Wooden Sticks 贪心

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

题目大意：题意是加工木材，如果后一个长度和重量都大于等于前一个，那么就不需要调器，否则需要重新调机器，时间要加一分钟，输出需要的最少时间。

思路：就是先按长度或者重量从小到大排序，然后再按另一种进行找需要调机器的最少次数。

代码如下：

#include<stdio.h>
#include<algorithm>
using namespace std;
struct note
{
    int l,w;
} e[10100];
bool  cmp(note x,note y)
{
    if(x.l==y.l)
        return x.w<y.w; //当长度相同时按重量排序，不写这个会出错
    else
        return x.l<y.l;
}
int main()
{
    int n;
    scanf("%d",&n);
    while(n--)
    {
        int m;
        scanf("%d",&m);
        int i,j;
        for(i=0; i<m; i++)
            scanf("%d%d",&e[i].l,&e[i].w);
        sort(e,e+m,cmp);           //用sort按长度排序
        int book[10100]={0},sum=0;
        for(i=0; i<m; i++)
        {

            if(book[i]) continue;   //如果已经用过了跳过
            int maxx=e[i].w;
            for(j=i+1; j<m; j++)    //找后面比前一个重量大的，
                                    //因为长度已经排好序了可以不用管。
            {
                if(!book[j]&&maxx<=e[j].w) 
                {
                    book[j]=1;
                    maxx=e[j].w;    //将找到的最新的重量赋给maxx
                }
            }
            sum++;
        }
        printf("%d\n",sum);  
    }
    return 0;
}