hdu 1051-Wooden Sticks

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

意思是给你几组木棍的长度和重量，从第二组开始，如果后边木棍数据的长度和重量大于前一组（不一定连续），则不加时间，否则加一分钟，求最少时间

比如（4 ,9）,（5 ,2）,（2,1）,（3,,5）,(1,4),先对其中一种数据排序（长度或重量都可以），比如对长度从小到大排序，

则数据变为（1,4），（2,1），（3,5），（4,9），（5,2），则满足条件的结果为【（1,4），（3,5），（4,9）】，【（2,1），（5,2）】，共需要2分钟。

#include <stdio.h>#include<string.h>int main (){    int a[5005],b[5005],c[5005];    int n,i,j;    int t,temp;    int num,sum;    int max1,max2;    scanf ("%d",&t);    while (t--)    {        num=1;        sum=0;        memset (c,0,sizeof(c));        scanf ("%d",&n);        for (i=0;i<n;i++)        {            scanf ("%d",&a[i]);            scanf("%d",&b[i]);        }              for (i=0;i<n;i++)            for(j=i;j<n;j++)        {            if(a[i]<a[j])            {                temp=a[i];a[i]=a[j];a[j]=temp;                temp=b[i];b[i]=b[j];b[j]=temp;            }        }        while(num)        {            num=0;            max1=max2=30005;            for(i=0;i<n;i++)            {                if(c[i]==0)                {                    if(max1>=a[i] && max2>=b[i])                    {                        max1=a[i];                        max2=b[i];                        c[i]=1;                    }                    num=1;                }            }            sum++;        }        printf("%d\n",sum-1);    }    return 0;}