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Method: Starting from the root node wide search, if you encounter the point should be deleted, and then a wide search to delete its sub-tree and mark, and then count the number of marks is the answer, the so-called technique is to start from the root node search, if encountered a node distance The code is as follows:#include #include#include#include#includeusing namespacestd;#defineN 100010#defineINF 1e9structnode{intTo,val;}; VectorVt[n];intA[n],n,vis[n],del[n],fa[n],mark[n];queueint>Que;queuei

Tag: Tin sig Code back destroys C + + second OID struct[Topic link]Https://codeforces.com/contest/507/problem/EAlgorithmFirst, the BFS to find the shortest 1 to the remaining points, N to the remaining points of the shortest, recorded as dista[] and distb[]Obviously, we only need to maximize the shortest road without being destroyed on the edge, may wish to use f[i] to show that now in the city I, Dista[i] + distb[i] = Dista[n], at most can also pass through several non-destructive edgesMemory S

(); Que.pop (); if(maps[now.x][now.y]=='T' now.col==0)returnNow.tim; Node NXT=Now ; Nxt.tim++;///consumes 1s per operationNxt.dir = (Now.dir +1)%4; if(OK (NXT)) {Que.push (NXT); Mark (NXT);} ///the first one turns to the right, notice that the turnNxt.dir= Now.dir-1;if(nxt.dir0) nxt.dir=3; if(OK (

AC automaton nude problem, really nude ...The first n to match the string to save, and then the M-mode string structure of the AC automata, and then asked to be matched to the string is not ...1#include 2#include string.h>3#include 4 using namespacestd;5 Const intmaxm=600006;6 7 Chars[100005][10005],word[10005];8 intnxt[maxm][ -],tail[maxm],f[maxm],size;9 intLAST[MAXM];Ten One intNewNode () { Amemset (Nxt[size],0,sizeof(

know the value of the result modulus 10007.Sample Input2 2ABsample Output100Positive solution: $AC $ automaton + $DP $.First we construct the $ac$ automata, and record each bit is possible to form a match, is to record each bit of $val$ and $last$ on the line.Set $f[i][j]$ said statistics to the text before the $i$ bit, the $j$ on the machine, there is no matching scheme number, remove can match the situation, direct vigorously transfer on the line.Finally, we subtract the number of mismatched

];inlinevoidReadintk) { intf=1; k=0;CharC=GetChar (); while(c'0'|| C>'9') c=='-' (f=-1), c=GetChar (); while(c'9' c>='0') k=k*Ten+c-'0', c=GetChar (); K*=F; } InlinevoidInsert () {intLen=strlen (s+1), now=0; for(intI=1, ch;i) { if(!tree[now].nxt[ch=s[i]-'A']) Tree[now].nxt[ch]=++Tott; now=Tree[now].nxt[ch]; } Cnt[now]=1;} InlinevoidGetfail () {int

[ednum] = Root [a];Root [A] = Ednum ++ ;} Int Spfa (){ Int I, top, cur, TMP, NXT;Top = 0 ; // No Source Vertex is added during initialization, and all the rooted vertices are directly added to the stack, which is also good. For (I = 1 ; I N; I ++ ){Dis [I] = INF; // Key1 If (Root [I] ! = - 1 ){Stack [ ++ Top] = I;CNT [I] ++ ;Mark [I] = 1 ;}} While (Top ){Cur = Stack [Top -- ];TMP = Root [

we all know, the suffix array is dejected and the constants are quite large. But Sam is good to be able to construct it online .First put the code:struct Sam{int pre[maxn],son[maxn][26],cnt,len,now,step[maxn],np,nq,p,q; SAM () {memset (pre,0,sizeof (pre)); Memset (Son,0,sizeof (son)); cnt=1;len=0;now=1;} void Extend (int nxt) {p=now;np=++cnt;step[np]=step[now]+1;now=np;while (p!son[p][nxt]) {son[p][

Leetcode-reverse Linked List IIReverse a linked list from position m to N. Do it in-place and in One-pass.For example:Given 1->2->3->4->5->NULL , m = 2 and n = 4,Return 1->4->3->2->5->NULL .Note:Given m, n satisfy the following condition:1 ≤ m ≤ n ≤length of list.1 /**2 * Definition for singly-linked list.3 * struct ListNode {4 * int val;5 * ListNode *next;6 * ListNode (int x): Val (x), Next (NULL) {}7 * };8 */9 classSolution {Ten Public: Onelistnode* Reversebetween (listnode* head,intMintN)

Links: http://www.spoj.com/problems/LCS/Test instructions two strings of LCSThere's really nothing to say, the first time. Let's turn over the online tutorial to see it again.Another found that Baidu internal users to communicate the use of pictures in the Baidu snapshot.#include #include#include#includeusing namespacestd;#defineMAXN 2510000#defineMaxt maxn*2structsam_node{intPnt,len; intnxt[ -]; voidPrint (intid=-1) {printf ("-----------------\ n"); printf ("Sam id:%d\n", id); printf ("Next: \

fourth? Can we place the second a in this position? We can see that the strings from the last character to the third a are aba, and the strings from the first character to the second a are also aba. Are they not the same? At this point, you should probably understand how KMP jumps. We can record an nxt array. nxt [I] indicates the length of the longest prefix Suffix from the first character to the first ch

(int K) { for (int i=head;i;i=a[i].point" if ((--k) ==1// find precursor {a[i].point =a[a[i].point].point; // break ; }}Traversing a linked list: for (int i=a[head].point;i;i=a[i].point) cout2, doubly linked listAs the name implies, a doubly linked list is a linked list with two directions. Unlike a one-way list, each node in a doubly linked list stores not only a pointer to the next node, but a pointer to the previous node. Its advantage is that access,

, but no fruitFirst, by the y-coordinate, we assume that this line is infinitely low, when the answer corresponds to the maximum number of points in the middle of the next two same colorAnd then this line up, one-time Delete rowMaintain the position of the previous and the latter of the same color with a doubly linked listDelete, delete this point from the tree array and the linked list, delete each one, and count the adjacent and adjacent answers before the point.And then you have to assign Y a

] = Ednum ++ ;} /* The function returns-1, indicating that there is a negative edge weight loop, which indicates that it is impossible.If the function returns-2, it indicates that the distance from the start point to the end point is uncertain. Here, we use dis [N] = inf to judge, because the start value is INF.Otherwise, DIS [N] is returned, where the maximum distance from the start point to the end point is stored. */ /* Module 3 */ Int Spfa (){ Int I, top, cur, TMP,