https://pintia.cn/problem-sets/994805342720868352/problems/994805472333250560
A table tennis club has N tables available to the public. The tables are numbered from 1 to N. For any pair of players, if there are some tables open when they arrive, they will be assigned to the available table with the smallest number. If all the tables are occupied, they will have to wait in a queue. It is assumed that every pair of players can play for at most 2 hours.
Your job is to count for everyone in queue their waiting time, and for each table the number of players it has served for the day.
One thing that makes this procedure a bit complicated is that the club reserves some tables for their VIP members. When a VIP table is open, the first VIP pair in the queue will have the priviledge to take it. However, if there is no VIP in the queue, the next pair of players can take it. On the other hand, if when it is the turn of a VIP pair, yet no VIP table is available, they can be assigned as any ordinary players.
Input Specification:
Each input file contains one test case. For each case, the first line contains an integer N (≤10000) - the total number of pairs of players. Then N lines follow, each contains 2 times and a VIP tag: HH:MM:SS - the arriving time, P - the playing time in minutes of a pair of players, and tag - which is 1 if they hold a VIP card, or 0 if not. It is guaranteed that the arriving time is between 08:00:00 and 21:00:00 while the club is open. It is assumed that no two customers arrives at the same time. Following the players' info, there are 2 positive integers: K (≤100) - the number of tables, and M (< K) - the number of VIP tables. The last line contains M table numbers.
Output Specification:
For each test case, first print the arriving time, serving time and the waiting time for each pair of players in the format shown by the sample. Then print in a line the number of players served by each table. Notice that the output must be listed in chronological order of the serving time. The waiting time must be rounded up to an integer minute(s). If one cannot get a table before the closing time, their information must NOT be printed.
Sample Input:
9
20:52:00 10 0
08:00:00 20 0
08:02:00 30 0
20:51:00 10 0
08:10:00 5 0
08:12:00 10 1
20:50:00 10 0
08:01:30 15 1
20:53:00 10 1
3 1
2
Sample Output:
08:00:00 08:00:00 0
08:01:30 08:01:30 0
08:02:00 08:02:00 0
08:12:00 08:16:30 5
08:10:00 08:20:00 10
20:50:00 20:50:00 0
20:51:00 20:51:00 0
20:52:00 20:52:00 0
3 3 2
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
struct person {
int arrive, start, time;
bool vip;
}tempperson;
struct tablenode {
int end = 8 * 3600, num;
bool vip;
};
bool cmp1(person a, person b) {
return a.arrive < b.arrive;
}
bool cmp2(person a, person b) {
return a.start < b.start;
}
vector<person> player;
vector<tablenode> table;
void alloctable(int personid, int tableid) {
if(player[personid].arrive <= table[tableid].end)
player[personid].start = table[tableid].end;
else
player[personid].start = player[personid].arrive;
table[tableid].end = player[personid].start + player[personid].time;
table[tableid].num++;
}
int findnextvip(int vipid) {
vipid++;
while(vipid < player.size() && player[vipid].vip == false) vipid++;
return vipid;
}
int main() {
int n, k, m, viptable;
scanf("%d", &n);
for(int i = 0; i < n; i++) {
int h, m, s, temptime, flag;
scanf("%d:%d:%d %d %d", &h, &m, &s, &temptime, &flag);
tempperson.arrive = h * 3600 + m * 60 + s;
tempperson.start = 21 * 3600;
if(tempperson.arrive >= 21 * 3600) continue;
tempperson.time = temptime <= 120 ? temptime * 60 : 7200;
tempperson.vip = ((flag == 1) ? true : false);
player.push_back(tempperson);
}
scanf("%d%d", &k, &m);
table.resize(k + 1);
for(int i = 0; i < m; i++) {
scanf("%d", &viptable);
table[viptable].vip = true;
}
sort(player.begin(), player.end(), cmp1);
int i = 0, vipid = -1;
vipid = findnextvip(vipid);
while(i < player.size()) {
int index = -1, minendtime = 999999999;
for(int j = 1; j <= k; j++) {
if(table[j].end < minendtime) {
minendtime = table[j].end;
index = j;
}
}
if(table[index].end >= 21 * 3600) break;
if(player[i].vip == true && i < vipid) {
i++;
continue;
}
if(table[index].vip == true) {
if(player[i].vip == true) {
alloctable(i, index);
if(vipid == i) vipid = findnextvip(vipid);
i++;
} else {
if(vipid < player.size() && player[vipid].arrive <= table[index].end) {
alloctable(vipid, index);
vipid = findnextvip(vipid);
} else {
alloctable(i, index);
i++;
}
}
} else {
if(player[i].vip == false) {
alloctable(i, index);
i++;
} else {
int vipindex = -1, minvipendtime = 999999999;
for(int j = 1; j <= k; j++) {
if(table[j].vip == true && table[j].end < minvipendtime) {
minvipendtime = table[j].end;
vipindex = j;
}
}
if(vipindex != -1 && player[i].arrive >= table[vipindex].end) {
alloctable(i, vipindex);
if(vipid == i) vipid = findnextvip(vipid);
i++;
} else {
alloctable(i, index);
if(vipid == i) vipid = findnextvip(vipid);
i++;
}
}
}
}
sort(player.begin(), player.end(), cmp2);
for(i = 0; i < player.size() && player[i].start < 21 * 3600; i++) {
printf("%02d:%02d:%02d ", player[i].arrive / 3600, player[i].arrive % 3600 / 60, player[i].arrive % 60);
printf("%02d:%02d:%02d ", player[i].start / 3600, player[i].start % 3600 / 60, player[i].start % 60);
printf("%.0f\n", round((player[i].start - player[i].arrive) / 60.0));
}
for(int i = 1; i <= k; i++) {
if(i != 1) printf(" ");
printf("%d", table[i].num);
}
return 0;
}
https://pintia.cn/problem-sets/994805342720868352/problems/994805458722734080
With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may give different price. You are asked to carefully design the cheapest route to go.
Input Specification:
Each input file contains one test case. For each case, the first line contains 4 positive numbers: Cmax (≤ 100), the maximum capacity of the tank; D (≤30000), the distance between Hangzhou and the destination city; Davg (≤20), the average distance per unit gas that the car can run; and N (≤ 500), the total number of gas stations. Then N lines follow, each contains a pair of non-negative numbers: Pi, the unit gas price, and Di (≤D), the distance between this station and Hangzhou, for i=1,⋯,N. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print the cheapest price in a line, accurate up to 2 decimal places. It is assumed that the tank is empty at the beginning. If it is impossible to reach the destination, print The maximum travel distance = X where X is the maximum possible distance the car can run, accurate up to 2 decimal places.
Sample Input 1:
50 1300 12 8
6.00 1250
7.00 600
7.00 150
7.10 0
7.20 200
7.50 400
7.30 1000
6.85 300
Sample Output 1:
749.17
Sample Input 2:
50 1300 12 2
7.10 0
7.00 600
Sample Output 2:
The maximum travel distance = 1200.00#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
const int inf = 99999999;
struct station {
double price, dis;
};
bool cmp1(station a, station b) {
return a.dis < b.dis;
}
int main() {
double cmax, d, davg;
int n;
scanf("%lf%lf%lf%d", &cmax, &d, &davg, &n);
vector<station> sta(n + 1);
sta[0] = {0.0, d};
for(int i = 1; i <= n; i++)
scanf("%lf%lf", &sta[i].price, &sta[i].dis);
sort(sta.begin(), sta.end(), cmp1);
double nowdis = 0.0, maxdis = 0.0, nowprice = 0.0, totalPrice = 0.0, leftdis = 0.0;
if(sta[0].dis != 0) {
printf("The maximum travel distance = 0.00");
return 0;
} else {
nowprice = sta[0].price;
}
while(nowdis < d) {
maxdis = nowdis + cmax * davg;
double minPriceDis = 0, minPrice = inf;
int flag = 0;
for(int i = 1; i <= n && sta[i].dis <= maxdis; i++) {
if(sta[i].dis <= nowdis) continue;
if(sta[i].price < nowprice) {
totalPrice += (sta[i].dis - nowdis - leftdis) * nowprice / davg;
leftdis = 0.0;
nowprice = sta[i].price;
nowdis = sta[i].dis;
flag = 1;
break;
}
if(sta[i].price < minPrice) {
minPrice = sta[i].price;
minPriceDis = sta[i].dis;
}
}
if(flag == 0 && minPrice != inf) {
totalPrice += (nowprice * (cmax - leftdis / davg));
leftdis = cmax * davg - (minPriceDis - nowdis);
nowprice = minPrice;
nowdis = minPriceDis;
}
if(flag == 0 && minPrice == inf) {
nowdis += cmax * davg;
printf("The maximum travel distance = %.2f", nowdis);
return 0;
}
}
printf("%.2f", totalPrice);
return 0;
}