PAT A1017 优先队列

这道题有点像优先队列的思想，简而言之就是挑选最小的入队处理，如果有多个队列就进行多个队列的处理；

借鉴的思想是采用记录每个队列中的任务完成时间，然后在读入任务的时候进行轮询，选择结束时间最小的那个队列，然后进行处理和等待时间的计算；

代码如下：

#include 
#include
#include
#include 
#include 
using namespace std;
struct node {
    int come, time;
} tempcustomer;
bool cmp1(node a, node b) {
    return a.come < b.come;
}
int main() {
    int n, k;
    scanf("%d%d", &n, &k);
    vector custom;
    for(int i = 0; i < n; i++) {
        int hh, mm, ss, time;
        scanf("%d:%d:%d %d", &hh, &mm, &ss, &time);
        int cometime = hh * 3600 + mm * 60 + ss;
        if(cometime > 61200) continue;
        tempcustomer = {cometime, time * 60};
        custom.push_back(tempcustomer);
    }
    sort(custom.begin(), custom.end(), cmp1);
    vector window(k, 28800);
    double result = 0.0;
    for(int i = 0; i < custom.size(); i++) {
        int tempindex = 0, minfinish = window[0];
        for(int j = 1; j < k; j++) {
            if(minfinish > window[j]) {
                minfinish = window[j];
                tempindex = j;
            }
        }
        if(window[tempindex] <= custom[i].come) {
            window[tempindex] = custom[i].come + custom[i].time;
        } else {
            result += (window[tempindex] - custom[i].come);
            window[tempindex] += custom[i].time;
        }
    }
    if(custom.size() == 0)
        printf("0.0");
    else
        printf("%.1f", result / 60.0 / custom.size());
    system("pause");
    return 0;
}