poj1275 Cashier Employment

Topic transmission

sol:

It's not easy to think of...

Let(S[i] (0 < I < 23) indicate that I have been appointed in the previous hour.

Then constraints are made according to the given and implicit conditions of the title:
\[ s[i]-s[i-8]≥need[i]\ (8≤i≤23)\\ sum-(s[i+16]-s[i])≥need[i]\ (0≤i≤7)\\ s[i-1]≤s[i]\ (0≤i≤23)\\ s[i+1]-s[i]≤have[i]\ (0≤i≤23)\\ \]
I'll sort it out and get it:
\[ s[i]≥s[i-8]+need[i]\ (8≤i≤23)\\ s[i]≥need[i]-sum+s[i+16]\ (0≤i≤7)\\ s[i]≥s[i-1]\ (0≤i≤23)\\ s[i]≥[i+1]-have[i]\ (0≤i≤23)\\ \]
However, it is noted that there is an extra \(s[23]) in the second inequality, which can not be done directly.

So consider the value of enumeration or dichotomy sum and then find out that the \ (s[23] should satisfy \\\\\\\\\\

So there is another limitation (let s[-1]=0): \ (s[23] (> sum+0=sum+s[-1]).

For convenience, add 1 to all subscriptions, and then ok.

code:

#include<queue>
#include<string>
#include<cstdio>
#include<cstring>
#define LL long long
#define DB double
#define RG register
#define IL inline 
using namespace std;

const int N=1003;

queue<int> q;
int n,l,r,mid,tot,t[33],s[33],R[N],inq[N],cnt[N],head[N];

struct EDGE{int next,to,v;}e[N*21];

IL int gi() {
    RG int x=0,p=1; RG char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-') p=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') x=x*10+(ch^48),ch=getchar();
    return x*p;
}

IL void make(int a,int b,int c) {e[++tot]=(EDGE){head[a],b,c},head[a]=tot;}

IL void New_Graph() {
    tot=0;
    memset(&e,0,sizeof(e));
    memset(head,0,sizeof(head));
}

IL int spfa(int val) {
    RG int i,x,y;
    memset(s,0xcf,sizeof(s));
    memset(inq,0,sizeof(inq));
    memset(cnt,0,sizeof(cnt));
    while(!q.empty()) q.pop();
    s[0]=0,cnt[0]=1,q.push(0);
    while(!q.empty()) {
        x=q.front(),inq[x]=0,q.pop();
        for(i=head[x];i;i=e[i].next)
            if(s[y=e[i].to]<s[x]+e[i].v) {
                s[y]=s[x]+e[i].v;
                if(++cnt[y]==24) return 0; 
                if(!inq[y]) inq[y]=1,q.push(y);
            }
    }
    return s[24]>=val;
}

int main()
{
    RG int i,T=gi();
    while(T--) {
        for(i=1;i<=24;++i) R[i]=gi();
        n=gi(),l=0,r=n+1;
        memset(t,0,sizeof(t));
        for(i=1;i<=n;++i) ++t[gi()+1];
        while(l<r) {
            mid=l+r>>1,New_Graph(),make(0,24,mid);
            for(i=1;i<=24;++i) {
                if(i>=9) make(i-8,i,R[i]);
                else make(i+16,i,R[i]-mid);
                make(i-1,i,0),make(i,i-1,-t[i]);
            }
            if(spfa(mid)) r=mid;
            else l=mid+1;
        }
        if(r==n+1) puts("No Solution");
        else printf("%d\n",r);
    }
    return 0;
}
// Differential constraints (good topic!)