pots acronym

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Poj3414--pots (BFS, record Path)

Pots Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10149 Accepted: 4275 Special Judge DescriptionYou are given the pots of the volume of A and B liters respectively. The following operations can be performed: Fill (i) Fill the pot i (1≤ i ≤ 2) from the tap; DROP (i) empty the pot I to the drain; Pour (i,

(POJ) 3414 Pots (wide search of output path)

Descriptionyou is given the pots, having the volume of A and B liters respectively. The following operations can be Performed:fill (i) FILL the pot I (1≤i≤2) fromThe tap;drop (i) empty the pot I to the drain; Pour (i,j) pour fromPot I to pot J; After ThisOperation either the pot J isFull (and there is some water leftinchThe pot i), or the pot I isempty (and all its contents has been moved to the pot J). Write a program to find the shortest possible se

POJ 3414--pots (bfs+ backtracking path)

Pots Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9963 Accepted: 4179 Special Judge DescriptionYou are given the pots of the volume of A and B liters respectively. The following operations can be performed: Fill (i) Fill the pot i (1≤ i ≤ 2) from the tap; DROP (i) empty the pot I to the drain; Pour (i,j

poj3414 Pots (BFS)

Pots Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 12198 Accepted: 5147 Special Judge DescriptionYou are given the pots of the volume of A and B liters respectively. The following operations can be performed: Fill (i) Fill the pot i (1≤ i ≤ 2) from the tap; DROP (i) empty the pot I to the drain; Pour (i,

POJ 3414 Pots (bfs+ backtracking)

Pots Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 11705 Accepted: 4956 Special Judge DescriptionYou are given the pots of the volume of A and B liters respectively. The following operations can be performed: Fill (i) Fill the pot i (1≤ i ≤ 2) from the tap; DROP (i) empty the pot I to the drain; Pour (i,

Poj 3414 -- pots -- BFS + backtracking path

Pots Time limit:1000 ms Memory limit:65536 K Total submissions:9963 Accepted:4179 Special Judge Description You are given two pots, having the volumeAAndBLiters respectively. The following operations can be saved med: Fill (I) fill the potI(1 ≤I≤ 2) from the tap; Drop (I) Empty the potITo the drain; Pour (I, j)

POJ 3414 Pots (bfs+ recursive printing)

Pots Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10255 Accepted: 4333 Special Judge DescriptionYou are given the pots of the volume of A and B liters respectively. The following operations can be performed: Fill (i) Fill the pot i (1≤ i ≤ 2) from the tap; DROP (i) empty the pot I to the drain; Pour (i,

POJ 3414 Pots (BFS) __ Search

Description You are are given two pots, having the volume of A and B liters respectively. The following operations can be Performed:fill (i) FILL the pot I (1≤i≤2) from the tap; DROP (i) empty the pot I to the drain; Pour (i,j) pour from pot I to pot J; After this operation either the pot J are full (and there may to some water left in the pot I), or the pot I is empty (and All its contents have been moved to the pot J). Write a program to find the s

POJ 3414 Pots violence, BFS difficulty: 1

http://poj.org/problem?id=3414Record bottle status, breadth first search#include   POJ 3414 Pots violence, BFS difficulty: 1

Bfs/poj 3414 Pots

); } } //DROP A 3333333333 if(now.a!=0) {Next=Now ; NEXT.A=0; next.b=now.b; if(!v[next.a][next.b]) {v[next.a][next.b]=1; Next.step=now.step+1; Next.way[next.step]=3; Que.push (next); } } //DROP B 444444444444 if(now.b!=0) {Next=Now ; NEXT.A=NOW.A; next.b=0; if(!v[next.a][next.b]) {v[next.a][next.b]=1; Next.step=now.step+1; Next.way[next.step]=4; Que.push (next); } } //pour (5555555555555) if(now.a!=0) {

Poj3414 pots (BFS)

Pouring Water. Give two cups A and B respectively. By connecting the water from the tap (each time it is filled with water), all the water in the cup is dumped, and the water in one cup is poured into the other three operations to make the water in one cup C. Solution: 6-entry BFs. The difficulty lies in how to save the path. #include Poj3414 pots (BFS)

Poj 3414 Pots (bfs)

Tags: poj BFS Link: poj 3414 Question: The capacity of two bottles a, B, and a target water volume C are given. You can perform the following operations on A and B: Fill (I) fills the bottle with water Drop (I) Empty the bottle I Pour (I, j) pour the water in bottle I into bottle J. After this operation, either the bottle J is filled with water or the bottle I is empty Perform at least several operations to make the water in A or B Package C and output the specific operation. Analysis: BFS can b

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-abbreviation-First letter abbreviation-Address-Text Direction--Long references-Short cited terms-citations, citations-Defining items tagsDefinition and Usage: tags indicate abbreviations or abbreviations, such as "WWW" or "NATO".By tagging

Funny recursive abbreviation (recursive acronym)

The number of recursive abbreviations is as follows: Recursive abbreviation is an abbreviation that recursively references itself in the full name. Because it is recursive, it cannot be accurately interpreted. The earliest example may be visa: Visa

Solution of POJ 3414 Pots BFS

Title Link: http://poj.org/problem?id=3414 Idea: Very simple, is the search can, just a cup of water in the state transfer when the code is a bit less good to write. Code: #include #include #include #include #include #include #include

Poj 3414 pots search BFS

Question;How to load water to C using only two containersAnalysis:BFS questions, but to record the path, referAlgorithmIntroduction to the BFS algorithm (I refer to other people'sProgram).Here is a brief summary: seeCodeText in # Include # Include

POJ 3414 Pots (BFS record Shortest path)

Topic links POJ3414 Topic Two bottles 1th and 2nd were given the capacity of A and b respectively, and a target water C, the following operations on two bottles:Fill (i): Fill the bottle IDrop (i): Pour the bottle I into the lightPour (i,j): Pour

POJ-3414 pots BFS + Memory path

This reminds me of the three Coke cans of hangdian. It seems that only the number of output times is enough, and all the paths need to be output here, in this way, BFs alone will have a lot of space redundancy in the state of retention. Therefore,

Poj 3414 pots

// The solution for poj1606 is the same, but it should be noted that impossible should be output when there is no solution for this question, and no matter which cup of water capacity meets the requirements, you can output it! # Include # include #

Poj 3414 pots [classic BFS]

# Include # include # include # include # include # include #include # define bug cout pre! = NULL) {DFS (t-> pre); cout op] k1 = c | CP-> k2 = c) {printf ("% d \ n ", CP-> steps); DFS (CP); Return ;}for (INT I = 0; I k2; break; // a full

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