nex 5tl

Number SequenceTime limit:10000/5000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 15052 Accepted Submission (s): 6597Problem Descriptiongiven-sequences of numbers:a[1], a[2], ..., a[n], and b[1], b[2], ..., b[m] (1 Inputthe first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is the numbers n and M (1 Outputfor Each test case, you should the output one line which only contain K described above. If N

of water over.In this topic also learned to customize the structure of the priority queue used. Refer to the blog http://blog.csdn.net/dooder_daodao/article/details/5761550It is necessary to define a CMP struct as a comparison when initializing the priority queue of the struct body.#include #include#include#include#includestring>#include#includeusing namespacestd;Const intinf=0x3f3f3f3f;Const intmaxn=1100;Const intmaxm=110000;intVAL[MAXN];intN,m,u,v,w,q,cc,ss,ee,tot;intdp[maxn][ the];intHEAD[MA

,root); Splay (x,ch[root][0]); ch[x][1]=z; fa[z]=x; update (x); update (root) ;Wuyi } the voidEraseintz) { - intX=pre (z), Y=nex (z); Splay (Y,root); Splay (x,ch[root][0]); ch[x][1]=0; fa[z]=0; update (x); update (root) ; Wu } - voidTopints) { About intX=pos[s]; Erase (x);intY=querynum (Root,1), z=Nex (y); add (y,z,x); $ } - voidBottomints) { - intX=pos[s]; Erase (x);intY=querynum (root,sz[root]

the second question: there are several ways to add Lotus leaf.First, what if the shortest path number to the end point is required? Just judge when the SPFA is slack:ifvalue == dist[nex]) { tot[nex] += tot[cur]; push();}ifvalue value; tot[nex] = tot[nex]; push();}But this is the "Shortest path number", not

1#include 2 using namespacestd;3 4 Const intMAXN = 1e6 +Ten;5 intNEX[MAXN];6 intS[MAXN], T[MAXN];7 8 voidGet_nex (intlm) {9 inti =0, j =-1; nex[0] = -1;Ten while(I lm) { One if(j = =-1|| T[J] = =T[i]) { Ai++; j + +; Nex[i] =J; - } - Elsej =Nex[j]; the } - } - - intKMP (intlnintlm) { + Get_nex (LM); - inti

root node is not evaluated at the time of calculation.Code:1#include 2 Const intm=100005;3 #defineSwap (x, y) t=x,x=y,y=t4 intt,cnt,ans,v[m1],dp[m],dep[m],hea[m1],nex[m1],p[m][ -];5 6 intRead ()7 {8 intx=0;CharCh=GetChar ();9 while(ch -|| Ch> $) ch=GetChar ();Ten while(ch> - ch -) x= (x1) + (x3) +ch- -, ch=GetChar (); One returnx; A } - - voidAddintXintY) {v[++cnt]=y,nex[cnt]=hea[x],hea[

#include #include#include#includeusing namespacestd;structe{intnext,c;}; Vector101];BOOLmark[101];intdis[101];intMain () {intn,m; while(Cin>>n>>m n!=0 m!=0){ intA,b,c; e temp; //Initialize for(intI=1; i) { edge[i].clear (); Dis[i]=-1; Mark[i]=false; } dis[1]=0; mark[1]=true; while(m--) {cin>>a>>b>>C; TEMP.C=C; Temp.next=A; Edge[b].push_back (temp); Temp.next=b; Edge[a].push_back (temp); } intnewp=1; for(intI=1; i){ for(intj=0

Node {int l,r;ll sum;int lazy;}; Node Tree_even[n * 4 + 55],tree_odd[n * 4 + 55];typedef struct NODE {int fro,to;int nex;}; NODE edge[2 * N + 55];int tot;int cnt;void Add (int u,int v) {Edge[tot].fro = U;edge[tot].to = V;edge[tot].nex = Head[u];he Ad[u] = tot++;} void Dfs (int u,int pre,int d) {Le[u] = ++cnt;for (int i=head[u];i!=-1;i=edge[i].nex) {int v = edge[

。Funny Sao Action Watch instance-command line clockwatch -t -n1 "date +%T|toilet"Enhanced Clockwatch -t -n1 "date +%T|toilet -fbigmono12"To record terminal operations:script scriptreplay录制命令:script -t 2>example.time -a example.txt播放命令:scriptreplay example.time example.txt两个文件可以随便命名 例如timing.log和output.session解释:1. -t是把时间数据输出到标准错误(standard error),这里使用 2>example.time 把数据重定向到example.time这个文件当中.2. -a 选项则指定输出录制的文件.3. 在录制过程中,使用 exit 结束录制过程.Hacker Empire Terminalsudo apt install cmatrix选项-a :异步滚动（默认）-

http://poj.org/problem?id=3041There are K asteroids in the N*n grid, where the asteroid I is (RI,CI), and now there is a powerful weapon that can blast a whole row or a whole array of asteroids to ashes with a luminous speed, and to use this weapon to destroy all asteroids requires at least a few luminous beams.Mainly the composition, will each line into a point, constitute a set of 1, each column also as a point, the composition of the set 2, each obstacle position coordinates will set 1 and se

the longest route to the hotel. Note that there must be no forks on the road and the road is backward. Direct dfs + pruning on the line, the graph is saved with set... Look at others using a one-dimensional array. Code: /** Author: illuz * Blog: http://blog.csdn.net/hcbbt* File: b.cpp* Create Date: 2013-10-02 00:17:29* Descripton: b */#include #include #include #include #include using namespace std;#define rep(i, n) for (int

:#include #include #include #include using namespace STD;inthead[60005],val[160005];inttot,to[160005],nex[160005];intNintAA,BB,CC; Queueint>Qintd[60005];intvis[60005];voidSPFA (intx) {memset(d,-1,sizeof(d)); d[x]=0; vis[x]=1; Q.push (x); while(!q.empty ()) {intNow=q.front (); Q.pop (); vis[now]=0; for(inti=head[now];i!=-1; I=nex[i]) {intV=to[i];if(d[v]==-1|| D[v]if(!vis[v]) {vis[v]=1;

characters in "AGCT", which is the DNA segments causing inherited disease.The last line of the that test case is a non-empty string of length not greater than-containing only characters in "AGCT", Which is the DNA to be repaired.The last test was followed by a line containing one zeros.Outputfor each test case, print a line containing the "test Case number" (beginning with 1) followed by theNumber of characters which need to be changed. If it ' s impossible to repair the given DNA, print-1.Samp