# "UVa 11093" Just Finish it Up (algorithmic efficiency + greed)

Source: Internet
Author: User

Test instructions: There are N petrol stations on the circular runway, numbered 1~n. The I petrol station can refuel AI gallons, from gas station I to the next station need bi gallons of gasoline. Ask the smallest gas station number that can be used as a starting point to go back to the starting point.

Solution: We combine the ai,bi of each gas station, and treat Ai-bi as the right CI of n points, indicating the remaining oil quantity after I. It is known that the first petrol station is sum{}+ci>=0,sum{} indicating the amount of oil remaining before opening from the starting point to I, sum{}>=0. Therefore, if sum{}+ci<0, then from the beginning of the enumeration from the starting point to I all points can not be used as a starting point, because at this time the sum{} is already with their starting point to reach the maximum value, there is no more oil residue, must not pass the point I. Enumerates i+1 as the starting point and continues to judge.

P.S. Because it is a ring, it is a bit of trouble to judge to walk a circle, we should be careful. I think I played a short, but not very beautiful. = =

`1#include <cstdio>2#include <cstdlib>3#include <cstring>4#include <algorithm>5#include <iostream>6#include <queue>7 using namespacestd;8 Const intn=100010;9 Ten intN; One intA[n],b[n]; A intMmax (intXintY) {returnX>y?x:y;} - intCheckintx) - { the     intH=0, t=x; -     BOOLtf=true; -      while(1) -     { +       if(T==X&AMP;&AMP;!TF)return-1; -h+=a[t]-B[t]; +       if(h<0)returnT; Atf=false; atT= (t+1)%N; -     } - } - intMain () - { -     intT; inscanf"%d",&T); -      for(intKase=1; kase<=t;kase++) to     { +scanf"%d",&n); -        for(intI=0; i<n;i++) scanf ("%d",&a[i]); the        for(intI=0; i<n;i++) scanf ("%d",&b[i]); *       intans=n,x=0, tmp,mx=-1; \$        while(1)Panax Notoginseng       { -         if(X&LT;=MX) Break; theTmp=check (x), mx=Mmax (mx,x); +         if(tmp==-1) {ans=x; Break;} A         Elsex=tmp+1; the       } +       if(ans!=n) printf ("Case%d:possible from station %d\n", kase,ans+1); -       Elseprintf"Case %d:not possible\n", Kase); \$     } \$     return 0; -}`

"UVa 11093" Just Finish it Up (algorithmic efficiency + greed)

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