Catch that Cow

Catch that Cow

Time limit:5000/2000 MS (java/others) Memory limit:32768/32768 K (java/others)

Total submission (s): 243 Accepted Submission (s): 88

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?

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

Outputline 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.

Sourceusaco Open Silver

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#include <bits/stdc++.h>#defineN 5000100#defineINF 0x3f3f3f3fusing namespacestd;intn,m;structnode{intX,val; Node () {} node (intAintb) {x=A; Val=C; }};BOOLVis[n];intBFs () {Queue<node>Q; Q.push (node (n,0)); Vis[n]=true;  while(!Q.empty ()) {Node Fr=Q.front ();        Q.pop (); //cout<<fr.x<<endl;        if(fr.x==m) {//cout<< "x=" <<fr.x<<endl;            returnFr.val; }        intans=fr.x+1; if(ans>=0&&ans<=100000&&!Vis[ans]) {Q.push (node (ans,fr.val+1)); Vis[ans]=true; } ans=fr.x-1; if(ans>=0&&ans<=100000&&!Vis[ans]) {Q.push (node (ans,fr.val+1)); Vis[ans]=true; } ans=fr.x*2; if(ans>=0&&ans<=100000&&!Vis[ans]) {Q.push (node (ans,fr.val+1)); Vis[ans]=true; }    }    //return-1;}intMain () {//freopen ("C:\\users\\acer\\desktop\\in.txt", "R", stdin);     while(SCANF ("%d%d", &n,&m)! =EOF) {memset (Vis,false,sizeofvis); printf ("%d\n", BFS ()); }    return 0;}

Catch that Cow