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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

DescriptionThe farmer John's farm suffered an earthquake. Some of the cattle sheds were damaged, but fortunately, all the road stalls between the cattle can be used. FJ's Farm has P (1 Input* Line 1th: Three spaces separate number: P, C, and N* 2nd. C+1 lines: Two spaces per line: a_i and B_i * c+2..c+n+1 line: One number per line: Report_jOutput* Line 1th: Number of cows that cannot return to the farm at least (including damaged barns).Sample Input4 3 11 22 33 43Sample Output3HINTThe Barn 2 was

: +End_index = Self.total_count + 1 AStart_index = self.total_count-self.pager_num + 1 the + ifSelf.current_page = = 1: -Prev ='' $ Else: $Prev =''% (Base_url, self.current_page-1,) - page_list.append (prev) - the forIinchRange (int (start_index), int (end_index)): - ifi = =Self.current_page:Wuyitemp =''%(Base_url, I, i) the Else: -temp =''%(Base_url, I, i) Wu page_list.append (temp) - About ifSelf.current_page = =Self.total_count: $

Test instructions: To a city to get money, to leave to spend money. There is cash at the beginning. The city is circular, ask from which start, can on the way any time money >=0;A start pointer I, a trailing pointer J. Indicates an interval. If the conditional ++j is met and the proceeds are added to the sum (the earnings may be negative). Non-conforming is ++i, and the return from sum proceeds until the sum>=0; interval length is n, the position of I is the result.Proof of correctness: Suppose

2#include 3#include Set>4#include 5#include 6#include 7#include 8#include 9#include string>Ten#include One#include A using namespacestd; - #defineINF 0x3f3f3f3f - the Charp[100010],s[100010]; - intnex[100010]; - intLens; - + void Get(Char*p) - { + intplen=strlen (p); Anex[0]=-1; at intk=-1, j=0; - while(J Plen) { - if(k==-1|| P[J] = =P[k]) { -++J; -++K; - if(P[j]! =P[k]) innex[j]=K; - Else tonex[j]=Nex

];isum; - } -l[sum]=r[sum-1]+1; r[sum]=en; - for(intI=l[sum];isum; in } - voidInsert (Const intAMP;X) {++b[x]; + +sumv[num[x]];} to voidDelete (Const intAMP;X) {--b[x];--sumv[num[x]];} + intNext (Const intx) - { the for(inti=x;iif(B[i])returni; * for(inti=num[x]+1; iif(Sumv[i]) $ for(intj=l[i];j)Panax Notoginseng if(B[j])returnJ; - return-1; the } + intMain () A { the R (n); R (m); + for(intI=1; ii) a[i]. Read (); - for(inti=n+1; ii) $ { $

integer for each test case:the lowest intersection number @ which a new fire station SH Ould is built so as to minimize the maximum distance from any intersection to the nearest fire station. Sample input 1 621 2 102 3 103 4 104 5 105 6 106 1 10 Sample output 5 Source University of Waterloo Local Contest 199 Uploaded by Zhang Yunzun Helples

Iteration:Code:/*** Definition for singly-linked list.* Public class ListNode {* int val;* ListNode Next;* ListNode (int x) {val = x;}* }*/public class Solution {Public ListNode reverselist (ListNode head) {if (head = = NULL) return null;ListNode tail = head.next;ListNode cur = head;while (tail!= null) {ListNode NEX = Tail.next;Tail.next = head;Cur.next = NEX;head = tail;tail =

1OutputNOInput4 7.. x.xx.. Xx.. X.X... X..X......2 21 6OutputYESTitle, every step, go through that place into a broken ice, if you go again will fall, you can not beat in situ to make the ice change state, and your final goal is not just to find the destination, and to make it into X and then re-enterOriginally I wanted to think about the situation, such as the destination is X or. , and then judge the condition is changed, and then count, mark is not a problem, and consider from the origin to

Struct node {int data; node * Next;}; // create a linked list and enter a number as the input. If the input is 0, the linked list ends (0 is excluded) node * creat () {node * head, * P, * s; int X, cycle = 1; head = (node *) malloc (sizeof (node); P = head; while (cycle) {cout // Calculate the length of the linked list int computlength (node * head) {node * P; int n = 0; If (Head = NULL) {return 0;} p = head; while (P! = NULL) {n ++; P = p-> next;} return N ;}// Print the data void printlist (no

Http://poj.org/problem? Id = 3256 Store the paths of N farms in an adjacent table, record the initial number of cattle in each farm, and query the number and number of cattle in each farm along the path DFS. Code:# Include # Include Using namespace STD; Int num [1001]; Int sum [1001]; Int head [1001]; Bool vis [1001]; Int X; Struct past { Int V, NEX; } Edge [2, 10001]; Void addedge (int u, int v ){ Edge [X]. V = V; Edge [X].

) scanf ("% s",)# Define SLD (a) scanf ("% LLD", )# Define PF (a) printf ("% d \ n",)# Define PPF (a, B) printf ("% d \ n", a, B)# Define SZ (ARR) (INT) A. Size ()# Define swap (A, B) A = a xor B; B = a xor B; A = a xor B;# Define read freopen ("in.txt", "r", stdin)# Define write freopen ("out.txt", "W", stdout)# Define max 1 # Define ESP 1e-5# Define lson L, M, RT # Define rson m + 1, R, RT Template Template Template Template Template Const int maxn = 55555;Int Loy [maxn Int L [maxn], R [maxn];

HDU 4411 Arrest fee stream Question link: Click the open link Question: Given n + 1 vertex ([0, n]) an undirected graph of m edge. The starting point is 0, and k people are initially at the starting point, Traversing the graph enables at least one person to walk each vertex and the I-1 must have been traversed when traversing the I point. To minimize the path and number of k people, k people should return to the starting point. Ideas: Cost flow, because for a person, the sequence of the person's