Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 19672    Accepted Submission(s): 7991

Problem Description

There is a pile of n wooden sticks. The length and weight of each stick are known in advance. The sticks are to be processed by a woodworking machine in one by one fashion. It needs some time, called setup time, for the machine to prepare processing a stick. The setup times are associated with cleaning operations and changing tools and shapes in the machine. The setup times of the woodworking machine are given as follows: 

(a) The setup time for the first wooden stick is 1 minute. 

(b) Right after processing a stick of length l and weight w , the machine will need no setup time for a stick of length l' and weight w' if l<=l' and w<=w'. Otherwise, it will need 1 minute for setup. 

You are to find the minimum setup time to process a given pile of n wooden sticks. For example, if you have five sticks whose pairs of length and weight are (4,9), (5,2), (2,1), (3,5), and (1,4), then the minimum setup time should be 2 minutes since there is a sequence of pairs (1,4), (3,5), (4,9), (2,1), (5,2).

 

 

Input

The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case consists of two lines: The first line has an integer n , 1<=n<=5000, that represents the number of wooden sticks in the test case, and the second line contains n 2 positive integers l1, w1, l2, w2, ..., ln, wn, each of magnitude at most 10000 , where li and wi are the length and weight of the i th wooden stick, respectively. The 2n integers are delimited by one or more spaces.

 

 

Output

The output should contain the minimum setup time in minutes, one per line.

 

 

Sample Input

3 5 4 9 5 2 2 1 3 5 1 4 3 2 2 1 1 2 2 3 1 3 2 2 3 1

 

 

Sample Output

2 1 3

 

题目大意：有个木材加工机器，第一个放入的木材要花1分钟加工，若接下来放入的木材的长度和重量都大于等于前一个放入的木材，则这个木材加工花费的时间为零，否则得花费一分钟时间来加工；以此类推，直到木材加工完为止。

思想：结构体二级排序，对其中一个属性贪心。

 

 

代码：

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 typedef struct wood { 5     int l; 6     int w; 7 }Wood; 8 int compare(const void * p1, const void * p2) { 9     Wood * a1 = (Wood *)p1;10     Wood * a2 = (Wood *)p2;11     if (a1->l != a2->l)12         return a1->l - a2->l;13     else14         return a1->w - a2->w;15 }16 int main(void)17 {18     int T;19     scanf("%d", &T);20     while (T--) {21         int n;22         scanf("%d", &n);23         Wood sticks[5000];24         bool vis[5000] = { 0 };25         for (int i = 0; i < n; i++) {26             scanf("%d %d", &sticks[i].l, &sticks[i].w);27         }28         qsort(sticks, n, sizeof(sticks[0]), compare);29         int time = 0;30         for (int i = 0; i < n; i++) {31             if (vis[i])32                 continue;33             int curw = sticks[i].w;34             for (int j = i + 1; j < n; j++) {35                 if (!vis[j] && sticks[j].w >= curw) {36                     curw = sticks[j].w;37                     vis[j] = 1;38                 }39             }40             time++;41         }42         printf("%d\n", time);43     } 　　　　return 0;44 }