Gear-wheelstime limit:1.0 Second
Memory limit:64 mb-arny! What is happened with coordinator? Bad working coordinator was the everlasting trouble of their spaceship. Arny already had been working under this trouble while he not very attentive and responsible mate just noticed the Breaka Ge. Judging by schematics The broken module of coordinator consists of the "set of special gears connected with each other in O Rder to transfer the traction from the kinetic generator to the lot of antenna driving engines. Despite the extraterrestrial origin of these gears, they is connected by usual terrestrial method:the cogs of one gear-w Heel get into the slots between cogs of another gear-wheel. So the rotation of the first gear-wheel are transmitted to the second Gear-wheel. After the multiple Arny ' s revisions, no unnecessary gears stayed in the coordinator. It means that there are no cycles in gears connection graph. The only problem are to check the-all the gears has right directions and speeds of rotation. Inputfirst Line of input Contains the number of gear-wheels in mechanism
N(1≤
N≤1000). The next
NLines contain the information about the gear-wheels.
I-th Line contains
K(1≤
K≤1000)-the number of cogs on the
I-th Gear-wheel followed by the list of gears, that is connected to the
I-th one. Zero ends the list. The last line of input contains the number of Gear-wheel, which is connected to the Kinetic-generator and the Rotation
V(1≤
V≤1000). This gear-wheel rotates in counter-clockwise direction. Outputoutput should contain
NLines. In the
I-th line There are a speed of rotation of
I-th Gear-wheel in the form of irreducible fraction. Numerator and denominator of this fraction should is separated by the sign '/'. If speed is negative, it's assumed that the gear-wheel rotates in clockwise direction (in this case the minus sign should Be displayed before numerator). Otherwise the Gear-wheel rotates in counter-clockwise direction. If speed equals zero than numerator should is equal 0 and denominator should is equal to 1. It is guaranteed this neither numerator nor denominator of all speeds would be greater than 106.Sample
| input |
Output |
410 2 3 020 1 040 1 4 0100 3 01 6 |
6/1-3/1-3/23/5 |
problem Author:Pavel Egorov "Analysis" gives you n gears, and then the following n lines, the first number of each line is the number of teeth gear, followed by his linked gear number, the last line to the starting position of the gear number and speed, the speed of the rest of the gear, with the simplest fraction expressed.
#include <iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<cmath>#include<string>#include<map>#include<stack>#include<queue>#include<vector>#defineINF 0x3f3f3f3f#defineMet (b) memset (a,b,sizeof a)#definePB Push_backtypedefLong Longll;using namespacestd;Const intN = 1e3+Ten;Const intM =124750+Ten;Const intmod=1e9+7;intn,m,k,tot=0, s,t,v;intHead[n],vis[n],speed[n],sum[n];structans{intF,s;} Ans[n];structedg{intTo,next;} edg[n*N];voidAddintUintv) {edg[tot].to=v;edg[tot].next=head[u];head[u]=tot++;}intgcdintXinty) { intxx=x,yy=y; X=Max (XX,YY); Y=min (yy,xx); if(x%y==0)returny; returnGCD (xy,y);}voidBFs () {Queue<int>Q; Q.push (s); Vis[s]=1; ans[s].f=t;ans[s].s=1; while(!Q.empty ()) { intU=q.front (); Q.pop ();//printf ("!!! %d%d%d\n ", U,ANS[U].F,ANS[U].S); System (" pause "); for(inti=head[u];i!=-1; i=Edg[i].next) { intv=edg[i].to; if(!Vis[v]) {Vis[v]=1; intG=GCD (ABS (SUM[U]*ANS[U].F), sum[v]*ans[u].s); ANS[V].F=-sum[u]*ans[u].f/g;ans[v].s=sum[v]*ans[u].s/G; Q.push (v); } } }}intMain () {Met (ans,0); Met (head,-1); intu,v; scanf ("%d",&N); for(intu=1; u<=n;u++) {scanf ("%d",&Sum[u]); while(~SCANF ("%d", &v) &&v) {Add (u,v); add (V,u); }} scanf ("%d%d",&s,&t); BFS (); for(intI=1; i<=n;i++){ if(ans[i].f>0) {printf ("%d/%d\n", ANS[I].F,ANS[I].S); }Else if(ans[i].f==0) {printf ("0/1\n"); }Else{printf ("-%d/%d\n",-ans[i].f,ans[i].s); } } return 0;}
URAL 1291 Gear-wheels (BFS)
