hdu 4122 Alice's mooncake shop(单调队列)

Alice's mooncake shopTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2559    Accepted Submission(s): 609


Problem Description
The traditional food of this festival is the mooncake. Chinese family members and friends will gather to admire the bright mid-autumn harvest moon, and eat mooncakes under the moon together. In Chinese, “round”(圆) also means something like “faultless” or “reuion”, so the roundest moon, and the round mooncakes make the Zhongqiu Festival a day of family reunion.

Alice has opened up a 24-hour mooncake shop. She always gets a lot of orders. Only when the time is K o’clock sharp( K = 0,1,2 …. 23) she can make mooncakes, and We assume that making cakes takes no time. Due to the fluctuation of the price of the ingredients, the cost of a mooncake varies from hour to hour. She can make mooncakes when the order comes,or she can make mooncakes earlier than needed and store them in a fridge. The cost to store a mooncake for an hour is S and the storage life of a mooncake is T hours. She now asks you for help to work out a plan to minimize the cost to fulfill the orders.InputOutputSample InputSample OutputSourceRecommend#include<algorithm>#include<iostream>#include<string.h>#include<sstream>#include<stdio.h>#include<math.h>#include<vector>#include<string>#include<queue>#include<set>#include<map>using namespace std;const int INF=0x3f3f3f3f;const int maxn=100010;const int days[2][13]={ {0,31,59,90,120,151,181,212,243,273,304,334,365} ,{0,31,60,91,121,152,182,213,244,274,305,335,366}};map<string,int> mp;struct node{ int r,tm;} ord[2510];int q[maxn],head,tail;int cost[maxn];bool isLeap(int y){ return (y%400==0)||(y%100!=0&&y%4==0);}void mpInit(){ mp.clear(); 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;}int getHour(string mon,int d,int y,int h){ int i,tm=0; for(i=2000;i<y;i++) tm+=365+isLeap(i); tm+=days[isLeap(y)][mp[mon]-1]; tm+=d; return tm*24+h-23;}int main(){ mpInit(); string mon; int y,d,h,T,n,m,i,now; __int64 ans,s; while((~scanf("%d%d",&n,&m))&&(m||n)) { ans=0; for(i=0;i<n;i++) { cin>>mon; scanf("%d%d%d%d",&d,&y,&h,&ord[i].r); ord[i].tm=getHour(mon,d,y,h); } scanf("%d%I64d",&T,&s); head=tail=now=0; for(i=1;i<=m;i++)//晕死。这里不能写成i<=m&&now<n因为要把cost读完。2B了。。。 { scanf("%d",cost+i);//cost[i]+(ord[now].tm-i)*s<=cost[q[tail-1]]+(ord[now].tm-q[tail-1])*s while(head<tail&&cost[i]-i*s<=cost[q[tail-1]]-q[tail-1]*s)//head<tail写前面防RE tail--; q[tail++]=i; while(i==ord[now].tm&&now<n)//这里必须写成while因为有订单时间相同的情况 { while(q[head]<ord[now].tm-T) head++; ans+=ord[now].r*(cost[q[head]]+(ord[now].tm-q[head])*s); now++; } } printf("%I64d\n",ans); } return 0;}