HDU 4122 Alice's mooncake shop

单调队列优化算法

最新推荐文章于 2019-10-04 03:06:20 发布

最新推荐文章于 2019-10-04 03:06:20 发布 · 136 阅读

本文介绍了一种使用单调队列优化动态规划问题的方法，并通过一个具体案例解释了如何实现及运用该算法解决特定问题。文章深入探讨了单调队列的构建、维护以及如何有效地更新答案。

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;
}