bfs windows

Title: Fzu 2150 Fire Game Give a m*n figure, ' # ' means lawn, '. ' Show the open space, and then you can choose to fire in any of the two lawn lattices, the fire every 1 s will spread to the surrounding four lattice, ask those two points to make burning all the lawn spend the least time. Analysis: This topic is a bit difficult to think about, but given the small data range, we can violently enumerate the points where any lawn is located, and then press the two points into the queue to

) { Lnode *temp = result->top->next; if (temp!= NULL) { result->top->next = temp->next; } Temp->next = NULL; return temp; } We'll give you the definition of the main function. int main (void) {/* MAX's value is 5 * defines a two-dimensional array * Input the value of the maze and initializes a stack/int maze[max][max]; int I, J; for (i = 0;i Finally when we output the stack path, we are back from the end to the beginning of the stack output, at this tim

Original title URL: http://nanti.jisuanke.com/t/11064 What is the river, for the big nails in the chess world, the lake is a matrix, his goal is to ride a horse in the rivers and lakes, from his position to go to the end. Of course, the big nailed horse also obeys the Chinese chess "The horse Goes Day" the rule, and in the matrix, there will be some obstacles, the horse can not jump to the obstacles, if the big nails in front of the horse is a barrier, that is, "horse legs", then he will not be

Very simple BFS. WA reason: or because rqnoj data input with GetChar () seems to be wrong, changed to scanf ("%s"), and then because did not notice starting from 1 ... scanf ("%s", Map[i]); Change into scanf ("%s", map[i][1]);It's AC. This simple search question can not be once AC is the reflection of the following ... #define _crt_secure_no_warnings #include

to the edges of the vertex. The vertex VJ adjacent to VI are linked to a single linked list called VI adjacency list. Each node in the adjacency list is composed of two domains: one is the neighboring point Domain (Adjvex), the ordinal number J (can be the subscript of the array unit of vertex VJ in the vertex table) used to store the vertex VJ adjacent to VI; the second is the chain field (next), which links the nodes in the adjacency list. The specific procedures are implemented as follows:

disappear (i.e., in this turn). Well, given the description of a maps, the positions of your tank and the target, how many turns would you take at least to Arrive there? Input the input consists of several test cases. The first line of all test case contains, integers m and n (2 Output for each test case, please output the turns to least in a separate line. If you can ' t arrive at the target, output "-1" instead. Sample Input 3 4 ybeb EERE sste 0 0 Sample Output 8Test instructions: How long

the target, how many turns would you take at least to Arrive there? Input The input consists of several test cases. The first line of all test case contains, integers m and n (2 Output For each test case, please output the turns to least in a separate line. If you can ' t arrive at the target, output "-1" instead.Sample Input 3 4YbebEERESste0 0Sample Output 8 title Link:http://poj.org/problem?id=2312 Test Instructions Calculate the shortest time from ' Y ' to ' t ', where ' S ' and ' R ' cann

target, how many turns would you take at least to Arrive there? Input The input consists of several test cases. The first line of all test case contains, integers m and n (2 Output For each test case, please output the turns to least in a separate line. If you can ' t arrive at the target, output "-1" instead.Sample Input 3 4YbebEERESste0 0Sample Output 8Source POJ Monthly, Lu Xiao Test Instructions:Give you a matrix of m rows n columns.Y represents the starting point, and T represents the en

description of a maps, the positions of your tank and the target, how many turns would you take at least to Arrive there? Input the input consists of several test cases. The first line of all test case contains, integers m and n (2 Output for each test case, please output the turns to least in a separate line. If you can ' t arrive at the target, output "-1" instead. Sample Input 3 4 ybeb EERE sste 0 0 Sample Output 8 Source POJ Monthly, Lu Xiao http://blog.csdn.net/furturerock/article/deta

here you should understand, there is no place to move the situation OH ~~~~~~) Ideas: Because there are a lot of points can be repeated, so we have to open three-dimensional array vis "" "" "to judge the weight, if one dimension is used to represent X, one dimension is used to represent Y, then the other one to indicate what. Of course, the number of steps to walk, but we do not know the total number of steps we can walk AH ~ so we here to the K to seek redundancy, as long as it is not 0, all i

Daily Punch-In (2/2) Click to open linkMain topic: A three-dimensional maze, given the starting point and end point, to find the shortest path. PS: The length and width of the height are not more than 30 Idea 1: First think of Dfs, only the first time to write three-dimensional DFS, feeling and two-dimensional nothing different. #include The first time I write the experience of the morning, cut a branch first. If the current TOT value has exceeded the previous optimal solution, then pass. Last

The storage structure of graphs This article focuses on the depth-first search (DFS) and breadth-first search (BFS), so it is no longer to introduce the basic concepts of the graph too much, but to get a general idea of some of the common storage structures in the diagram below. Adjacency Matrix The adjacency matrix can be used both to store the graph without direction and to store the graph. In fact, the structure uses a two-dimensional array (adj

Topic links POJ3278 Topic Enter two number n,k (0≤\leqn,k≤\leq100000) to find at least a few legal operations from N to K. Legal actions include: + 1,-1, * *. Analysis Consider each number of 0~100000 as a state, there are 3 ways to transfer the state, only from N start with the BFS expansion, see the expansion to K is the first layer, Time complexity O (3*100000). (similar to the shortest path in the maze)PS: This problem in OJ with g++ submitted on

calculation of time complexity is not very clear. My calculation of the time complexity of this problem is: Obviously the time complexity of the two for Loop is O ((m*n) ^2) =10^4. Then calculate the time complexity of the BFS: obviously each layer has a for loop, the loop volume is 4, the maximum layer is diagonal, 10 layers, so the total complexity of 4^10, in the front of the complexity, obviously timed out. But this is not the case, we can think,

Dating with Girls (2) Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 2656 Accepted Submission (s): 741 Problem Description If You had solved the problem Dating with girls (1). I Think you can solve this problem too. This problem was also about dating with girls. Now you are in a maze and the girl you want to date with is also in the maze. If you can find the girl and then you can date with the girl. Else the girl would date with other boy

DFS uses recursive step-by-step heuristics to select the shortest path in all paths Code: 0 is the road, 1 is the wall #include BFS Use the queue one layer at a time to take the shortest Code: #include Print out Path Input 0 for road, 1 for Wall, 5*5 map//from upper left corner to bottom right #include

Transmission DoorTest instructions: Is given a starting point and an end point, each can go to four directions up to the K-step, each walk spent a second, asked to go to the end of the minimum number of seconds required.Idea: Direct BFS will definitely t, we need to add a few pruning, one is when we go to some point, the point has been reached, and at this time than from the current point of time to go to this point is shorter, then do not go, because

1 1 1 1 1 0 31 1 1 1 1 1 1 11 1 1 1 1 1 1 10 0 Sample Output 4-11320-1 -1 Test instructions: On a n*m map, there are 5 numbers. 0 means not to walk here, 1 means to go here, 3 is the Ant's Cave, 2 is where the ant starts to walk, and 4 says there is food here. An ant to go home, it initially only 6HP, and each step will be reduced by one, when HP is 0 o'clock will die, when it arrives at the food can be restored full hp and the food can be used indefinitely. Ask ants at least how many steps to

case contains the integers N and M indicate the size of the board. Then goes-N line, each line with M character shows the board. "#" indicates the grass. You can assume this there is at least one grid which are consisting of grass in the board.1 OutputFor each case, output of the case number first, if they can play the more special (Hentai) game (fire all the grass), output The minimal time they need to wait after they set fire, otherwise just output-1. See the sample, input and output for more

HDU-5040-Instrusive (BFS + priority queue), instrusiveProblem DescriptionThe legendary mercenary Solid Matt gets a classic mission: infiltrate a military base.The military base can be seen as an N * N grid. Matt's target is in one of the grids and Matt is now in another grid.In normal case, Matt can move from a grid to one of the four neighbor grids in a second. But this mission is not easy.Around the military base there are fences, Matt can't get out