HDU 4122 Alice's mooncake shop // RMQ 线段树

题目描述

HDU 4122 Alice’s mooncake shop

解题思路

题目大意:
有一家24小时营业的月饼店, 会连续营业m个小时, 且月饼每个小时的单价会浮动.在第i个小时会有一份订单.订单可以现做,也可以提前做好保存在冰箱里,(放在冰箱里每小时会花费一定的费用,且月饼有保质期为T).问在满足所有订单的前提下,最少的制作费用是多少?

抽象出来就是,对于第i个小时的订单. 查询区间 [i-T, i] 这段时间内的制作最小值即可.

参考代码

#include <iostream>#include <cstdio>#include <string>#include <map>#define lson rt<<1#define rson rt<<1|1#define mid ((l+r)>>1)using namespace std;const int MAX_N = 2510;const int MAX_NN = 100010;const int inf = 0x7fffffff;typedef __int64 ll;map<string, int> mp;struct Order {    int year, day, hour, num, h;    string month;} order[MAX_N];struct SegTree {    ll cost;} node[MAX_NN << 2];int mon[13] = {0,31,28,31,30,31,30,31,31,30,31,30,31};int price[MAX_NN], k;inline int max(int a, int b){return a>b?a:b;}inline int min(int a, int b){return a<b?a:b;}inline bool isleap(int year) {return ((year%4 == 0 && year%100 != 0) || year%400 == 0);}inline void pushup(int rt) {node[rt].cost = min(node[lson].cost, node[rson].cost);}//月份映射void init() {    k = 1;    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;}// 建树void build(int l, int r, int rt) {    if (l == r) {        node[rt].cost = price[k++];        return ;    }    build(l, mid, lson);    build(mid+1, r, rson);    pushup(rt);}// 查询区间[L, R]的最小值ll query(int L, int R, int l, int r, int rt) {    if (L <= l && r <= R) return node[rt].cost;    ll ret = inf;    if (L <= mid)   ret = min(ret, query(L, R, l, mid, lson));    if (R > mid)    ret = min(ret, query(L, R, mid+1, r, rson));    return ret;}// 根据输入的日期转换成小时int cal_hour(int year, string month, int day, int hour) {     mon[2] = (isleap(year) ? 29 : 28);    int ans = hour + (day-1) * 24 + 1;    for (int i = mp[month]-1; i >= 1; --i)        ans += mon[i] * 24;    year--;    while (year >= 2000) {        ans += (365 * 24 + (isleap(year) ? 24 : 0));        year--;    }    return ans;}int main() {    int n, m, S, T;    while (~scanf("%d %d", &n, &m) && (n || m)) {        init();        for (int i = 1; i <= n; ++i) {            cin >> order[i].month;            scanf("%d %d %d %d", &order[i].day, &order[i].year, &order[i].hour, &order[i].num);            order[i].h = cal_hour(order[i].year, order[i].month, order[i].day, order[i].hour);        }        scanf("%d %d", &T, &S);        for (int i = 1; i <= m; ++i) {            scanf("%d", &price[i]);            price[i] += S * (m - i);             // price[i]表示的是从第i个小时保存到第m个小时(即最后一个小时)的花费             //(即 price[i] = 第i个小时的花费 + 保存在冰箱(m-i)个小时的花费)        }        build(1, m, 1);        ll ans = 0;        for (int i = 1; i <= n; ++i) {            int l = max(1, order[i].h - T), r = order[i].h;            ll p = query(l, r, 1, m, 1) - (m - order[i].h) * S;            // p还要减去后面的是因为订单在第i个小时就要了            // 那么,第i个小时之后都不用保存在冰箱了,要减去那部分的花费            ans += p * order[i].num;        }        printf("%I64d\n", ans);    }    return 0;}