bfs disease

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

following notation:> [*] Marks a wall, into which you can not move;> [.] Marks an empty space, into which you can move;> [@] Marks the initial position of the adventurer;> [> [A]-[J] marks the jewels. Outputresults shocould be directed to standard output. start each case with "case #:" on a single line, where # Is the case number starting from 1. two consecutive cases shoshould be separated by a single blank line. no blank line shocould beProduced after the last test case. If the adventurer can

a line containing "Poor Angel has to stay in the prison all his life." Sample Input 7 8#.#####.#. A #... R.#... # X.....#..#.##...##...#.............. Sample output 13 This is to say that angle has been captured and her friends have helped her .. X is an obstacle to his enemies, but angle's friends are strong enough to destroy all enemies, but while at the same time eliminating the enemy, It takes one second for him to save the angle .. This question uses