add a1017

PATAdvanced/1017.c

@@ -0,0 +1,114 @@
+/**
+ * 1017. Queueing at Bank (25)
+ * 
+ * Suppose a bank has K windows open for service. There is a yellow line in 
+ * front of the windows which devides the waiting area into two parts. All the 
+ * customers have to wait in line behind the yellow line, until it is his/her 
+ * turn to be served and there is a window available. It is assumed that no 
+ * window can be occupied by a single customer for more than 1 hour.
+ * 
+ * Now given the arriving time T and the processing time P of each customer, you
+ * are supposed to tell the average waiting time of all the customers.
+ * 
+ * Input Specification:
+ * 
+ * Each input file contains one test case. For each case, the first line 
+ * contains 2 numbers: N (<=10000) - the total number of customers, and K 
+ * (<=100) - the number of windows. Then N lines follow, each contains 2 times: 
+ * HH:MM:SS - the arriving time, and P - the processing time in minutes of a 
+ * customer. Here HH is in the range [00, 23], MM and SS are both in [00, 59]. 
+ * It is assumed that no two customers arrives at the same time.
+ * 
+ * Notice that the bank opens from 08:00 to 17:00. Anyone arrives early will 
+ * have to wait in line till 08:00, and anyone comes too late (at or after 
+ * 17:00:01) will not be served nor counted into the average.
+ * 
+ * Output Specification:
+ * 
+ * For each test case, print in one line the average waiting time of all the 
+ * customers, in minutes and accurate up to 1 decimal place.
+ * Sample Input:
+ * 
+ * 7 3
+ * 07:55:00 16
+ * 17:00:01 2
+ * 07:59:59 15
+ * 08:01:00 60
+ * 08:00:00 30
+ * 08:00:02 2
+ * 08:03:00 10
+ * 
+ * Sample Output:
+ * 
+ * 8.2
+ **/
+
+#include <stdio.h>
+#include <stdlib.h>
+
+typedef struct Customer{
+    int start, len;
+}Customer;
+
+/* Compare two Customer stucture by start time */
+int cmp(const void *a, const void *b)
+{
+    Customer c1 = *(Customer*)a;
+    Customer c2 = *(Customer*)b;
+    return c1.start - c2.start;
+}
+
+int main()
+{
+    int N, K, earliest, i;
+    int HH, MM, SS;
+    int wait_time = 0, queue_time[100] = {0};
+    Customer customers[10000], *p;
+    
+    scanf("%d %d", &N, &K);
+    for(int i = 0; i < N; i++)
+    {
+        scanf("%d:%d:%d %d", &HH, &MM, &SS, &customers[i].len);
+        /* Relative time to 08:00:00 */
+        customers[i].start = SS + 60 * (MM + 60 * (HH - 8));
+        customers[i].len *= 60;
+    }
+    
+    qsort(customers, N, sizeof(Customer), cmp);
+    
+    for(i = 0; i < N; i++)
+    {
+        p = customers + i;
+        
+        /* Find the queue number which will finish next */
+        earliest = 0;
+        for(int i = 0; i < K; i++)
+            if(queue_time[i] < queue_time[earliest])
+                earliest = i;
+        
+        /* later than 17:00:00 */
+        if(p->start > (17 - 8) * 3600)
+            break;
+        /* processing time longer than one hour */
+        if(p->len > 3600)
+            p->len = 3600;
+        
+        /* increase total waiting time and modify the time of each queue */
+        if(p->start < queue_time[earliest])
+        {
+            wait_time += queue_time[earliest] - p->start;
+            queue_time[earliest] += p->len;
+        }
+        else
+        {
+            queue_time[earliest] = p->start + p->len;
+        }
+    }
+    
+    if(i)
+        printf("%.1f", wait_time / 60.0 / i);
+    else
+        printf("0.0");
+    
+    return 0;
+}