HDU 4122 Alice's mooncake shop

         单调队列，裸的！！坑死了，说好的“All the orders are sorted by the time in increasing order. 呢，我就当成严格上升的序列了，于是就各种错。测试数据是有重复元素的！！！

 

//#pragma comment(linker, "/STACK:1024000000,1024000000")

#include<algorithm>

#include<iostream>

#include<cstring>

#include<fstream>

#include<sstream>

#include<vector>

#include<string>

#include<cstdio>

#include<bitset>

#include<queue>

#include<stack>

#include<cmath>

#include<map>

#include<set>

#define FF(i, a, b) for(int i=a; i<b; i++)

#define FD(i, a, b) for(int i=a; i>=b; i--)

#define REP(i, n) for(int i=0; i<n; i++)

#define CLR(a, b) memset(a, b, sizeof(a))

#define debug puts("**debug**")

#define LL long long

#define PB push_back

#define MP make_pair

#define eps 1e-10

using namespace std;



const int N = 3333;

const int M = 111111;



int order[N], tim[N], pay[M];

char month[15][5] = {"zero", "Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"};

int hav[15] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

char mo[5];

int p[M];

int n, m, t, s, y, h, r, d;

map<string, int> mp;



void pre()

{

    int i;

    mp.clear();

    for(i = 1; i <= 12; i ++)

    {

        mp[month[i]] = i;

    }

}



int get_tm(int m, int d, int y, int h)

{

    int ret = 0, i;

    for(i = 0; i < y; i ++)

    {

        if(i % 4 == 0) ret += 366 * 24;

        else ret += 365 * 24;

    }

    for(i = 1; i < m; i ++)

    {

        ret += hav[i] * 24;

        if(i == 2 && y % 4 == 0) ret += 24;

    }

    ret += (d - 1) * 24 + h;

    return ret;

}



void input()

{

    int i, j;

    for(i = 0; i < n; i ++)

    {

        scanf("%s%d%d%d%d", mo, &d, &y, &h, &r);

        order[i] = r;

        tim[i] = get_tm(mp[mo], d, y - 2000, h);

    }

    scanf("%d%d", &t, &s);

    for(i = 0; i < m; i ++)

    {

        scanf("%d", &pay[i]);

    }

}



void slove()

{

    int l = 0, r = 0, i = 0, j = 0;

    LL ans = 0;

    p[r ++] = 0;

    while(i < m)

    {



        while(pay[i] <= pay[p[r - 1]] + (i - p[r - 1]) * s && r > l) r --;

        p[r ++] = i;

        while(i - p[l] > t) l ++;

        while(j < n && i == tim[j])ans += (pay[p[l]] + (i - p[l]) * (LL)s) * (LL)order[j], j ++;

        i ++;

    }

    printf("%I64d\n", ans);

}



int main()

{

    //freopen("input.txt", "r", stdin);

    pre();

    while(scanf("%d%d", &n, &m), n + m)

    {

        input();

        slove();

    }

    return 0;

}