Wooden Sticks(杭电1051)

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Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12157    Accepted Submission(s): 5036


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
213
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct st{int l,w;}data[10050];int cmp(st a,st b)  //按木头长度从长到短排序,若长度一样则按质量从大到小排序。 {if(a.l!=b.l)return a.l>b.l;elsereturn a.w>b.w;}int a[10050];int main(){int T;scanf("%d",&T);while(T--){int i,j,n,k,sum=0;scanf("%d",&n);for(i=0;i<n;i++){scanf("%d %d",&data[i].l,&data[i].w);}sort(data,data+n,cmp);memset(a,0,sizeof(a));for(i=0;i<n;i++)   //将遍历过的且满足条件的木头做标记,用sum统计所需要的次数。 {if(a[i])continue;k=data[i].w;for(j=i+1;j<n;j++){if(k>=data[j].w&&!a[j]){k=data[j].w;a[j]=1;}}sum++;}printf("%d\n",sum);}}


  


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