacent to the le

, Num[i]) SFI (a), F[i].push_back (a); } Repu (I,1, n +1) scanf ("%d%d", xx[i], Yy[i]); Repu (i,1, n +1) Repu (J,1+ I, n +1) {W[m]= (Xx[i]-xx[j]) * (Xx[i]-xx[j]) + (Yy[i]-yy[j]) * (Yy[i]-Yy[j]); U[M]=i; V[M]=J; //printf ("%d%d%d%lf\n", M, I, J, W[m]);m++; } Repu (I,1, n +1) P[i] =i; Repu (i,0, m) R[i] =i; Sort (R, R+m, CMP); LL Minn=Kruskal1 (); Sort (KB, KB+ N-1, CMP); //printf ("%lld\n", Minn); intLim =1Q; ll cc; Repu (i,1, Lim) {CC=0; Repu (J,1, n +1) P[j] =J; Repu (J,0, q)if((1

Test instructionsGive the diagram, rotate in the direction of the a-h, so that the median 8 grid numbers are the same. The minimum number of rotations is required and the sequence of operation sequences is as small as possible.Analysis:Save 24 squares In one-dimensional array. A two-dimensional array represents 7 squares in each direction. Ida* pruning is the number of letters that repeat the most letters in 8-8 squares >maxd.Code:　　#include #include #include #include using namespace Std;int a[2