POJ 1275 - Cashier Employment
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Description
The manager has provided you with the least number of cashiers needed for every one-hour slot of the day. This data is given as R(0), R(1), ..., R(23): R(0) represents the least number of cashiers needed from midnight to 1:00 A.M., R(1) shows this number for duration of 1:00 A.M. to 2:00 A.M., and so on. Note that these numbers are the same every day. There are N qualified applicants for this job. Each applicant i works non-stop once each 24 hours in a shift of exactly 8 hours starting from a specified hour, say ti (0 <= ti <= 23), exactly from the start of the hour mentioned. That is, if the ith applicant is hired, he/she will work starting from ti o???clock sharp for 8 hours. Cashiers do not replace one another and work exactly as scheduled, and there are enough cash registers and counters for those who are hired.
You are to write a program to read the R(i) ???s for i=0..23 and ti ???s for i=1..N that are all, non-negative integer numbers and compute the least number of cashiers needed to be employed to meet the mentioned constraints. Note that there can be more cashiers than the least number needed for a specific slot.
Input
Output
If there is no solution for the test case, you should write No Solution for that case.
Sample Input
1
1 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
5
0
23
22
1
10
Sample Output
1
Solution:
??????????????????????????????????????????????????????????????????????????????
?????????????????????????????????????????????????????????????????????????????????????????????R(i) (0 <= i < 24)?????????i???????????????i+1???????????????????????????????????????????????????N(N <= 1000)???????????????????????????????????????????????????????????????????????????????????? ti????????????i???????????????????????????????????????ti????????????????????????8?????????????????????????????????????????????????????????????????????????????????i????????????????????????????????????R(i)?????????????????????????????????????????????
???????????????????????????????????????
???????????????????????????????????????????????????????????????????????????????????????????????????????????????
???????????????????????????$s[i]$???????????????$i+1$????????????????????????????????????
????????????1?????????$i>=7$,???$s[i] - s[i-8] geq R[i]$???
????????????2?????????$0 leq i < 7$??????????????????$s[23] - s[i+16] + s[i] geq R[i]$???
????????????3????????????????????????????????????????????????????????????$i$????????????????????????????????????$b[i]$,??????????????????$i$???$0 leq s[i] - s[i-1] leq b[i]$???
????????????????????????????????????????????????????????????????????????$3$???$s$??????????????????????????????????????????????????????????????????
???????????????????????????????????????????????????????????????????????????????????????$s[23]$??????????????????????????????????????????????????????????????????$s[i] - s[i+16]geq R[i] - s[23]$,???????????????????????????????????????$s[23]$??????????????????????????????????????????????????????????????????????????????$s[23]$??????????????????????????????????????????$0$???$n$??????$s[23]$,?????????????????????????????????????????????????????????????????????????????????????????????????????????.
?????????????????????s$[-1] = 0$?????????????????????????????????????????????????????????????????????????????????$s[0] = 0$???
????????????????????????????????????$s[24]$????????????????????????????????????????????????????????????
?????????
1 #include<bits/stdc++.h> 2 #define il inline 3 #define For(i,a,b) for(int (i)=(a);(i)<=(b);(i)++) 4 using namespace std; 5 const int N=2005,M=26,inf=23333333; 6 int T,n,a[M],dis[M],tot[M],net[N],to[N],cnt,h[M],w[N]; 7 bool vis[M]; 8 queue<int>q; 9 10 il int gi(){ 11 int a=0;char x=getchar(); 12 while(x<???0???||x>???9???)x=getchar(); 13 while(x>=???0???&&x<=???9???)a=(a<<3)+(a<<1)+x-48,x=getchar(); 14 return a; 15 } 16 17 il void add(int u,int v,int c){to[++cnt]=v,net[cnt]=h[u],h[u]=cnt,w[cnt]=c;} 18 19 il void init(int x){ 20 cnt=0; 21 memset(vis,0,sizeof(vis)); 22 memset(h,0,sizeof(h)); 23 memset(dis,-0x3f,sizeof(dis)); 24 while(!q.empty())q.pop(); 25 For(i,0,23) add(i,i+1,0),add(i+1,i,-tot[i]); 26 For(i,7,23) add(i-7,i+1,a[i]); 27 add(0,24,x),add(24,0,-x); 28 For(i,0,6) add(i+17,i+1,a[i]-x); 29 } 30 31 il bool check(int x){ 32 init(x); 33 q.push(0);dis[0]=0; 34 while(!q.empty()){ 35 int u=q.front();q.pop();vis[u]=0; 36 if(u==24&&dis[u]>x)return 0; 37 for(int i=h[u];i;i=net[i]) 38 if(dis[to[i]]<dis[u]+w[i]){ 39 dis[to[i]]=dis[u]+w[i]; 40 if(!vis[to[i]])q.push(to[i]),vis[to[i]]=1; 41 } 42 } 43 return dis[24]==x?1:0; 44 } 45 46 il void solve(){ 47 int l=0,r=n+1,mid,ans=inf; 48 while(l<=r){ 49 mid=l+r>>1; 50 if(check(mid))r=mid-1,ans=mid; 51 else l=mid+1; 52 } 53 if(ans>n)printf("No Solution "); 54 else printf("%d ",ans); 55 } 56 57 int main(){ 58 T=gi(); 59 while(T--){ 60 memset(tot,0,sizeof(tot)); 61 For(i,0,23) a[i]=gi(); 62 n=gi(); 63 int u; 64 For(i,1,n)u=gi(),tot[u]++; 65 solve(); 66 } 67 return 0; 68 }
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