hdu 1051 Wooden Sticks

Wooden Sticks

                                                                Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    先排序，如果长度相等，则重量从小到大排；否则，长度从小到大排。
   
   
   
   
   
   
   
   

    
    
    
    然后选择第一个没有用过的木头依次向后找，用掉所有可以用的木头，然后返回第一个没有用过的木头继续找。
   
   
   
   
   
   
   
   

    
    
    
    AC代码：
   
   
   
   
   
   
   
   

    
    
    
    
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std; struct stick { int l,w;  //l为长度，w为重量
}a[5005]; bool cmp(stick a1,stick a2) { if(a1.l==a2.l) return a1.w<a2.w;  //长度相等，重量按从小到大排序
 else return a1.l<a2.l;  //长度不等，则长度按从小到大排序
 } int main() { int T,n,i,p,j,vis[5005],flag,count;
    scanf("%d",&T); while(T--) {
        scanf("%d",&n);
        memset(vis,0,sizeof(vis)); //设置标记，标记是否访问过这组数据
 for(i=0;i<n;i++)
           scanf("%d%d",&a[i].l,&a[i].w);
        sort(a,a+n,cmp);
        vis[0]=1;  //先访问第一组
        p=0;  //记录第一个没有用过的木头
        count=0;  //记录次数
 while(p!=n) {
            count++; for(i=p+1,j=p,flag=1;i<n;i++) { if(vis[i]) //已访问，则访问下一组数据
 continue; if(a[i].l>=a[j].l&&a[i].w>=a[j].w)  //不用重新设置
 {
                   vis[i]=1;  //标记已访问
                   j=i;  //更新要比较的数据
 } else { if(flag) {
                        p=i;  //只记录第一个没有用的木头
                        flag=0; } } } if(flag)  //都用过了
 break; }
        printf("%d\n",count); } return 0; }