1065 Wooden Sticks

Wooden Sticks

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11769		Accepted: 4803

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 ( 9 , 4 ) , ( 2 , 5 ) , ( 1 , 2 ) , ( 5 , 3 ) , and ( 4 , 1 ) , then the minimum setup time should be 2 minutes since there is a sequence of pairs ( 4 , 1 ) , ( 5 , 3 ) , ( 9 , 4 ) , ( 1 , 2 ) , ( 2 , 5 ) .

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 2n 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

Taejon 2001

 

//大水题 #include<cstdio> #include<algorithm> #include<cstring> using namespace std; struct T { int w,l; }a[5010]; bool cmp(T a,T b) { if(a.l==b.l) return a.w<b.w; return a.l<b.l; } bool vis[5010]; int main() { int cas; scanf("%d",&cas); while(cas--) { int n; scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d%d",&a[i].l,&a[i].w); sort(a,a+n+1,cmp); memset(vis,0,sizeof(vis)); int num=1;//从num开始找 int w,l; int count=0,tt=0; while(tt<n) { vis[num]=1; tt++; w=a[num].w; l=a[num].l; bool flag=false; for(int i=num+1;i<=n;i++) if(!vis[i]&&a[i].l>=l&&a[i].w>=w) { w=a[i].w; l=a[i].l; vis[i]=1; tt++; } else if(!vis[i]&&!flag) { flag=true; num=i; } count++; } printf("%d/n",count); } return 0; }