题目大意：一个月饼店开m个小时（24小时营业），只在整点做月饼，做月饼的能力非常强。现在只需要考虑成本的问题。给m个cost值，cost[i]表示第i个小时做1个月饼的代价。
再给n个时间，从2000年1月1日0时开始计算。表示订单的截止时间。当然为了节约成本，可以提前趁成本不高的时候做月饼。但是月饼有保质期，t天，月饼放冰箱保存也需要代价，一天花费s。现在求完成n个订单最小代价。
思路：
水题嘛，写写写T^T
#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <queue>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
template <class T>
inline bool rd(T &ret) {
char c; int sgn;
if(c=getchar(),c==EOF) return 0;
while(c!='-'&&(c<'0'||c>'9')) c=getchar();
sgn=(c=='-')?-1:1;
ret=(c=='-')?0:(c-'0');
while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
ret*=sgn;
return 1;
}
template <class T>
inline void pt(T x) {
if (x <0) {
putchar('-');
x = -x;
}
if(x>9) pt(x/10);
putchar(x%10+'0');
}
int MON[2][12]={31,28,31,30,31,30,31,31,30,31,30,31,\
31,29,31,30,31,30,31,31,30,31,30,31
};
int YEAR[3000];//year[i]表示 到第i年结束时的小时数
bool run(int x){if(x%400 == 0)return true; return (x%100) && x%4==0;}
void init(){
memset(YEAR, 0, sizeof YEAR);
for(int i = 2000; i < 3000; i++)
{
YEAR[i] = YEAR[i-1];
YEAR[i] += 8760;
if(run(i)) YEAR[i] += 24;
}
for(int i = 0; i < 2; i++)
for(int j = 0; j < 12; j++)
MON[i][j] *= 24;
}
int find_yue(string str){
if(str=="Jan")return 1;
if(str=="Feb")return 2;
if(str=="Mar")return 3;
if(str=="Apr")return 4;
if(str=="May")return 5;
if(str=="Jun")return 6;
if(str=="Jul")return 7;
if(str=="Aug")return 8;
if(str=="Sep")return 9;
if(str=="Oct")return 10;
if(str=="Nov")return 11;
if(str=="Dec")return 12;
return -100000000;
}
const int N = 50010;
int n, m, s, t, cost[100010];
struct Buy{
int tim, r;
}buy[2555];
void input(){
string mon;
for(int i = 1, day, year, hour; i <= n; i++)
{
cin>>mon; rd(day); rd(year); rd(hour); rd(buy[i].r);
buy[i].tim = YEAR[year-1];
int cur = run(year);
int yue = find_yue(mon) -1;
for(int j = 0; j < yue; j++)
buy[i].tim += MON[cur][j];
buy[i].tim += 24*(day-1);
buy[i].tim += hour;
}
rd(t); rd(s);
for(int i = 0; i < m; i++)rd(cost[i]);
// for(int i = 1; i <= n; i++)printf("(%d,%d)\n", buy[i].tim, buy[i].r);
}
struct node {
int id;
ll v;
node() {}
node(int _id, ll _v) {
id = _id; v = _v;
}
friend bool operator < (const node& j, const node& i) {
if (i.v!=j.v)
return i.v<j.v;
else
return i.id<j.id;
}
};
priority_queue<node>Q;
void work() {
ll ans = 0, x = 0;
while (!Q.empty()) Q.pop();
int idx = 1;
for (int i = 0; i < m; ++i) {
Q.push(node(i, cost[i]-x));
while (!Q.empty() && i-Q.top().id>t)
Q.pop();
while (idx<=n&&buy[idx].tim==i) {
ans += (ll)buy[idx].r * (cost[Q.top().id] + (ll)(i-Q.top().id)*s);
//ans += (ll)buy[idx].r * (Q.top().v+x);
++idx;
}
if (idx == n+1)
break;
x += s;
}
while (idx<=n) {
while (!Q.empty() && buy[idx].tim-Q.top().id>t)
Q.pop();
ans += (ll)buy[idx].r * (cost[Q.top().id] + (ll)(buy[idx].tim-Q.top().id)*s);
++idx;
}
pt(ans);
putchar('\n');
}
int main() {
init();
while(cin>>n>>m, n+m) {
input();
work();
}
return 0;
}
/*
1 4
Jan 1 2000 7 10
3 10
1
20
200
10
*/
