POJ 3278 Catch that Cow wide search

HDU 2717 Catch that Cow, topic link

Time limit:5000/2000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 11466 Accepted Submission (s): 3551

Problem Description

Farmer John had been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0≤n≤100,000) on a number line and the cow are at a point K (0≤k≤100,000) on the same number Line. Farmer John has modes of transportation:walking and teleporting.

• WALKING:FJ can move from any point x to the points X-1 or X + 1 inch a single minute
• TELEPORTING:FJ can move from any point X to the point 2xX in a single minute.

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

Input

Line 1:two space-separated integers:n and K

Output

Line 1:the least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

Sample Input

5 17

Sample Output

4

Hint
The fastest-Farmer John to reach the fugitive cow are to move along the following PATH:5-10-9-18-17, which takes 4 Minutes.

Source

Usaco Open Silver

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Test instructions: give you two positions K and N to find the minimum number of steps from K to N.
There are three ways of changing from K: k+1,k-1,k*2.

Queue wide search, a vis array to see if it has been accessed, a step array to record the number of steps.

#include <stdio.h>#include <queue>#include <algorithm>#include <string>#include <iostream>#include <string.h>#include <math.h>#define MAXN 100005using namespace STD;intVIS[MAXN];//Whether the tag has been accessed;intSTEP[MAXN];//Record the number of steps; Queue <int >Q;intSerch (intKintN) {intflag=0, a,mark=0;memset(Vis,0,sizeof(VIS));//array clear 0;    memset(Step,0,sizeof(step));//array clear 0;Q.push (k);//Put k in the queue;vis[k]=1;//marked as 1;     while(!q.empty ())//When the array is not empty;{A=q.front ();//Assign the first value of the queue to A;Q.pop ();//delete queue first;         for(intI=1; i<=3; i++) {if(i==1) flag=a+1;Else if(i==2) flag=a-1;Elseflag=a*2;if(flag>100000|| flag<0)//cross-border judgment;                Continue;if(!vis[flag])//If marked as 0;{step[flag]=step[a]+1;//The current number of steps equals the last step of +1;vis[flag]=1;//marked as 1;Q.push (flag);//queue;                if(flag==n)//If the current value is n;                    returnStep[flag];//Returns the number of steps;}        }    }return 0;//If you are submitting C + +, do not return a value;    //But if it is g++, a return 0 must be required;    //This is a compiler problem, the specific is not very clear;}intMain () {intN,k;//k Chase N;     while(~scanf("%d%d", &k,&n))//Here I write K chase N;{if(k>n)//If k>n, can only through the constant reduction of one get, that is, k-n step;            printf("%d\n", k-n);Else            printf("%d\n", Serch (k,n)); }return 0;}

Before about that return 0,wa, good pit, changed the compiler on a.

T^t, just asked the voyage seniors about that C + + submission, g++ submit different questions, results seniors also do not know, I was the first time to encounter this problem, good worry Mulberry ~ ~
But seniors also said have not seen, he said only in this OJ met is double to use%.f good pit oj~~qaq

POJ 3278 Catch that Cow wide search