【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[1010];
int i,n,k,t,l,r,mid,ans,tot,T;
int v[30],R[30],dis[30],head[30];

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[30];
        static int cnt[30];
        tot = 0;
        memset(head,0,sizeof(head));
        memset(inq,false,sizeof(inq));
        memset(cnt,0,sizeof(cnt));
        memset(dis,255,sizeof(dis));
        add(0,24,x);
        for (i = 1; i <= 24; i++) add(i,i-1,-v[i]);
        for (i = 1; i <= 24; i++) add(i-1,i,0);
        for (i = 8; i <= 24; i++) add(i-8,i,R[i]);
        for (i = 1; i <= 7; i++) add(i+16,i,R[i]-x);
        while (!q.empty()) q.pop();
        q.push(0);
        dis[0] = 0;
        inq[0] = true;
        cnt[0] = 1;
        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] > 24) return false;
                                }
                        }
                }
        }
        return dis[24] == x;
}

int main() 
{
        
        scanf("%d",&T);
        while (T--)
        {
                memset(v,0,sizeof(v));
                for (i = 1; i <= 24; i++) scanf("%d",&R[i]);
                scanf("%d",&k);
                for (i = 1; i <= k; i++)
                {
                        scanf("%d",&t);
                        v[t+1]++;        
                }        
                l = 1; r = k; 
                ans = -1;
                while (l <= r)
                {
                        mid = (l + r) >> 1;
                        if (spfa(mid)) 
                        {
                                r = mid - 1;
                                ans = mid;
                        } else
                                l = mid + 1;
                }
                if (ans == -1) printf("No Solution
");
                else printf("%d
",ans);
        }
        
        return 0;
    
}