Wooden Sticks
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9937 Accepted Submission(s): 4081
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
Source
采用的是,常规的方法。。。。
先进行排序,人选一个作为第一优先级...
比如以第一个聚类...以重量作为依据排序..
2 1
5 2
1 4
3 5
4 9
序列排好了...然后那样每次拿掉一条递增的序列,并统计即可!
代码: View Code
1 #include2 #include 3 #include 4 #include 5 using namespace std; 6 typedef struct 7 { 8 int we; 9 int le;10 int tag;11 }go;12 int cmp(go const a, go const b)13 { 14 /*升序*/15 if(a.le==b.le)16 {17 if(a.we st;27 go tem;28 scanf("%d",&t);29 while(t--)30 {31 st.clear();32 scanf("%d",&n);33 while(n--)34 {35 scanf("%d%d",&tem.le,&tem.we);36 tem.tag=true;37 st.push_back(tem);38 }39 sort(st.begin(),st.end(),cmp);40 vector ::iterator it,tem;41 cnt=0;42 bool falg;43 while(!st.empty())44 {45 falg=true;46 for(it=st.begin(); it!=st.end();it++)47 {48 if(it->tag&&falg) tem=it;49 if(it->tag&&tem->we<=it->we)50 {51 tem=it;52 it->tag=false;53 falg=false;54 }55 }56 if(!falg) cnt++;57 else break;58 }59 printf("%d\n",cnt);60 }61 return 0;62 }