PTA 7-5 银行排队问题之单队列多窗口服务

假设银行有K个窗口提供服务，窗口前设一条黄线，所有顾客按到达时间在黄线后排成一条长龙。当有窗口空闲时，下一位顾客即去该窗口处理事务。当有多个窗口可选择时，假设顾客总是选择编号最小的窗口。

本题要求输出前来等待服务的N位顾客的平均等待时间、最长等待时间、最后完成时间，并且统计每个窗口服务了多少名顾客。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int MAXN = 1010;
const int INF = 0x3f3f3f3f;
struct Data {
int x, y;
}data[MAXN];
struct People {
int start, finish;
};
struct Window {
int cnt;
People people[MAXN];
Window() {
cnt = 0;
people[0].start = people[1].start = -1;
}
};
int main() {
int n; scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%d %d", &data[i].x, &data[i].y);
if (data[i].y > 60) data[i].y = 60;
}
int m; scanf("%d", &m);
Window window[m + 5];
int sumwaittime = 0, maxwaittime = 0, maxfinishtime = 0;
for (int i = 1; i <= n; i++) {
bool ok = true;
int minval = INF, pos = -1;
for (int j = 1; j <= m; j++) {
Window &w = window[j];
if (w.people[w.cnt].finish <= data[i].x) {
int &cnt = window[j].cnt; ++cnt;
People &p = window[j].people[cnt];
p.start = data[i].x, p.finish = data[i].x + data[i].y;
maxfinishtime = max(maxfinishtime, p.finish);
ok = false;
break;
}
if (minval > w.people[w.cnt].finish) {
minval = w.people[w.cnt].finish;
pos = j;
}
}
if (ok) {
int &cnt = window[pos].cnt; ++cnt;
People &p = window[pos].people[cnt];
p.start = data[i].x, p.finish = minval + data[i].y;
sumwaittime += minval - p.start;
maxwaittime = max(maxwaittime, minval - p.start);
maxfinishtime = max(maxfinishtime, p.finish);
}
}
printf("%.1lf %d %d\n", 1.0*sumwaittime/n, maxwaittime, maxfinishtime);
printf("%d", window[1].cnt);
for (int i = 2; i <= m; i++) printf(" %d", window[i].cnt);
printf("\n");
return 0;
}