hdu1051 Wooden Sticks(贪心+排序，逻辑)

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26095    Accepted Submission(s): 10554

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

有若干组木棍：木棍的长度为I 重为W   现在用机器加工这些木头： 规则如下：

1；加工第一组木头会花去1分钟
2：设加工的下一组木头的长度为 I'  重量为W‘   如果   I=<I' 且 W=<W'  ,则加工这根木头不需要花费时间。

问，所花费的最少时间是多少？

可以先按长度和重量从小到大排序，长度相同按重量从小到大排

每次从最前面没选过的开始，计算有多少种递增子序列，加一点技巧会更清楚，具体看代码中间的while部分

 #include<bits/stdc++.h>
 using namespace std;
 struct node
 {
     int l,w,flag;
 }a[];
 void init()
 {
     for(int i=;i<;i++)
     {
         a[i].l=;a[i].w=;a[i].flag=;
     }
 }
 bool cmp(node x,node y)
 {
     if(x.l==y.l)return x.w<y.w;
     else return x.l<y.l;
 }
 int main()
 {
     int t;
     while(~scanf("%d",&t))
     {
         while(t--)
         {
             init();
             int n;
             scanf("%d",&n);
             for(int i=;i<n;i++)
             {
                 scanf("%d %d",&a[i].l,&a[i].w);
             }
             sort(a,a+n,cmp);
 //            for(int i=0;i<n;i++)
 //            {
 //                printf("%d %d\n",a[i].l,a[i].w);
 //            }
             int ans=,j=;
             while(j<n)//判断有没有全部选完 ，核心部分
             {
                 ans++;
                 int tempx=,tempy=;
                 for(int i=;i<n;i++)
                 {
                     if(a[i].l>=tempx&&a[i].w>=tempy&&a[i].flag)//flag==1表示没有被选过
                     {
                         a[i].flag=;//选过的标记，下次就不选了
                         j++;
                         tempx=a[i].l;
                         tempy=a[i].w;
                     }
                 }
             }
             printf("%d\n",ans);
         }
     }
     return ;
 }