PAT 1026 Table Tennis (30分)

蛮复杂的一道模拟题
需要注意的点挺多的：
1.优先级问题（vip桌子的安排）
2.playtime规定在2h之内，超过2h按2h计算
3.servingtime等于或大于21点的不输出

#include <iostream>
#include <stdio.h>
#include<stdlib.h>
#include <string>
#include <algorithm>
#include <vector>
#include <unordered_map>
#include <string.h>
#include <map>
#include <set>using namespace std;
int N, Pt,h,m,s,tag, K, M, ind,pi,vi;typedef struct player
{int atime;int ptime;int stime;int wtime;
};
player P[10010];
player V[10010];typedef struct table {int index;int stime;int players;int tag;
};
table T[110];bool cmp1(player a,player b)
{return a.atime < b.atime;
}bool cmp2(player a, player b)
{return a.stime < b.stime;
}bool cmp3(table a, table b)
{if(a.stime != b.stime)return a.stime<b.stime;else return a.index < b.index;
}bool cmp4(table a, table b)
{return a.index < b.index;
}vector <int> vip_table;int main()
{#ifdef  LIUFILE *ss;freopen_s(&ss, "1.txt", "r", stdin);
#endif cin >> N;int open = 8 * 3600;int close = 21 * 3600;char c;vi = 0; pi = 0;for (int i = 0; i < N; i++){cin >> h >> c >> m >> c >> s;cin >> Pt>>tag;if (tag == 1){V[vi].atime= h * 3600 + m * 60 + s;V[vi].ptime = min(120*60,Pt*60);vi++;}else {P[pi].atime= h * 3600 + m * 60 + s;P[pi].ptime = min(120*60,Pt*60);pi++;}}sort(P, P + pi, cmp1);//按player到达时间排序sort(V, V + vi, cmp1);cin >> K >> M;for (int i = 0; i < K; i++){if (i < M) {cin >> ind;T[ind - 1].tag = 1;//数组中编号为0-K-1vip_table.push_back(ind-1);}T[i].index = i + 1;T[i].players = 0;T[i].stime = open;}sort(vip_table.begin(), vip_table.end());//安排前K个顾客中的vip若N<K,则安排前N个int posv = 0,posp=0;int otable = 0;int vtable = 0;for (int t = 0; t < K; t++){if (posv + posp == N)break;if (posv < vi&&(posp==pi||V[posv].atime < P[posp].atime)){//是vip先检测有无空闲的vip桌子if(vtable<M){V[posv].stime = V[posv].atime;V[posv].wtime = 0;T[vip_table[vtable]].stime = V[posv].stime + V[posv].ptime; if (V[posv].stime < close) {T[vip_table[vtable]].players++;}vtable++;}else {while (T[otable].tag==1) {++otable;}V[posv].stime = V[posv].atime;V[posv].wtime = 0;T[otable].stime = V[posv].stime + V[posv].ptime;if (V[posv].stime < close) {T[otable].players++;}otable++;}posv++;}else if (posp < pi){while(T[otable].tag == 1&& (vtable==M||otable < vip_table[vtable]))otable++;if (T[otable].tag == 1) vtable++;P[posp].stime = P[posp].atime;P[posp].wtime = 0;T[otable].stime = P[posp].stime + P[posp].ptime;if (P[posp].stime < close) {T[otable].players++;}otable++;posp++;}}while (posp + posv < N)//顾客没有安排完{sort(T, T + K, cmp3);if (T[0].tag == 1&&posv<vi && V[posv].atime <= T[0].stime)//是vip桌子{//看队列中有没有vipV[posv].stime = T[0].stime;V[posv].wtime = (V[posv].stime - V[posv].atime+30) / 60;if (V[posv].stime < close) {T[0].stime += V[posv].ptime;T[0].players++;}posv++;}else{//给顺次下一位顾客安排桌子if (posp<pi&&(posv==vi||P[posp].atime < V[posv].atime)){  //一般顾客P[posp].stime = max(P[posp].atime,T[0].stime);P[posp].wtime = (P[posp].stime-P[posp].atime+30)/60;T[0].stime = P[posp].stime + P[posp].ptime;if (P[posp].stime < close) {T[0].players++;}posp++;}else{V[posv].stime = max(V[posv].atime,T[0].stime);V[posv].wtime = (V[posv].stime - V[posv].atime+30) / 60;T[0].stime = V[posv].stime + V[posv].ptime;if (V[posv].stime < close) {T[0].players++;}posv++;}}}sort(P, P + pi, cmp2);//按player服务时间排序sort(V, V + vi, cmp2);posp = posv = 0;while (posp+posv<N){if (posp<pi&&(posv == vi || P[posp].stime < V[posv].stime)){if (P[posp].stime < close) {printf("%02d:%02d:%02d %02d:%02d:%02d %d\n",P[posp].atime / 3600, (P[posp].atime % 3600) / 60, P[posp].atime % 60,P[posp].stime / 3600, (P[posp].stime % 3600) / 60, P[posp].stime % 60,P[posp].wtime);}++posp;}else {if (V[posv].stime < close) {printf("%02d:%02d:%02d %02d:%02d:%02d %d\n",V[posv].atime / 3600, (V[posv].atime % 3600) / 60, V[posv].atime % 60,V[posv].stime / 3600, (V[posv].stime % 3600) / 60, V[posv].stime % 60,V[posv].wtime);}++posv;}           }sort(T, T + K, cmp4);for (int t = 0; t < K; t++){  if (t == 0) cout << T[t].players;else cout << ' ' << T[t].players;}cout << endl;return 0;
}

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