poj——1275 Cashier Employment 差分约束系统

Cashier Employment

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

分析：

  
  
  
  

   
   
   
   
 
    约束条件分析,为方便分析,我们把讨论区间由[0,23]改为[1,24] 
    ：  
   
   
   
   
   

     设apply[i] 为来应聘的在第i个小时开始工作的总人数 
   
   
   
   
   

     设R[i] 为第i个小时至少需要的人数 
   
   
   
   
   

     设X[i] 为实际雇佣的在第i个小时开始工作的总人数 
   
   
   
   
   

     设S[i] = X[1] + … + X[i]   (第1到第i个小时一共雇佣的总人数) 
   
  
  
  
  
约束条件：

由于0<=x[i]<=apply[i]，于是

1> s[i]-s[i-1]>=0; 2> s[i]-s[i-1]<=apply[i] <=> s[i-1]-s[i]>=-apply[i];

由于题目限制：

3> s[i]-s[i-8]>=R[i](8<=i<=24) 4> s[i]-s[i+16]>=R[i]-s[24] (i<8) (*) 

(*)不等式中，s[24]（也是最后要求的）为变量，于是需要二分答案；

二分答案为p后，注意假设了s[24]==p；

于是 5> s[24]-s[0]>=p && s[0]-s[24]<= -p;

注意，这一条必须加入图中一起跑，不能最后单独判定，因为s[24]不是一个独立的变量，与其他变量有关。

最后，求最小，跑最长路即可。

代码如下：

<pre name="code" class="cpp">#include<cstdio>      
#include<iostream>    
#include<cstring>      
#include<queue>      
#include<vector>       
    
#define LL long long      
#define CLEAR(XXX) memset((XXX),0,sizeof(XXX))   
   
using namespace std;        
const int inf=1e9;      
const int maxn=1000+5,maxm=100005;      
  
int n,apply[30],R[30];      
inline void _read(int &x){      
    char ch=getchar(); bool mark=false;      
    for(;!isdigit(ch);ch=getchar())if(ch=='-')mark=true;      
    for(x=0;isdigit(ch);ch=getchar())x=x*10+ch-'0';      
    if(mark)x=-x;      
}      
struct Edge{      
    int from,to,w;      
    Edge(int from,int to,int w):from(from),to(to),w(w){}    
};   
  
vector<Edge> edge;      
int last[maxm],Next[maxm];      
int dist[maxn];      
int cnt[maxn];      
bool vis[maxn];   
      
struct SPFA{      
    int  n,m;          
    void init(int n){      
        this->n = n; m=0;      
        CLEAR(last);  CLEAR(Next);      
        edge.clear();      
        edge.push_back(Edge(0,0,0));      
    }      
    void add_edge(int from,int to,int dist){      
        edge.push_back(Edge(from,to,dist));      
        m=edge.size()-1;      
        Next[m]=last[from];      
        last[from]=m;      
    }      
    bool solve(int s){      
        int i;      
        CLEAR(vis); CLEAR(cnt);        
        for(i=1;i<=n;i++) dist[i]= -inf;      
        dist[s]=0;    
        vis[s]=true;cnt[s]++;      
        queue <int> q;   
        q.push(s);      
        while(!q.empty()){      
            int x=q.front();      
            q.pop();vis[x]=false;     
            for(i=last[x];i;i=Next[i]){      
                Edge& e=edge[i];      
                if(dist[e.from]+e.w>dist[e.to]){   //最少，用最长路   
                    dist[e.to]=dist[e.from]+e.w;      
                    if(!vis[e.to]){      
                        cnt[e.to]++;      
                        if(cnt[e.to]==n+1)return false;      
                        q.push(e.to) ;      
                        vis[e.to]=true;      
                    }      
                }       
            }      
        }      
        return true;      
    }      
    void answer(){      
        for(int i=1;i<=n;i++)      
            if(dist[i]>=inf)printf("NoPath\n");      
            else printf("%d\n",dist[i]);      
    }      
};  
SPFA solver;  
bool check(int p){  //二分ans，就是s[24]；  
    int i;   
    solver.init(24);  
    for(i=1;i<=24;i++){  
        solver.add_edge(i-1,i,0); //1> s[i]-s[i-1]>=0;  
        solver.add_edge(i,i-1,-apply[i]);  
         //2> s[i]-s[i-1]<=apply[i] <=> s[i-1]-s[i]>=-apply[i];  
        solver.add_edge(0,i,0);  
        if(i>=8)solver.add_edge(i-8,i,R[i]); //3> s[i]-s[i-8]>=r[i];  (8<=i<=24)  
        else solver.add_edge(i+16,i,R[i]-p);  
        //4> s[i]-s[i+16]>=r[i]+p  
    }  
    solver.add_edge(0,24,p);  // 5> dist[24]-dist[0]==p; 
	solver.add_edge(24,0,-p); 
    return solver.solve(0);

}       
int main(){     
     //freopen("ans.out","w",stdout);   
     int t,i,x,j,l,r,y;  
     _read(t);  
     while(t--){  
        CLEAR(R); CLEAR(apply);  
        for(i=1;i<=24;i++)_read(R[i]);  
        _read(n);    
        for(i=1;i<=n;i++){  
            _read(x);  apply[x+1]++;  
        }  
        l=0;r=n;  
        while(l<=r){  
            int mid=(l+r)>>1;  
            check(mid)? (r=mid-1):(l=mid+1);  
         }  
         if(check(l)&&l<=n)printf("%d\n",l);  
         else puts("No Solution");  
     }  
     return 0;  
}     

another version

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
const int inf=-1e9;
using namespace std;
int dis[5005],next[100005],last[100005];
bool vis[5005];
int cnt[5005];
int num[25];
int r[25];
int m=0;
int n=24;
struct edge{
	int from,to,len;
	edge(){}
	edge(int a,int b,int c){from=a;to=b;len=c;}
};
edge line[100005];
inline void read(int &x){  
    char t;  
    bool mark=false;  
    for(;t=getchar(),t<'0'||t>'9';) if(t=='-') mark=1;  
    for(x=t-'0',t=getchar();'0'<=t&&t<='9';x=x*10+t-'0',t=getchar());  
    x=mark?-x:x;  
}
void add_edge(int from,int to,int len){
	m++;
	next[m]=last[from];
	last[from]=m;
	line[m]=edge(from,to,len);
}
void check_clear(){
	memset(next,0,sizeof(next));
	memset(last,0,sizeof(last));
	memset(dis,0,sizeof(dis));
	memset(line,0,sizeof(line));
}
void init_clear(){
	memset(next,0,sizeof(next));
	memset(last,0,sizeof(last));
	memset(dis,0,sizeof(dis));
	memset(line,0,sizeof(line));
	memset(r,0,sizeof(r));
}
bool spfa(int s){
	queue<int>q;
	memset(vis,0,sizeof(vis));
	memset(cnt,0,sizeof(cnt));
	int i,j,k,t;
	for(i=1;i<=n;i++)dis[i]=inf;
	vis[s]=true;
	dis[s]=0;
	cnt[s]++; 
	q.push(s);
	while(q.size()){
		t=q.front();q.pop();vis[t]=false;
		for(i=last[t];i;i=next[i]){
			if(dis[line[i].to]<dis[line[i].from]+line[i].len){
				dis[line[i].to]=dis[line[i].from]+line[i].len;
				if(vis[line[i].to]==false){
					cnt[line[i].to]++;
					if(cnt[line[i].to]>=n+1){
						return false;
					}
					q.push(line[i].to);
					vis[line[i].to]=true;
				}
			}
		}
	}
	return true;
}
bool check(int mid){
	check_clear();
	bool flag;
	int i,j,k;
	m=0;
	for(i=1;i<=24;i++)add_edge(i-1,i,0);
	for(i=1;i<=24;i++)add_edge(i,i-1,-num[i]);
	for(i=8;i<=24;i++)add_edge(i-8,i,r[i]);
	for(i=1;i<=7;i++)add_edge(i+16,i,r[i]-mid);
	for(i=1;i<=24;i++)add_edge(0,i,0);
	add_edge(0,24,mid);
	add_edge(24,0,-mid);
	flag=spfa(0);
	//if(flag==false)cout<<m id<<" fuck"<<endl;
	//else cout<<mid<<" "<<dis[24]<<" "<<dis[0]<<endl;
	if(flag==false)return false;
	//if(dis[n]>mid)return false;
	else return true;
}
int main(){
	int h;
	cin>>h;
	while(h--){
		//s[i]-s[i-1]>=0
		//s[i-1]+num[i]>=s[i]
		//s[i]-s[i-8]>=num[i]  (8=<i<=24)
		//s[24]-s[i+16]+s[i]>=r[i]   (1<=i<=7)
		init_clear();
		int minn=1,maxn,tot;
		int i,j,k,x;
		m=0;
		for(i=1;i<=24;i++)scanf("%d",&r[i]);
		scanf("%d",&tot);
		for(i=1;i<=tot;i++){
			scanf("%d",&x);
			num[x+1]++;
		}
		maxn=tot;
		while(minn<=maxn){
			int mid=(minn+maxn)/2;
			if(check(mid))maxn=mid-1;
			else minn=mid+1;
		}
		if(check(minn)==false)cout<<"No Solution"<<endl;
		else cout<<minn<<endl;
	}
}