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,n3,dp[N][N];char s1[N],s2[N],s3[N*2];int dfs(int i,int j,int k){ if(k==n3) return 1; if(dp[i][j]) return 0; dp[i][j]=1; if(s1[i]==s3[k]) { if(dfs(i+1,j,k+1)) return 1; } if(s2[j]==s3[k]) { if(dfs(i,j+1,k+1)) return 1; } return 0;}int main(){ int T,cnt=1; scanf("%d",T); while

http://poj.org/problem?id=1564The problem uses DFS but to pay attention to the heavy, can not output duplicate answersThere are two kinds of de-weight methods in the code that are markedThe first if (a[i]!=a[i-1]) means that if the number A[i] is the same as the previous number, then there is no need to put that number in the same position of the record array. For example: 4 3 3 2 The composition and the second position of the 7,b array is 3, then the

Test instructions: Given a structure with a root tree and the color of each point at the beginning, there are two actions 1: coloring the subtree of a given point 2: How many colors are in the subtree of the given pointIt's easier to think of Dfs sequence + line tree to doDFS sequence is a long time ago to see the Bilibili on the video of the study of a tree through the DFS number to the number of each poin

Topic linksThe simple application of Dfs, the more cumbersome is to process the input of the English alphabet. Use and check set can also do (but the author has not mastered and check set, before only used once, later learned to back up)1#include 2#include 3 using namespacestd;4 intm,n,typ[ A][4]={{1,1,0,0},{1,0,1,0},{0,1,0,1},{0,0,1,1},{1,0,0,1},{0,1,1,0},{1,1,1,0},{1,1,0,1},{0,1,1,1},{1,0,1,1},{1,1,1,1},{0,0,0,0}};5 intdir[4][2]={{-1,0},{0,-1},{0,1}

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5938Test instructions: Give a 2020 string of the maximum length, you want to divide the numbers into 55 segments, then fill in the ' + ', '-', ' * ', '/', and ask for the maximum result you can get.To make this result the largest, as far as possible to let the number of the left side of the largest, only need to care about the left of the first bit is a number or the last one is a number, followed by the enumeration * and/the position of the

HDU-2259-Continuous Same Game (2) (BFS + DFS + simulation), dfsbfsProblem DescriptionAfter repeated attempts, LL finds the greedy strategy is very awful in practice. even there is no apparent evidence to proof it is better than a random one. so he has to drop this strategy and try to discover a better one.InputThere are 100 test cases. each test case begins with two integers n, m (5 OutputFor each test case, first output a single line containing the n

Hdu 1445 dfs pruning, hdu1445 The meaning of the question is relatively simple; the focus is on pruning. # Include Using namespaceStd; IntN,Num[70],Mark[70],K,Flash; IntDfs(IntS, IntIi, IntT, IntC) {If (T=C){Flash=1; Return1;} If (Flash) Return0; If (S=K){Dfs(0,1,T+1,C);// Return 1;} Else {intI; (I=Ii;IN;I++) {If (S+Num[I]>K) Continue; if (Mark[I]) Continue;Mark[I] =1;

1011 simply change it. # Include Using Namespace STD; # define Max 64 Int Sticks [Max]; Bool Used [Max]; Int Sticknum, Plen, N; Bool Compare ( Int A, Int B ){ Return A> B ;} // Match from beginindex. The next step is to match the stick of matchlen. hasmatch stick has been matched before. Bool DFS (Int Beginindex, Int Matchlen, Int Hasmatch ){ If (Matchlen = 0) {hasmatch ++; If (Hasmatch = 4 ){ Return True ;} For (Beginindex = 0; used [begininde

The DFS implementation steps are as follows: ① Access vertex v and mark that V has accessed ② Find the first vertex W of V ③ If W exists, the execution continues; otherwise, the algorithm ends. ④ If W is not accessed, use DFS to recursively access W ⑤ Search for the next adjacent node of V, mark it as W, and go to step ③ For DFS, the access order is A B D C E Th

Http://poj.org/problem? Id = 3009 In fact, at the beginning, we wanted to use dir as a parameter and search based on the map condition, but it was processed with while in recursion. It was hard to process the backtracking of map from 0 to 1. The code that hurts is tangled .. Code:# Include # Include # Define min (A, B) A> B? B: # Define Max 1E + 6 Int Tur [4] [2] = {0, 1, 0,-1, 1, 0,-1, 0 }; Int map [21] [21]; Int W, H, ANS, Sx, Sy, ex, ey; Bool check (int x, int y ){ If (x Return true; } Void

# Include Using namespace STD;Int CNT = 0;Int flag = 0;Int to [400007], NEX [400007], vis [100007], head [100007];Void add (int A, int B) {// The head Insertion Method of the linked list. When the NEX array is opened to next, it will be compiled incorrectly.To [++ CNT] = B;Nex [CNT] = head [a];Head [a] = CNT;}Void DFS (int A, int B ){For (INT I = head [a]; I! = 0; I = NEX [I]) {// traverse the adjacent contacts of this VertexInt v = to [I];If (V = B)

Label: style blog HTTP Io OS AR for SPQuestion connection: answer questions Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1427 Train of Thought: simple DFS, DFS (sum, next, p) indicates that the calculated value is sum, the calculated value in parentheses is next, and the currently used card subscript is P, in fact, we divide the two values out of and in brackets into sum and next for processing. Intuitio

HDU-2258-Continuous Same Game (1) (DFS)Problem DescriptionContinuous Same Game is a simple game played on a grid of colored blocks. groups of two or more connected (orthogonally, not diagonally) blocks that are the same color may be removed from the board. when a group of blocks is removed, the blocks above those removed ones fall down into the empty space. when an entire column of blocks is removed, all the columns to the right of that column shift t

BZOJ 3083 remote country DFS sequence tree Chain Division Given a tree with a root tree, each vertex has a weight and provides three operations: 1. Change x to the root node. 2. Change the weights of all vertices from x to y to v. 3. Ask the minimum value of the vertex weight in the subtree of x. Tree link splitting is not discussed! However, because it is the minimum weight of the subtree to be queried, we choose

sum (intx) {return(1ll1LL;} + -ll DFS (ll Cur,ll CNT,intstop) the { * //printf ("cur:%d cnt:%d\n", cur,cnt); $ if(cur = = M)returnCNT;Panax Notoginseng inti =0; - while(Cur-sum (i) > M) i++; the + if(Cur-sum (i) = = M)returnCNT +i; ALL up = M-max (0ll,cur-sum (i)); theLL res = i + max (0ll,up-stop); + returnMin (Cnt+res, DFS (cur-sum (i-1), CNT + i,stop+1)); - } $ $ intMain () -

1#include 2#include 3#include string.h>4 5typedefstructNode6 {7 intx, y;8 }node;9 Ten Const intMAX =10000; One Const intN = the; A Const intdir[4][2] = { {-1,0}, {0,1}, {1,0}, {0,-1} };//Move Direction - BOOLVisit[n][n];//Tag Array - CharBoard[n][n];//Chess Board the intTurn[n][n];//The turn array is used to record the minimum number of turns known on the Board . -Node origin, destination;//Start, End - intMaxturncount;//the maximum number of turns the road fetish can withstand - intM, N;//t