【POJ 1275】 Cashier Employment

【题目链接】

点击打开链接

【算法】

设Ti为第i小时有多少个出纳员开始工作，Vi表示第i小时有多少个来应聘的出纳员

那么，有 :

1. 0 <= Ti <= Vi

2. Ti + Ti-1 + Ti-2 + Ti-3 + Ti-4 + Ti-5 + Ti-6 + Ti-7 >= Ri

令Si = T1 + T2 + T3 + ... Ti

则 :

1. Si >= Si-1

2. Si - Si-1 <= Vi

3. Si >= Si-8 + Ri( 8 <= i <= 24)

4. Si>= Si+16 - S24 +Ri (1 <= i <= 7)

所以，我们可以枚举S24,用差分约束系统判断是否可行，枚举可以二分

【代码】

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;

struct Edge
{
        int to,w,nxt;
} e[];
int i,n,k,t,l,r,mid,ans,tot,T;
int v[],R[],dis[],head[];

inline void add(int u,int v,int w)
{
        tot++;
        e[tot] = (Edge){v,w,head[u]};
        head[u] = tot;
}
inline bool spfa(int x)
{
        int i,cur,to,w;
        queue<int> q;
        static bool inq[];
        static int cnt[];
        tot = ;
        memset(head,,sizeof(head));
        memset(inq,false,sizeof(inq));
        memset(cnt,,sizeof(cnt));
        memset(dis,,sizeof(dis));
        add(,,x);
        for (i = ; i <= ; i++) add(i,i-,-v[i]);
        for (i = ; i <= ; i++) add(i-,i,);
        for (i = ; i <= ; i++) add(i-,i,R[i]);
        for (i = ; i <= ; i++) add(i+,i,R[i]-x);
        while (!q.empty()) q.pop();
        q.push();
        dis[] = ;
        inq[] = true;
        cnt[] = ;
        while (!q.empty())
        {
                cur = q.front();
                q.pop();
                inq[cur] = false;
                for (i = head[cur]; i; i = e[i].nxt)
                {
                        to = e[i].to;
                        w = e[i].w;
                        if (dis[cur] + w > dis[to])
                        {
                                dis[to] = dis[cur] + w;
                                if (!inq[to])
                                {
                                        inq[to] = true;
                                        q.push(to);
                                        cnt[to]++;
                                        if (cnt[to] > ) return false;
                                }
                        }
                }
        }
        return dis[] == x;
}

int main()
{

        scanf("%d",&T);
        while (T--)
        {
                memset(v,,sizeof(v));
                for (i = ; i <= ; i++) scanf("%d",&R[i]);
                scanf("%d",&k);
                for (i = ; i <= k; i++)
                {
                        scanf("%d",&t);
                        v[t+]++;
                }
                l = ; r = k;
                ans = -;
                while (l <= r)
                {
                        mid = (l + r) >> ;
                        if (spfa(mid))
                        {
                                r = mid - ;
                                ans = mid;
                        } else
                                l = mid + ;
                }
                if (ans == -) printf("No Solution\n");
                else printf("%d\n",ans);
        }

        return ;

}