PAT1026 Table Tennis (30)(排队模拟)
又是排队模拟,PAT很喜欢出排队模拟题,这题巨复杂,反正我是没想出来,基本的思路就是分为桌子为VIP和选手为VIP一共4种情况讨论,太复杂。抄了一下别人的代码,其中的几个难点写了注释。
#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#define INF 0xfffffff
using namespace std;
struct person {
int arrive, start, time;
bool vip;
}tempPerson;
struct tableNode {
int end = 8 * 3600, num;
bool vip;
};
vector<person> player;
vector<tableNode> table;
void alloctable(int personID, int tableID) {
if (player[personID].arrive <= table[tableID].end) //选手到达时前面还有人在占用桌子
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, m, k, vipTable;
cin >> 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; //这里需要加,为了配合后面table[index].end>= 21 * 3600脱出
if (tempPerson.arrive >= 21 * 3600) continue;//也就是说此时无法在9点前占台,最后将不会打印。
tempPerson.time = tempTime<=120?tempTime*60:7200;
tempPerson.vip = ((flag == 1 ? true : false));
player.push_back(tempPerson);
}
scanf("%d%d", &m, &k);
table.resize(m + 1);
for (int i = 0; i < k; i++) {
cin >> vipTable;
table[vipTable].vip = true;
}
sort(player.begin(), player.end(), [](const person &a, const person &b) {
return a.arrive < b.arrive;
});
int i = 0, vipID = -1;
vipID = findNextVip(vipID); //指向第一个VIP选手
while (i < player.size()) {
int index = -1, minEndTime = INF;
for (int j = 1; j <= m; j++) {
if (table[j].end < minEndTime) {
minEndTime = table[j].end;
index = j;
}
}
if (table[index].end >= 21 * 3600) //已经超过营业时间
break;
if (player[i].vip == true && i < vipID) { //有VIP选手插队时,该处情况就是第i个选手是已经
i++; //插队进入的VIP选手,所以直接跳过
continue;
}
if (table[index].vip == true) { //桌子是VIP桌
if (player[i].vip == true) { //当前选手是VIP
alloctable(i, index);
if (i == vipID) //当前VIP选手占台,所以不再是第一个VIP
vipID = findNextVip(vipID);
i++;
}else {
if (vipID < player.size() && player[vipID].arrive <= table[index].end) {
alloctable(vipID, index); //此时队列中第一个VIP选手已经在等待
vipID = findNextVip(vipID); //选取该VIP选手占台
//i++; //此时当前的选手被VIP选手插队,所以不能i++
}else { //不然该选手就失去占台机会
alloctable(i, index);
i++;
}
}
}else {
if (player[i].vip == false) {
alloctable(i, index);
i++;
}
else { //桌子不为VIP桌而选手为VIP选手
int vipIndex = -1, minVipEndTime = INF;
for (int j = 1; j <= m; j++) {
if (table[j].vip == true && minVipEndTime > table[j].end) {
minVipEndTime = table[j].end;
vipIndex = j;
}
}
if (vipIndex != -1 && table[vipIndex].end <= player[i].arrive) {
alloctable(i, vipIndex); //对于当前VIP选手有其他VIP台空出来,进入VIP台
if (i == vipID)
vipID = findNextVip(vipID);
i++;
}else {
alloctable(i, index); //对于当前VIP选手有没有其他VIP台空出来,进入普通台
if (i == vipID)
vipID = findNextVip(vipID);
i++;
}
}
}
}
sort(player.begin(), player.end(), [](const person &a, const person &b) {
return a.start < b.start;
});
for (int 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((double)((player[i].start - player[i].arrive) / 60.0)));
}
for (int i = 1; i <= m; i++) {
if (i != 1)
printf(" ");
printf("%d", table[i].num);
}
return 0;
}