hdu 4122 單調隊列或線段樹

/**
hdu 4122 單調隊列
題目大意:給定n個時刻，在每一個時刻都要生產ai個月餅。給出m個可以生產的時刻，每個時刻單個生產費用爲bi，單個月餅可儲存T時間，單位時間費用爲S
         問如何安排生產花費最少
解題思路：用單調隊列維護一個點之前所有點的最小花費（爲生產費+儲存費）
特別注意：n個時刻可能有重複的
*/
#include <string.h>
#include <algorithm>
#include <iostream>
#include <stdio.h>
#include <map>
using namespace std;
typedef long long LL;

map<string,int>mp;
int n,m,time[3050],sum[15],S,T,cost[100050],q[100050];
LL num[3000];

void init()
{
    mp["Jan"]=1,mp["Feb"]=2,mp["Mar"]=3,mp["Apr"]=4,mp["May"]=5,mp["Jun"]=6,mp["Jul"]=7,mp["Aug"]=8;
    mp["Sep"]=9,mp["Oct"]=10,mp["Nov"]=11,mp["Dec"]=12;
    sum[0]=0,sum[1]=31,sum[2]=sum[1]+28,sum[3]=sum[2]+31,sum[4]=sum[3]+30,sum[5]=sum[4]+31,sum[6]=sum[5]+30;
    sum[7]=sum[6]+31,sum[8]=sum[7]+31,sum[9]=sum[8]+30,sum[10]=sum[9]+31,sum[11]=sum[10]+30;
}

int get(int y,int m,int d,int t)
{
    int ans=0;
    for(int i=2000; i<y; i++)
    {
        if((i%4==0&&i%100!=0)||i%400==0)
            ans+=366;
        else
            ans+=365;
    }
    if((y%4==0&&y%100!=0)||y%400==0)
    {
        ans+=sum[m-1];
        if(m-1>=2)ans++;
    }
    else
    {
        ans+=sum[m-1];
    }
    ans+=(d-1);
    return ans*24+t;
}

int main()
{
    init();
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0)break;
        for(int i=0; i<n; i++)
        {
            int year,day,t;
            char mon[10];
            scanf("%s%d%d%d%I64d",mon,&day,&year,&t,&num[i]);
            time[i]=get(year,mp[mon],day,t);
            //printf("->%d\n",time[i]);
        }
        scanf("%d%d",&T,&S);
        int tail=0,head=0,k=0;
        LL cnt=0;
        for(int i=0; i<m; i++)
        {
            scanf("%d",&cost[i]);
            while(head<tail&&cost[q[tail-1]]+S*(i-q[tail-1])>=cost[i])tail--;
            q[tail++]=i;
            while(i==time[k])
            {
                while(head+1<tail&&(i-q[head]>T))head++;
                //printf("%d==>>%d\n",i,q[head]);
                cnt+=num[k]*(cost[q[head]]+S*(i-q[head]));
                k++;
            }
        }
        printf("%I64d\n",cnt);
    }
    return 0;
}
/**
2 10
Jan 1 2000 2 10
Jan 1 2000 9 10
5 2
20
20
20
10
10
8
7
9
5
10
0 0
*/

思路分析：

ans = segma( num[]*(cost[] + (i-j)*s) )

整理一下會發現式子就是  

cost[]-j*s + i*s 

對於每一個訂單，我們把i拿出來分析

所以也就用cost - j*s 建樹。

然後在儲存期間找到最小的花費就行了。
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#define lson num<<1,s,mid
#define rson num<<1|1,mid+1,e
#define maxn 2555
#define maxm 100005
#define inf 0x3f3f3f3f
using namespace std;
typedef long long LL;

int n,m;
int days[2][13]={{0,31,28,31,30,31,30,31,31,30,31,30,31},
                {0,31,29,31,30,31,30,31,31,30,31,30,31}};
string tab[] = {"","Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec"};
LL tre[maxm<<2];

int getmonth(string x)
{
    for(int i=1;i<=12;i++)
    if(x==tab[i])return i;
}
bool leap(int x)
{
    if(((x%4==0) && x%100!=0) || x%400==0)return true;
    return false;
}
LL gethour(int month,int day,int year,int hour)
{
    LL res=day-1;
    int is=leap(year);
    for(int i=1;i<month;i++)res+=days[is][i];
    for(int i=2000;i<year;i++)
    res+=365+leap(i);
    res*=24;
    res+=hour+1;
    return res;
}
void build(int num,int s,int e)
{
    tre[num]=inf;
    if(s==e)return;
    int mid=(s+e)>>1;
    build(lson);
    build(rson);
}
void update(int num,int s,int e,int pos,LL val)
{
    if(s==e)
    {
        tre[num]=val;
        return;
    }
    int mid=(s+e)>>1;
    if(pos<=mid)update(lson,pos,val);
    else update(rson,pos,val);
    tre[num]=min(tre[num<<1],tre[num<<1|1]);
}
LL query(int num,int s,int e,int l,int r)
{
    if(l<=s && r>=e)
    {
        return tre[num];
    }
    int mid=(s+e)>>1;
    if(r<=mid)return query(lson,l,r);
    else if(l>mid)return query(rson,l,r);
    else return min(query(lson,l,mid),query(rson,mid+1,r));
}

string tmp;
LL num[maxn];
LL cost[maxm];
LL time[maxm];
int main()
{
    while(cin>>n>>m)
    {
        if(n==0 && m==0)break;
        for(int i=1;i<=n;i++)
        {
            int d,y,h,Num;
            cin>>tmp;
            cin>>d>>y>>h>>Num;
            num[i]=Num;
            time[i]=gethour(getmonth(tmp),d,y,h);
        }
        LL S,T;
        build(1,1,m);
        cin>>T>>S;
        for(int i=1;i<=m;i++)
        {
            cin>>cost[i];
            cost[i]-=i*S;
            update(1,1,m,i,cost[i]);
        }
        LL ans=0;
        for(int i=1;i<=n;i++)
        {
            if(time[i]>m)break;
            ans+=num[i]*(query(1,1,m,max(1LL,time[i]-T+1),time[i])+time[i]*S);
        }
        cout<<ans<<endl;
    }
    return 0;
}