ZOJ 1025:Wooden Sticks

ZOJ 1025:Wooden Sticks

Wooden Sticks

Time Limit: 2 Seconds       Memory Limit: 65536 KB

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

Output for the Sample Input

2  1  3

Source: Asia 2001, Taejon (South Korea)

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#include #include #include using namespace std; struct mood { int w; int l; } d[5005]; bool cmp(mood a,mood b) { if(a.l==b.l) return a.w>n; while(n--) { int m; cin>>m; for(int i=0; i>d[i].l>>d[i].w; sort(d,d+m,cmp); memset(k,-1,sizeof(k)); int flag=0; for(int i=0; i=k[j]) { k[j]=d[i].w; if(j+1>flag)flag=j+1; break; } cout<

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