uva telemedicine

to perform m operations on each of these boxes for one of the x, Y reverse order of the two to the left of Y.After the operation is complete, all odd digits of the original box sequence number and;Direct simulation will definitely time out with the list in the STL also time-out can only use the array itself to simulate a doubly linked list le[i],ri[i] respectively, the box to the left of the first box and the right box in the ordinal code has a stare#include Copyright notice: This article Bo M

)); ECNT=0; while(m--){ intU,v,w; scanf"%d%d%d",u,v,W); Addedge (--u,--v,w); Addedge (V,U,W); } Dijkstra (--s,--e,d1); Dijkstra (E,S,D2); intPick =-1, p2, ans =D1[e]; intK scanf"%d",j); for(inti =0; i ){ intU,v,w; scanf"%d%d%d",u,v,W); if(D1[--u] + W v]) {ans= D1[u] + W +D2[v]; Pick= u; P2 =v; }Else if(D2[u] + W D1[v]) {ans= D2[u] + W +D1[v]; Pick= V; P2 =u; } } intu; if(~pick) {Stackint>Stk; U=pick; Stk.push (U); while(U! =S) {

(inti =1; I 0; for(intU =0; U ){ intV0 = Sccno[u]; wei[v0]++; for(inti = Head[u]; ~i; i =Nxt[i]) { intV1 =Sccno[to[i]]; if(V0! = v1) G[v0].push_back (v1), deg[v1]++; } }}intDP[MAXN];inttopo () {memset (DP,-1,sizeof(DP)); Queueint>Q; for(inti =1; I ){ if(!deg[i]) Q.push (i), dp[i] =Wei[i]; } while(Q.size ()) {intU =Q.front (); Q.pop (); for(inti =0; I int) G[u].size (); i++){ intv =G[u][i]; DP[V]= Max (dp[v],dp[u]+Wei[v]); if(--deg[v] = =0) Q.push (v

Test instructions: Similar to the 8 queen question, except that each lattice on the board has a score and the Queen's position is the one that can be obtained. Or the rule of the 8 Queen's question, and finally get the best possible score.Idea: 8 Queen problem solution, after 8 Queens to judge the score on the line.In the If statement judgment content that piece unexpectedly still write wrong once, not satisfied ~You can also use vis[3][2*8]; array markers, and then quicklyCode:#include

Test instructions: n planets, given a one-dimensional coordinate of each point, can delete m points, so that all remaining points to the center of the remaining points of the sum of the square and the smallest, to find the minimum value;Idea: Ruler method, each maintenance n-m point; Long Long is read in with%LLD, but codeblocks compile will cause problems, so use cin>>;#include #include#include#include#includeusing namespacestd;intt,n,m;Long Longs1,s2;Long Longnum[50500];Doublesum,mid,cnt;intMa

Title: Give a string of O and x (length 1-80), statistical score.The score for each o is the current number of consecutive o, for example, the ooxxoxxooo score is 1+2+0+0+1+0+0+1+2+3Problem-solving ideas: Using a variable term records the current O scores, if there is O, then term+1, if X appears, then term=0;Then use a sum to record the sum, no time add term can/*UVa 1585 Score---water problem*/#include#includeConst intMAXN = -;intMain () {intT; Char

Title: Two people take turns in the n*n of the parallelogram lattice into black and white two-color pieces,If the black side can give the creation of a continuous line from 1~n lines then black wins, otherwise white wins.Analysis: Graph theory, search. Use DFS or FloodFill solution to find solutions that can reach the low end from the top.Description: Target 600 question ╮(╯▽╰)╭.#include UVa 260-il gioco dell ' X

Manacher (); - getd (); About for(intI=0; i) $ { - for(intj=0; J) -Dp[i]=min (Dp[i], (d[i][j]==0?1:d p[d[i][j]-1]+1)); - d[i].clear (); A } +printf"%d\n", dp[len-1]); the } -}Then see LRJ code, Memory search, some violence. Before I heard that the VIS does not have to be initialized every time, this is the first time to see how to write.int Kase; { .... if return p[i][j]; = Kase; ....} int Main () { ... .. 0 sizeof (Vis)); for

2 3 | 6 7 5...4 3 1 2 | 7 6 54 3 2 1 | 7 6 5So each number needs to be exchanged for (even) 0, ..., (N/2)-1, (N/2) -1,...,2,1,0(odd) 0, ..., (N/2), (N/2) -1,...,2,1,0Sum can be."Code": The code may be slightly cumbersome. I can simplify it.1#include 2#include 3 using namespacestd;4 intMain ()5 {6 intT scanf"%d", T);7 while(t--)8 {9 intN, ans =0; scanf"%d", n);Ten if(n%2) One { A intnum = n/2; -Ans = (1+ (num-1)) * (num-1); -Ans + =num; the

Topic PortalTest instructions: Professors give students classes, there are n themes, each subject has TI time, the class has two restrictions: 1. Each topic can only be finished in one lesson, can not be separated in a multi-class session; 2. Must be in the order of the topic, not upset. A lesson l time, if the early class, according to the amount of time, students have dissatisfaction. Ask at least a few lessons to finish the topic, if the output of a variety of scenarios with the least degree

][MAXN]; in intCOU[MAXN][MAXN]; - intN, M, K; to voidinit () { + for(inti =1; I ) - for(intj =2; J ) theCOU[I][J] + = cou[i][j-1]; * for(inti =2; I ) $ for(intj =1; J )Panax NotoginsengCOU[I][J] + = cou[i-1][j]; - } the intCount_num (intR1,intC1,intR2,intC2) { + intA = COU[R2][C1-1]; A intb = cou[r1-1][C2]; the intc = cou[r1-1][C1-1]; + returnCOU[R2][C2]-A-B +C; - } $ intDP (intR1,intC1,intR2,intC2) { $ int cnt =DP[R1][C1][R2][C2]; - if(cnt = I

Topic PortalTest instructions: Test instructions is difficult to understand, is a thief in M days from the city 1 flew to the city N minimum cost, entered is each city flew to other cities of the flight.Analysis: Dp[i][j] represents the thief's first day in the city J's minimum cost. State transfer equation: dp[i][j] = min (Dp[i-1][k] + cost[k][j][t%day]) T represents the cost of an aircraft flying to J on T-day KHarvest:Code:/************************************************* author:running_time

; + if(IS[i]; A if(IB.s[i]; theC.s.push_back (%BASE); +g=x/BASE; - } $ returnC; $ } - BOOL operatorConstbiginteger b)Const{ - if(S.size ()!=b.s.size ())returnS.size () b.s.size (); the for(intI=s.size ()-1; i>=0; i--){ - if(S[i]!=b.s[i])returnS[i]>B.s[i];Wuyi } the return false; - } Wu BOOL operator!=(Constbiginteger b)Const{ - returnb This|| * Thisb; About } $ }; -ostreamoperator out

Main topic: There are n items, each item has m characteristics, each item may have or not each feature, now assume that an item, by asking certain characteristics to determine the item, ask the maximum number of times you can determine the item.Different strategies for further questioning may be taken after each inquiry, depending on the answer.With D[S][S0], the answer is S0 (that is, the item has S0 in s), at least the number of times it needs to be asked. Enumerates the feature completion rec

I wrote the code myself.#include #include using namespace Std;int main (){int N,a[100],b[100],q,flag;int k=1;while (CIN>>N>>Q){for (int i=0;i{cin>>a[i];}for (int j=0;j{cin>>b[j];}Sort (a,a+n);for (int j=0;j{flag=0;coutfor (int i=0;i{if (A[i]==b[j]){coutflag=1;Break}}if (flag==0){cout}}}return 0;}The code written on the book#include #include using namespace Std;int main (){int n,a[100],q,x;int k=1;while (CINGT;GT;NGT;GT;Q){for (int i=0;i{cin>>a[i];}Sort (a,a+n);coutwhile (q--){cin>>x;int P=lower_

intDr[] = {0,0, -1,1, -1, -1,1,1};Const intDc[] = {-1,1,0,0, -1,1, -1,1};Const intMOD = 1e9 +7;Const DoublePI = ACOs (-1.0);Const DoubleEPS = 1e-8;Const intMAXN =1000000+Ten;Const intMaxt =10000+Ten;using namespacestd;CharS[maxn];mapint>MP;intCnt//Total leaf nodesvoidDfsintStintEtintDeep ) { if(S[st] = ='['){ intTMP =0; for(inti = st +1; I i) { if(S[i] = ='[') ++tmp; if(S[i] = =']') --tmp; if(!tmp s[i] = =',') {DFS (St+1I1, Deep +1); DFS (i+1, et-1, Deep +1); } }

Test instructions: Given a string, it can be divided into several palindrome strings at least.Analysis: Dp[i] means that the first I characters can be divided into several palindrome, dp[i] = min{1 + dp[j-1] | J-i is a palindrome}.The code is as follows:#pragma COMMENT (linker, "/stack:1024000000,1024000000") #include UVa 11584 Partitioning by palindromes (simple DP)

Split the coin.01 Backpack for SUM/2, SUM-2*DP[SUM/2]#include #include#includeusing namespacestd;intvalue[100000],dp[100000];intMain () {intN,m,sum,sum1; CIN>>N; while(n--) {cin>>m; Sum=0; for(intI=1; i) {cin>>Value[i]; Sum+=Value[i]; } sum1=sum/2; Memset (DP,0,sizeof(DP)); for(intI=1; i) for(intj=sum1;j>=value[i];j--) Dp[j]=max (dp[j],dp[j-value[i]]+Value[i]); cout2*dp[sum1]Endl; } return 0;}UVA 562 Dividing coins (01 backpack)

Path Recording method: If DP[J-VALUE[I]]+VALUE[I]>DP[J] Description took this thing, the mark is 1,For loop flag, found to be 1, printed out, and the capacity of the backpack to reduce, and then in the secondary capacity to look for signs;#include #include#includeusing namespacestd;intvalue[ -],dp[10001],s[ -][10001];intMain () {intn,m; while(cin>>m) {cin>>N; for(intI=1; i) Cin>>Value[i]; Memset (DP,0,sizeof(DP)); memset (s),0,sizeof(s)); for(intI=1; i) for(intj=m;j>=value[i];j--)

Title: PortalTest instructions: To find the number of n greatest common divisor, violence will not time out, the difficulty in no number control input.Solution: Enter it in a special way.#include #include#include#includeusing namespacestd;intgcdintAintb) { if(b==0)returnA; returnGCD (b,a%b);}intMain () {intT; CIN>>T; GetChar (); while(t--) { CharC; intdata[ the],cnt=0; while((C=getchar ())! ='\ n') { if(c>='0'c'9') {ungetc (C,stdin);//returns the character C t