HDU1051切割木头 ACM_队列_贪心算法

Wooden Sticks Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16938Accepted Submission(s): 6910

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 //思路：题目大概就是讲一个木头放入一个机器切割需要1分钟，如果后面放入的木头高度宽度均小于前者，则不要时间，否则一分钟 #include #include #include #include #include#includeusing namespace std; struct node {int x,y,flag; }; int main() {int n,t; struct node w[10001],temp; while(scanf("%d",&t)!=EOF){while(t--){scanf("%d",&n); for(int i=0; iw[j].x)//x从小到大排列{temp=w[i]; w[i]=w[j]; w[j]=temp; }else if(w[i].x==w[j].y&&w[i].x>w[j].y)//如果x相同，看y排{temp=w[i]; w[i]=w[j]; w[j]=temp; }}}int time=0; for(int i=0; i