Cashier Employment（poj1275

Cashier Employment

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 9205		Accepted: 3564

Description

A supermarket in Tehran is open 24 hours a day every day and needs a number of cashiers to fit its need. The supermarket manager has hired you to help him, solve his problem. The problem is that the supermarket needs different number of cashiers at different times of each day (for example, a few cashiers after midnight, and many in the afternoon) to provide good service to its customers, and he wants to hire the least number of cashiers for this job. 

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

The first line of input is the number of test cases for this problem (at most 20). Each test case starts with 24 integer numbers representing the R(0), R(1), ..., R(23) in one line (R(i) can be at most 1000). Then there is N, number of applicants in another line (0 <= N <= 1000), after which come N lines each containing one ti (0 <= ti <= 23). There are no blank lines between test cases.

Output

For each test case, the output should be written in one line, which is the least number of cashiers needed. 
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

Source

Tehran 2000

[Submit]   [Go Back]   [Status]   [Discuss]

　　

　　题意：在一家超市里，每个时刻都需要有营业员看管，R(i) (0 <= i < 24)表示从i时刻开始到i+1时刻结束需要的营业员的数目，现在有N(N <= 1000)个申请人申请这项工作，并且每个申请者都有一个起始工作时间 ti，如果第i个申请者被录用，那么他会从ti时刻开始连续工作8小时。现在要求选择一些申请者进行录用，使得任何一个时刻i，营业员数目都能大于等于R(i)。求出至少需要录用多少营业员。无解输出“No Solution”

　　差分约束。

　　设f[i]表示前i个小时所需的营业员个数（题目中为0->23,我们看做1->24,因为要留0作为初始值）。根据每个时刻所需的营业员可以得出f[i]-f[i-8]>=r[i]（i>8）,f[i]+f[24]-f[i+16]>=r[i]（i<=8）.

　　隐含条件：0 <= f[i]-f[i-1] <= b[i]  （b[i]表示有几个申请人从第i天开始工作）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<bitset>
#define LL long long
#define RI register int
using namespace std;
const int INF = 0x7ffffff ;
const int N = 1000 + 10 ;
const int T = 24 + 2 ;
const int M = 1000 + 10 ;

inline int read() {
    int k = 0 , f = 1 ; char c = getchar() ;
    for( ; !isdigit(c) ; c = getchar())
      if(c == '-') f = -1 ;
    for( ; isdigit(c) ; c = getchar())
      k = k*10 + c-'0' ;
    return k*f ;
}
struct Edge {
    int to, next, val ;
}e[M] ;
int n, cnt ; int hh[T], num[T], f[T], dis[T], head[T], r[T], b[T] ;
inline void add_edge(int x,int y,int z) {
    e[++cnt].to = y, e[cnt].next = head[x], head[x] = cnt, e[cnt].val = z ;
}

inline bool spfa() {
    for(int i=1;i<=24;i++) dis[i] = INF ; dis[0] = 0 ; memset(num,0,sizeof(num)) ;
    deque<int>q ; q.push_back(0) ; bitset<T>inq ; inq[0] = 1 ;
    while(!q.empty()) {
        int x = q.front() ; q.pop_front() ; num[x]++ ; if(num[x] > 24) return 0 ;
        for(int i=head[x];i;i=e[i].next) {
            int y = e[i].to ;
            if(dis[y] > dis[x]+e[i].val) {
                dis[y] = dis[x]+e[i].val ;
                if(!inq[y]) {
                    inq[y] = 1 ;
                    if(!q.empty() && dis[y] < dis[q.front()]) q.push_front(y) ;
                    else q.push_back(y) ;
                }
            }
        }
        inq[x] = 0 ;
    }
    return 1 ;
}
inline bool check(int ff) {
    cnt = 1 ; memset(head,0,sizeof(head)) ;
    for(int i=9;i<=24;i++) add_edge(i-8,i,-r[i-1]) ;    // 边的权值全部取反 
    for(int i=1;i<=8;i++) add_edge(i+16,i,-r[i-1]+ff) ;
    for(int i=1;i<=24;i++) add_edge(i,i-1,b[i]), add_edge(i-1,i,0) ;
    add_edge(0,24,-ff), add_edge(24,0,ff) ;
    if(!spfa()) return 0 ;
    if(-dis[24] == ff) return 1 ; return 0 ;
}
inline void solve() {
    for(int i=1;i<=24;i++) r[i] = read() ;
    n = read() ;
    for(int i=1;i<=n;i++) b[read()+1]++ ;  // 注意，我们改变了对于时间的定义，所以这里要+1 
    int L = 0, R = n, ans = INF ;
    while(L <= R) {
        int mid = (L+R)>>1 ;
        if(check(mid)) ans = min(ans,mid), R = mid-1 ;
        else L = mid+1 ;
    }
    if(ans < INF) printf("%d
",ans) ;
    else printf("No Solution
") ; 
}

int main() {
    int t = read() ;
    while(t--) solve() ;
    return 0 ;
}