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;}